Question

NCERT Solutions Class 11 Maths Chapter 10 Straight Lines is one of the important chapters of mathematics. We have provided solutions for the entire chapter in in-depth detail.

These NCERT Solutions Class 11 is prepared by the subject matter experts at Sarthaks.

• Straight Lines – a straight line is an infinite one-dimensional figure which has no width. Straight-line has no curves in it. A straight line can be vertical, horizontal, or slanted.

• Two-dimensional geometries – shape or figure on a two-dimensional plane that have only length and width. Two-dimensional figures have two faces. Example: circle, triangle, square, rectangle, pentagon.

• Shifting of origin – during shifting of origin we choose a new point on the coordinate plane which acts as a new origin of the plane. Now new distance is measured from the new origin.

• Slope of a line and angle between two lines – the angle between two lines helps us to quantify the relationship between two lines. The angle between two lines is the measure of inclination of two lines with each other.

• Various forms of equations of a line –

Slope (m) of a non-vertical line passing through the points (x1 , y1 ) and (x2, y2 )	m=(y2-y1)/(x2-x1), x1≠x2
Equation of a horizontal line	y = a or y=-a
Equation of a vertical line	x=b or x=-b
Equation of the line passing through the points (x1 , y1 ) and (x2, y2 )	y-y1= [(y2-y1)/(x2-x1)]×(x-x1)
Equation of line with slope m and intercept c	y = mx+c
Equation of line with slope m makes x-intercept d.	y = m (x – d).
Intercept the form of the equation of a line	(x/a)+(y/b)=1
The normal form of the equation of a line	x cos α+y sin α = p

• Parallel to axis – a straight line that makes no angle with the axis such lines are parallel to the axis. The distance of the line from the axis is always constant.

• Slope-intercept form - in slope intercept form the equation of line is y = mx + c.

• General equation of a line – for the first-degree equation of two variables the general equation of a line is represented as Ax + By + C = 0, where A, B is not equal to 0 and A, B and C are constant which belongs to real numbers.

Our NCERT Solutions Class 11 Maths has a unique approach to the solutions in an organized manner.

Answer 1

NCERT Solutions Class 11 Maths Chapter 10 Straight Lines

1. Draw a quadrilateral in the Cartesian plane, whose vertices are (–4, 5), (0, 7), (5, –5) and (–4, –2). Also, find its area.

Answer:

Let ABCD be the given quadrilateral with vertices A (–4, 5), B (0, 7), C (5, –5), and D (– 4, –2). 

Then, by plotting A, B, C, and D on the Cartesian plane and joining AB, BC, CD, and DA, the given quadrilateral can be drawn as

To find the area of quadrilateral ABCD, we draw one diagonal, say AC. 

Accordingly, area (ABCD) = area (∆ABC) + area (∆ACD) 

We know that the area of a triangle whose vertices are (x₁, y₁), (x₂, y₂), and (x₃, y₃) is

Therefore, area of ∆ABC

Area of ∆ACD

Thus, area (ABCD) 

 

2. The base of an equilateral triangle with side 2a lies along they y-axis such that the mid point of the base is at the origin. Find vertices of the triangle.

Answer:

Let ABC be the given equilateral triangle with side 2a. 

Accordingly, AB = BC = CA = 2a 

Assume that base BC lies along the y-axis such that the mid-point of BC is at the origin. 

i.e., BO = OC = a, where O is the origin.

Now, it is clear that the coordinates of point C are (0, a), while the coordinates of point B are (0, –a). 

It is known that the line joining a vertex of an equilateral triangle with the mid-point of its opposite side is perpendicular. 

Hence, vertex A lies on the y-axis.

On applying Pythagoras theorem to ∆AOC, we obtain 

(AC)² = (OA)² + (OC)² 

⇒ (2a)² = (OA)² + a² 

⇒ 4a² – a² = (OA)² 

⇒ (OA)² = 3a²

⇒ OA = √3a

∴Coordinates of point A = \((\pm\sqrt3a,0)\)

Thus, the vertices of the given equilateral triangle are (0, a), (0, –a), and \((\sqrt3a,0)\) or (0, a), (0, –a), and \((-\sqrt3a,0)\).

3. Find the distance between P(x₁,y₁) and Q(x₂, y₂) when: 

(i) PQ is parallel to the y-axis, 

(ii) PQ is parallel to the x-axis.

Answer:

The given points are P(x₁,y₁) and Q(x₂, y₂)

(i) When PQ is parallel to the y-axis, x₁ = x₂. 

In this case, distance between P and Q

(ii) When PQ is parallel to the x-axis, y₁ = y₂. 

In this case, distance between P and Q

4. Find a point on the x-axis, which is equidistant from the points (7, 6) and (3, 4).

Answer:

Let (a, 0) be the point on the x axis that is equidistant from the points (7, 6) and (3, 4).

On squaring both sides, we obtain a² – 14a + 85 = a² – 6a + 25 

⇒ –14a + 6a = 25 – 85

⇒ –8a = –60

⇒ \(a = \frac{60}{8}=\frac{15}2\)

Thus, the required point on the x-axis is \(\left(\frac{15}{2},0\right).\)

5. Find the slope of a line, which passes through the origin, and the mid-point of the line segment joining the points P (0, –4) and B (8, 0).

Answer:

The coordinates of the mid-point of the line segment joining the points P (0, –4) and B (8, 0) are 

It is known that the slope (m) of a non-vertical line passing through the points (x₁, y₁) and (x₂, y₂) is given by

Therefore, the slope of the line passing through (0, 0) and (4, –2) is

Hence, the required slope of the line is \(-\frac12\).

6. Without using the Pythagoras theorem, show that the points (4, 4), (3, 5) and (–1, –1) are the vertices of a right angled triangle.

Answer:

The vertices of the given triangle are A (4, 4), B (3, 5), and C (–1, –1). 

It is known that the slope (m) of a non-vertical line passing through the points (x₁, y₁) and (x₂, y₂) is given by

\(\therefore\) Slope of AB (m₁) = \(\frac{5-4}{3-4}= -1\)

Slope of BC (m₂) = \(\frac{-1-5}{-1-3}=\frac{-6}{-4}=\frac32\)

Slope of CA (m₃) = \(\frac{4+1}{4+1}=\frac55=1\)

It is observed that m₁m₃ = –1 

This shows that line segments AB and CA are perpendicular to each other 

i.e., the given triangle is right-angled at A (4, 4). 

Thus, the points (4, 4), (3, 5), and (–1, –1) are the vertices of a right-angled triangle.

7. Find the slope of the line, which makes an angle of 30° with the positive direction of y-axis measured anticlockwise.

Answer:

If a line makes an angle of 30° with the positive direction of the y-axis measured anticlockwise, then the angle made by the line with the positive direction of the x-axis measured anticlockwise is 90° + 30° = 120°.

 

Thus, the slope of the given line is tan 120° = tan (180° – 60°) = –tan 60° = \(-\sqrt3\)

8. Find the value of x for which the points (x, –1), (2, 1) and (4, 5) are collinear.

Answer:

If points A (x, –1), B (2, 1), and C (4, 5) are collinear, then 

Slope of AB = Slope of BC

Thus, the required value of x is 1.

Answer 2

9. Without using distance formula, show that points (–2, –1), (4, 0), (3, 3) and (–3, 2) are vertices of a parallelogram.

Answer:

Let points (–2, –1), (4, 0), (3, 3), and (–3, 2) be respectively denoted by A, B, C, and D.

Slope of AB = \(\frac{0+1}{4+2}=\frac16\)

Slope of CD = \(\frac{2-3}{-3-3}=\frac{-1}{-6}=\frac16\) 

⇒ Slope of AB = Slope of CD 

⇒ AB and CD are parallel to each other.

Now, slope of BC = \(\frac{3-0}{3-4}=\frac{3}{-1}=-3\)

Slope of AD = \(\frac{2+1}{-3+2}=\frac3{-1}=-3\)

⇒ Slope of BC = Slope of AD 

⇒ BC and AD are parallel to each other. 

Therefore, both pairs of opposite sides of quadrilateral ABCD are parallel. 

Hence, ABCD is a parallelogram. 

Thus, points (–2, –1), (4, 0), (3, 3), and (–3, 2) are the vertices of a parallelogram.

10. Find the angle between the x-axis and the line joining the points (3, –1) and (4, –2).

Answer:

The slope of the line joining the points (3, –1) and (4, –2) is 

Now, the inclination (θ) of the line joining the points (3, –1) and (4, – 2) is given by tan θ = –1 

⇒ θ = (90° + 45°) = 135° 

Thus, the angle between the x-axis and the line joining the points (3, –1) and (4, –2) is 135°.

11. The slope of a line is double of the slope of another line. If tangent of the angle between them is 1/3, find the slopes of the lines.

Answer:

Let m1 and m be the slopes of the two given lines such that m1 = 2m. 

We know that if θ is the angle between the lines l₁ and l₂ with slopes m₁ and m₂, then

\(tan \,\theta=\left|\frac{m_2-m_1}{1+m_1m_2}\right|\)

It is given that the tangent of the angle between the two lines is \(\frac13\).

Case I

If m = –1, then the slopes of the lines are –1 and –2. 

If m = \(-\frac12\), then the slopes of the lines are \(-\frac12\) and -1.

Case II 

If m = 1, then the slopes of the lines are 1 and 2.

If m = \(\frac12\), then the slopes of the lines are \(\frac12\) and 1.

Hence, the slopes of the lines are –1 and –2 or \(-\frac12\) and –1 or 1 and 2 or \(\frac12\) and 1.

12. A line passes through (x₁, y₁) and (h, k). If slope of the line is m, show that k – y₁ = m (h – x₁).

Answer:

The slope of the line passing through (x₁, y₁) and (h, k) is \(\frac{k-y_1}{h-x_1}\). 

It is given that the slope of the line is m.

Hence, k – y₁ = m (h – x₁)

13. If three point (h, 0), (a, b) and (0, k) lie on a line, show that \(\frac ah +\frac bk =1\).

Answer:

If the points A (h, 0), B (a, b), and C (0, k) lie on a line, then 

Slope of AB = Slope of BC

On dividing both sides by kh, we obtain

14. Consider the given population and year graph. Find the slope of the line AB and using it, find what will be the population in the year 2010?

Answer:

Since line AB passes through points A (1985, 92) and B (1995, 97), its slope is

Let y be the population in the year 2010. Then, according to the given graph, line AB must pass through point C (2010, y). 

\(\therefore\) Slope of AB = Slope of BC

Thus, the slope of line AB is \(\frac12\) while in the year 2010, the population will be 104.5 crores.

15. Write the equations for the x and y-axes.

Answer:

The y-coordinate of every point on the x-axis is 0. 

Therefore, the equation of the x-axis is y = 0. 

The x-coordinate of every point on the y-axis is 0. 

Therefore, the equation of the y-axis is y = 0.

16. Find the equation of the line which passes through the point (–4, 3) with slope \(\frac12\).

Answer:

We know that the equation of the line passing through point (x₀,y₀), whose slope is m, is

(y - y₀) = m(x - x₀)

Thus, the equation of the line passing through point (–4, 3), whose slope is \(\frac12\), is 

17. Find the equation of the line which passes though (0, 0) with slope m.

Answer:

We know that the equation of the line passing through point (x₀,y₀), whose slope is m, is (y - y₀) = m(x - x₀).

Thus, the equation of the line passing through point (0, 0), whose slope is m, is 

(y – 0) = m(x – 0)

 i.e., y = mx

18. Find the equation of the line which passes though (2, 2√3) and is inclined with the x-axis at an angle of 75°.

Answer:

The slope of the line that inclines with the x-axis at an angle of 75° is m = tan 75°

We know that the equation of the line passing through point (x₀, y₀) whose slope is m, is

(y - y₀) = m(x - x₀)

Thus, if a line passes though (2, 2√3) and inclines with the x-axis at an angle of 75°, then the equation of the line is given as

19. Find the equation of the line which intersects the x-axis at a distance of 3 units to the left of origin with slope –2.

Answer:

It is known that if a line with slope m makes x-intercept d, then the equation of the line is given as y = m(x – d) 

For the line intersecting the x-axis at a distance of 3 units to the left of the origin, d = –3. 

The slope of the line is given as m = –2 

Thus, the required equation of the given line is y = –2 [x – (–3)] y = –2x – 6 i.e., 2x + y + 6 = 0

Answer 3

20. Find the equation of the line which intersects the y-axis at a distance of 2 units above the origin and makes an angle of 30° with the positive direction of the x-axis.

Answer:

It is known that if a line with slope m makes y-intercept c, then the equation of the line is given as y = mx + c

Here, c = 2 and m = tan 30° = \(\frac1{\sqrt3}\).

Thus, the required equation of the given line is

21. Find the equation of the line which passes through the points (–1, 1) and (2, –4).

Answer:

It is known that the equation of the line passing through points (x₁, y₁) and (x₂, y₂) is

Therefore, the equation of the line passing through the points (–1, 1) and (2, –4) is

22. Find the equation of the line which is at a perpendicular distance of 5 units from the origin and the angle made by the perpendicular with the positive x-axis is 30°

Answer:

If p is the length of the normal from the origin to a line and ω is the angle made by the normal with the positive direction of the x-axis, then the equation of the line is given by x cos ω + y sin ω = p. 

Here, p = 5 units and ω = 30° 

Thus, the required equation of the given line is x 

cos 30° + y sin 30° = 5

23. The vertices of ∆PQR are P (2, 1), Q (–2, 3) and R (4, 5). Find equation of the median through the vertex R.

Answer:

It is given that the vertices of ∆PQR are P (2, 1), Q (–2, 3), and R (4, 5). 

Let RL be the median through vertex R. 

Accordingly, L is the mid-point of PQ. 

By mid-point formula, the coordinates of point L are given by \(\left(\frac{2-2}{2},\frac{1+3}{2}\right)=(0,2)\)

It is known that the equation of the line passing through points (x₁, y₁) and (x₂, y₂) is

Therefore, the equation of RL can be determined by substituting (x₁, y₁) = (4, 5) and (x₂, y₂) = (0, 2).

Hence,

Thus, the required equation of the median through vertex R is 3x – 4y + 8 = 0.

24. Find the equation of the line passing through (–3, 5) and perpendicular to the line through the points (2, 5) and (–3, 6).

Answer:

The slope of the line joining the points (2, 5) and (–3, 6) is \(m=\frac{6-5}{-3-2}=\frac1{-5}\)

We know that two non-vertical lines are perpendicular to each other if and only if their slopes are negative reciprocals of each other. 

Therefore, slope of the line perpendicular to the line through the points (2, 5) and (–3, 6)

Now, the equation of the line passing through point (–3, 5), whose slope is 5, is

25. A line perpendicular to the line segment joining the points (1, 0) and (2, 3) divides it in the ratio 1:n. Find the equation of the line.

Answer:

According to the section formula, the coordinates of the point that divides the line segment joining the points (1, 0) and (2, 3) in the ratio 1: n is given by

The slope of the line joining the points (1, 0) and (2, 3) is

\(m = \frac{3-0}{2-1}=3\)

We know that two non-vertical lines are perpendicular to each other if and only if their slopes are negative reciprocals of each other.

Therefore, slope of the line that is perpendicular to the line joining the points (1, 0) and (2, 3) = \(-\frac1m=-\frac13\)

Now, the equation of the line passing through \(\left(\frac{n+2}{n+1},\frac3{n+1}\right)\) and whose slope is \(-\frac13\) is given by

26. Find the equation of a line that cuts off equal intercepts on the coordinate axes and passes through the point (2, 3).

Answer:

The equation of a line in the intercept form is

\(\frac xa+\frac yb =1\)       .......(i)

Here, a and b are the intercepts on x and y axes respectively. 

It is given that the line cuts off equal intercepts on both the axes. This means that a = b. 

Accordingly, equation (i) reduces to

\(\frac xa+\frac yb =1\)

⇒ x + y = a           ......(ii)

Since the given line passes through point (2, 3), equation (ii) reduces to 

2 + 3 = a 

⇒ a = 5

On substituting the value of a in equation (ii), we obtain 

x + y = 5, which is the required equation of the line

27. Find equation of the line passing through the point (2, 2) and cutting off intercepts on the axes whose sum is 9.

Answer:

The equation of a line in the intercept form is

 \(\frac xa+\frac yb =1\)       .....(i)

Here, a and b are the intercepts on x and y axes respectively. 

It is given that 

a + b = 9 

⇒ b = 9 – a      … (ii) 

From equations (i) and (ii), we obtain

\(\frac xa+\frac y{9-a}=1\)      .....(iii)

It is given that the line passes through point (2, 2). Therefore, equation (iii) reduces to

If a = 6 and b = 9 – 6 = 3, then the equation of the line is

If a = 3 and b = 9 – 3 = 6, then the equation of the line is

28. Find equation of the line through the point (0, 2) making an angle \(\frac{2\pi}{3}\) with the positive x-axis. Also, find the equation of line parallel to it and crossing the y-axis at a distance of 2 units below the origin.

Answer:

The slope of the line making an angle \(\frac{2\pi}{3}\) with the positive x-axis is 

m = tan (\(\frac{2\pi}{3}\)) = -√3

Now, the equation of the line passing through point (0, 2) and having a slope -√3 is

The slope of line parallel √3x + y - 2 = 0 is -√3 to line.

It is given that the line parallel to line √3x + y - 2 = 0 crosses the y-axis 2 units below the origin i.e., it passes through point (0, –2).

Hence, the equation of the line passing through point (0, –2) and having a slope -√3 is

Answer 4

29. The perpendicular from the origin to a line meets it at the point (– 2, 9), find the equation of the line.

Answer:

The slope of the line joining the origin (0, 0) and point (–2, 9) is \(m_1=\frac{9-0}{-2-0}=-\frac92\)

Accordingly, the slope of the line perpendicular to the line joining the origin and point (– 2, 9) is

Now, the equation of the line passing through point (–2, 9) and having a slope m₂ is

30. The length L (in centimetre) of a copper rod is a linear function of its Celsius temperature C. In an experiment, if L = 124.942 when C = 20 and L = 125.134 when C = 110, express L in terms of C.

Answer:

It is given that when C = 20, the value of L is 124.942, whereas when C = 110, the value of L is 125.134. 

Accordingly, points (20, 124.942) and (110, 125.134) satisfy the linear relation between L and C. 

Now, assuming C along the x-axis and L along the y-axis, we have two points i.e., (20, 124.942) and (110, 125.134) in the XY plane. 

Therefore, the linear relation between L and C is the equation of the line passing through points (20, 124.942) and (110, 125.134).

which is the required linear relation 

31. The owner of a milk store finds that, he can sell 980 litres of milk each week at Rs 14/litre and 1220 litres of milk each week at Rs 16/litre. Assuming a linear relationship between selling price and demand, how many litres could he sell weekly at Rs 17/litre?

Answer:

The relationship between selling price and demand is linear. 

Assuming selling price per litre along the x-axis and demand along the y-axis, we have two points i.e., (14, 980) and (16, 1220) in the XY plane that satisfy the linear relationship between selling price and demand. 

Therefore, the linear relationship between selling price per litre and demand is the equation of the line passing through points (14, 980) and (16, 1220).

When x = Rs 17/litre,

Thus, the owner of the milk store could sell 1340 litres of milk weekly at Rs 17/litre.

32. P (a, b) is the mid-point of a line segment between axes. Show that equation of the line is \(\frac xa +\frac y b= 2\)

Answer:

Let AB be the line segment between the axes and let P (a, b) be its mid-point.

Let the coordinates of A and B be (0, y) and (x, 0) respectively. 

Since P (a, b) is the mid-point of AB,

Thus, the respective coordinates of A and B are (0, 2b) and (2a, 0). 

The equation of the line passing through points (0, 2b) and (2a, 0) is

On dividing both sides by ab, we obtain

Thus, the equation of the line is \(\frac xa +\frac y b= 2\).

33. Point R (h, k) divides a line segment between the axes in the ratio 1:2. Find equation of the line.

Answer:

Let AB be the line segment between the axes such that point R (h, k) divides AB in the ratio 1: 2.

Let the respective coordinates of A and B be (x, 0) and (0, y). 

Since point R (h, k) divides AB in the ratio 1: 2, according to the section formula,

Therefore, the respective coordinates of A and B are \((\frac{3h}{2},0)\)and (0,3k).

Now, the equation of line AB passing through points \((\frac{3h}{2},0)\) and (0,3k) is

Thus, the required equation of the line is 2kx + hy = 3hk

34. By using the concept of equation of a line, prove that the three points (3, 0), (–2, –2) and (8, 2) are collinear.

Answer:

In order to show that points (3, 0), (–2, –2), and (8, 2) are collinear, it suffices to show that the line passing through points (3, 0) and (–2, –2) also passes through point (8, 2). 

The equation of the line passing through points (3, 0) and (–2, –2) is

 

It is observed that at x = 8 and y = 2, 

L.H.S. = 2 × 8 – 5 × 2 = 16 – 10 = 6 = R.H.S.

Therefore, the line passing through points (3, 0) and (–2, –2) also passes through point (8, 2). 

Hence, points (3, 0), (–2, –2), and (8, 2) are collinear.

35. Reduce the following equations into slope-intercept form and find their slopes and the y intercepts. 

(i) x + 7y = 0

(ii) 6x + 3y – 5 = 0

(iii) y = 0

Answer:

(i) The given equation is x + 7y = 0.

It can be written as

 \(y=-\frac17x+0\)     ........(1)

This equation is of the form y = mx + c, where \(m =-\frac17\) and c = 0. 

Therefore, equation (1) is in the slope-intercept form, where the slope and the y-intercept are \(-\frac17\) and 0 respectively.

(ii) The given equation is 6x + 3y – 5 = 0.

It can be written as

This equation is of the form y = mx + c, where m = -2 and c = \(\frac53\).

Therefore, equation (2) is in the slope-intercept form, where the slope and the y-intercept are –2 and \(\frac53\) respectively.

(iii) The given equation is y = 0.

It can be written as 

y = 0.x + 0 … (3) 

This equation is of the form y = mx + c, where m = 0 and c = 0.

Therefore, equation (3) is in the slope-intercept form, where the slope and the y-intercept are 0 and 0 respectively.

Answer 5

36. Reduce the following equations into intercept form and find their intercepts on the axes. 

(i) 3x + 2y – 12 = 0 

(ii) 4x – 3y = 6 

(iii) 3y + 2 = 0.

Answer:

(i) The given equation is 3x + 2y – 12 = 0.

It can be written as

This equation is of the form \(\frac xa+\frac yb=1\) ,

where a = 4 and b = 6.

Therefore, equation (1) is in the intercept form, where the intercepts on the x and y axes are 4 and 6 respectively.

(ii) The given equation is 4x – 3y = 6. 

It can be written as

This equation is of the form \(\frac xa+\frac yb=1\), where a = \(\frac32\) and b = –2.

Therefore, equation (2) is in the intercept form, where the intercepts on the x and y axes are 3/2 and –2 respectively. 

(iii) The given equation is 3y + 2 = 0. 

It can be written as

3y = -2

The equation is of the form \(\frac xa +\frac yb=1\), where a = 0 and b = \(-\frac23\)

Therefore, equation (3) is in the intercept form, where the intercept on the y-axis \(-\frac23\) is and it has no intercept on the x-axis.

37. Reduce the following equations into normal form. Find their perpendicular distances from the origin and angle between perpendicular and the positive x-axis.

(i) x - √3y + 8 = 0

(ii) y – 2 = 0 

(iii) x – y = 4

Answer:

(i) The given equation is x - √3y + 8 = 0.

It can be reduced as:

On dividing both sides by

\(\sqrt{(-1)^2+(\sqrt3)^2}=\sqrt4=2\) , we obtain

Equation (1) is in the normal form. 

On comparing equation (1) with the normal form of equation of line x 

cos ω + y sin ω = p, we obtain ω = 120° and p = 4. 

Thus, the perpendicular distance of the line from the origin is 4, while the angle between the perpendicular and the positive x-axis is 120°.

(ii) The given equation is y – 2 = 0.

It can be reduced as 0.x + 1.y = 2

On dividing both sides by \(\sqrt{0^2+1^2}=1\), we obtain 0.x + 1.y = 2 

⇒ x cos 90° + y sin 90° = 2 … (1) 

Equation (1) is in the normal form. 

On comparing equation (1) with the normal form of equation of line x cos ω + y sin ω = p, we obtain ω = 90° and p = 2. 

Thus, the perpendicular distance of the line from the origin is 2, while the angle between the perpendicular and the positive x-axis is 90°.

(iii) The given equation is x – y = 4. 

It can be reduced as 1.x + (–1) y = 4

On dividing both sides by \(\sqrt{1^2+(-1)^2}=\sqrt2\), we obtain

Equation (1) is in the normal form.

On comparing equation (1) with the normal form of equation of line 

x cos ω + y sin ω = p, we obtain ω = 315° and p = 2√2. 

Thus, the perpendicular distance of the line from the origin is 2√2, while the angle between the perpendicular and the positive x-axis is 315°.

38. Find the distance of the point (–1, 1) from the line 12(x + 6) = 5(y – 2).

Answer:

The given equation of the line is 12(x + 6) = 5(y – 2). 

⇒ 12x + 72 = 5y – 10 

⇒ 12x – 5y + 82 = 0 … (1) 

On comparing equation (1) with general equation of line Ax + By + C = 0, we obtain 

A = 12, B = –5, and C = 82. 

It is known that the perpendicular distance (d) of a line Ax + By + C = 0 from a point (x₁, y₁) is given by 

The given point is (x1, y1) = (–1, 1). 

Therefore, the distance of point (–1, 1) from the given line

39. Find the points on the x-axis, whose distances from the line \(\frac x3 +\frac y4=1\) are 4 units.

Answer:

The given equation of line is

\(\frac x3 +\frac y4=1\) 

or, 4x + 3y - 12 = 0      ......(1)

On comparing equation (1) with general equation of line Ax + By + C = 0,

we obtain 

A = 4, B = 3, and C = –12. 

Let (a, 0) be the point on the x-axis whose distance from the given line is 4 units. 

It is known that the perpendicular distance (d) of a line Ax + By + C = 0 from a point (x₁, y₁) is given by 

Therefore,

Thus, the required points on the x-axis are (–2, 0) and (8, 0).

40. Find the distance between parallel lines 

(i) 15x + 8y – 34 = 0 and 15x + 8y + 31 = 0 

(ii) l (x + y) + p = 0 and l (x + y) – r = 0

Answer:

It is known that the distance (d) between parallel lines Ax + By + C₁ = 0 and Ax + By + C₂ = 0 is given by

\(d = \frac{|C_1-C_2|}{\sqrt{A^2+B^2}}\) .

(i) The given parallel lines are 15x + 8y – 34 = 0 and 15x + 8y + 31 = 0. 

Here, A = 15, B = 8, C₁ = –34, and C₂ = 31. 

Therefore, the distance between the parallel lines is

(ii) The given parallel lines are l (x + y) + p = 0 and l (x + y) – r = 0. 

lx + ly + p = 0 and lx + ly – r = 0 

Here, A = l, B = l, C₁ = p, and C₂ = –r. 

Therefore, the distance between the parallel lines is

41. Find equation of the line parallel to the line 3x – 4y + 2 = 0 and passing through the point (–2, 3).

Answer:

The equation of the given line is

which is of the form y = mx + c 

\(\therefore\) Slope of the given line = 3/4

It is known that parallel lines have the same slope.

\(\therefore\) Slope of the other line = m = \(\frac34\)

Now, the equation of the line that has a slope of \(\frac34\) and passes through the point (–2, 3) is

Answer 6

42. Find equation of the line perpendicular to the line x – 7y + 5 = 0 and having x intercept 3.

Answer:

The given equation of line is x – 7y + 5 = 0.

Or, y = \(\frac17x+\frac57\)

which is of the form y = mx + c 

∴Slope of the given line = \(\frac17\)

The slope of the line perpendicular to the line having a slope of \(\frac17\) is m 

 \(=\frac{1}{\left(\frac{1}{7}\right)}= -7\) 

The equation of the line with slope –7 and x-intercept 3 is given by y 

= m (x – d) 

⇒ y = –7 (x – 3) 

⇒ y = –7x + 21 

⇒ 7x + y = 21

43. Find angles between the lines √3x + y = 1 and x + √3y = 1

Answer:

The given lines are √3x + y = 1 and x + √3y = 1.

y = -√3x + 1    ......(1)

and

y = \(-\frac{1}{\sqrt3}x+\frac1{\sqrt3} \)    .......(2)

The slope of line (1) is m₁ = -√3 , while the slope of line (2) is m₂ = \(-\frac1{\sqrt3}\)

The acute angle i.e., θ between the two lines is given by

Thus, the angle between the given lines is either 30° or 180° – 30° = 150°.

44. The line through the points (h, 3) and (4, 1) intersects the line 7x – 9y – 19 = 0, at right angle. Find the value of h.

Answer:

The slope of the line passing through points (h, 3) and (4, 1) is

The slope of line 7x – 9y – 19 = 0 or 

It is given that the two lines are perpendicular.

Thus, the value of h is \(\frac{22}9\).

45. Prove that the line through the point (x₁, y₁) and parallel to the line Ax + By + C = 0 is A (x –x₁) + B (y – y₁) = 0.

Answer:

The slope of line Ax + By + C = 0 or 

It is known that parallel lines have the same slope. 

∴ Slope of the other line = m = \(-\frac AB\)

The equation of the line passing through point (x₁, y₁) and having a slope m = \(-\frac AB\) is 

Hence, the line through point (x₁, y₁) and parallel to line Ax + By + C = 0 is A (x –x₁) + B (y – y₁) = 0

46. Two lines passing through the point (2, 3) intersects each other at an angle of 60°. If slope of one line is 2, find equation of the other line.

Answer:

It is given that the slope of the first line, m1 = 2. 

Let the slope of the other line be m₂.

The angle between the two lines is 60°.

The equation of the line passing through point (2, 3) and having a slope of \(\frac{(2-\sqrt3)}{(2\sqrt3+1)}\)

is given by

In this case, the equation of the other line is (√3 - 2)x + (2√3 + 1)y = -1 + 8√3.

Case II: m₂ = \(\frac{(2-\sqrt3)}{(2\sqrt3+1)}\)

The equation of the line passing through point (2, 3) and having a slope of \(\frac{(2-\sqrt3)}{(2\sqrt3+1)}\) is

In this case, the equation of the other line is (2 + √3)x + (2√3 - 1)y = 1 + 8√3

Thus, the required equation of the other line is (√3 - 2)x + (2√3 + 1)y = -1 + 8√3 

or (2 + √3)x + (2√3 - 1)y = 1 + 8√3.

47. Find the equation of the right bisector of the line segment joining the points (3, 4) and (– 1, 2).

Answer:

The right bisector of a line segment bisects the line segment at 90°.

The end-points of the line segment are given as A (3, 4) and B (–1, 2).

Accordingly, mid-point of AB = \((\frac{3-1}{2},\frac{4+2}2)=(1,3)\)

Slope of AB = \(\frac{2-4}{-1-3}=\frac{-2}{-4}=\frac12\)

∴ Slope of the line perpendicular to AB = \(-\frac{1}{(\frac12)}=-2\) 

The equation of the line passing through (1, 3) and having a slope of –2 is 

(y – 3) = –2 (x – 1) 

y – 3 = –2x + 2 

2x + y = 5 

Thus, the required equation of the line is 2x + y = 5.

48. Find the coordinates of the foot of perpendicular from the point (–1, 3) to the line 3x – 4y – 16 = 0.

Answer:

Let (a, b) be the coordinates of the foot of the perpendicular from the point (–1, 3) to the line 3x – 4y – 16 = 0.

Slope of the line joining (–1, 3) and (a, b), m₁ = \(\frac{b-3}{a+1}\)

Slope of the line 3x – 4y – 16 = 0 or y = \(\frac34x-4,\) m₂ = \(\frac34\) 

Since these two lines are perpendicular, m₁m₂ = –1

Point (a, b) lies on line 3x – 4y = 16.

∴ 3a – 4b = 16 … (2) 

On solving equations (1) and (2), we obtain

a = \(\frac{68}{25}\) and b = \(-\frac{49}{25}\)

Thus, the required coordinates of the foot of the perpendicular are \(\left(\frac{68}{25},-\frac{49}{25}\right)\).

49. The perpendicular from the origin to the line y = mx + c meets it at the point (–1, 2). Find the values of m and c.

Answer:

The given equation of line is y = mx + c. 

It is given that the perpendicular from the origin meets the given line at (–1, 2). 

Therefore, the line joining the points (0, 0) and (–1, 2) is perpendicular to the given line.

∴ Slope of the line joining (0, 0) and (–1, 2) = \(\frac{2}{-1}=-2\)

The slope of the given line is m.

∴ m x -2 = -1           [The two lines are perpendicular]

⇒ m = \(\frac12\)

Since point (–1, 2) lies on the given line, it satisfies the equation y = mx + c.

Thus, the respective values of m and c are \(\frac12\) and \(\frac52\).

Answer 7

50. If p and q are the lengths of perpendiculars from the origin to the lines x cos θ – y sin θ = k cos 2θ and x sec θ+ y cosec θ = k, respectively, prove that p² + 4q² = k

Answer:

The equations of given lines are 

x cos θ – y sinθ = k cos 2θ ……………… (1)

 x secθ + y cosec θ= k ……………….… (2) 

The perpendicular distance (d) of a line Ax + By + C = 0 from a point (x₁, y₁) is given by

On comparing equation (1) to the general equation of line i.e., Ax + By + C = 0, we obtain 

A = cosθ, B = –sinθ, and C = –k cos 2θ. 

It is given that p is the length of the perpendicular from (0, 0) to line (1).

On comparing equation (2) to the general equation of line i.e., Ax + By + C = 0, we obtain 

A = secθ, B = cosecθ, and C = –k. 

It is given that q is the length of the perpendicular from (0, 0) to line (2).

From (3) and (4), we have

Hence, we proved that p² + 4q² = k².

51. In the triangle ABC with vertices A (2, 3), B (4, –1) and C (1, 2), find the equation and length of altitude from the vertex A.

Answer:

Let AD be the altitude of triangle ABC from vertex A. 

Accordingly, 

AD \(\perp\) BC

The equation of the line passing through point (2, 3) and having a slope of 1 is 

(y – 3) = 1(x – 2) 

⇒ x – y + 1 = 0 

⇒ y – x = 1 

Therefore, equation of the altitude from vertex A = y – x = 1. 

Length of AD = Length of the perpendicular from A (2, 3) to BC The equation of BC is

The perpendicular distance (d) of a line Ax + By + C = 0 from a point (x1, y1) is given by

On comparing equation (1) to the general equation of line Ax + By + C = 0, we obtain A = 1, B = 1, and C = –3.

∴ Length of AD 

 

Thus, the equation and the length of the altitude from vertex A are y – x = 1 and √2 units respectively.

53. If p is the length of perpendicular from the origin to the line whose intercepts on the axes are a and b, then show that:

\(\frac{1}{p^2}=\frac{1}{a^2}+\frac1{b^2}\)

Answer:

It is known that the equation of a line whose intercepts on the axes are a and b is

The perpendicular distance (d) of a line Ax + By + C = 0 from a point (x₁, y₁) is given by

On comparing equation (1) to the general equation of line Ax + By + C = 0, we obtain 

A = b, B = a, and C = –ab. 

Therefore, if p is the length of the perpendicular from point (x₁, y₁) = (0, 0) to line (1), we obtain

On squaring both sides, we obtain

Hence, we showed that \(\frac{1}{p^2}=\frac{1}{a^2}+\frac1{b^2}\) .

53. Find the values of k for which the line is (k - 3)x - (4 - k²)y + k² - 7k + 6 = 0

(a) Parallel to the x-axis, 

(b) Parallel to the y-axis, 

(c) Passing through the origin.

Answer:

The given equation of line is 

(k – 3) x – (4 – k²) y + k² – 7k + 6 = 0 … (1) 

(a) If the given line is parallel to the x-axis, then 

Slope of the given line = Slope of the x-axis 

The given line can be written as 

(4 – k²) y = (k – 3) x + k² – 7k + 6 = 0

which is of the form y = mx + c.

Slope of the given line = \(\frac{(k-3)}{(4-k^2)}\)

Slope of the x-axis = 0

Thus, if the given line is parallel to the x-axis, then the value of k is 3.

(b) If the given line is parallel to the y-axis, it is vertical. Hence, its slope will be undefined. 

The slope of the given line is \(\frac{(k-3)}{(4-k^2)}\) .

Now, \(\frac{(k-3)}{(4-k^2)}\) is undefined at k² = 4

k² = 4 

⇒ k = ±2 

Thus, if the given line is parallel to the y-axis, then the value of k is ±2.

(c) If the given line is passing through the origin, then point (0, 0) satisfies the given equation of line.

Thus, if the given line is passing through the origin, then the value of k is either 1 or 6.

54. Find the values of θ and p, if the equation x cosθ + y sinθ is the normal form of the line √3x + y + 2 = 0.

Answer:

The equation of the given line is √3x + y + 2 = 0. 

This equation can be reduced as 

√3x + y + 2 = 0. 

⇒ - √3x - y = 2

On dividing both sides by \(\sqrt{(-\sqrt3)^2+(-1)^2}=2\) , we obtain

Comparing equation (1) to , we obtain

Since the values of sin θ and cos θ are negative,

Thus, the respective values of θ and p are \(\frac{7\pi}{6}\) and 1.

Answer 8

55. Find the equations of the lines, which cut-off intercepts on the axes whose sum and product are 1 and –6, respectively.

Answer:

Let the intercepts cut by the given lines on the axes be a and b. 

It is given that 

a + b = 1 ……………… (1) 

ab = –6 ……………….. (2) 

On solving equations (1) and (2), we obtain 

a = 3 and b = –2 or a = –2 and b = 3 

It is known that the equation of the line whose intercepts on the axes are a and b is

Case I: a = 3 and b = –2 

In this case, the equation of the line is –2x + 3y + 6 = 0, i.e., 2x – 3y = 6. 

Case II: a = –2 and b = 3 

In this case, the equation of the line is 3x – 2y + 6 = 0, i.e., –3x + 2y = 6. 

Thus, the required equation of the lines are 2x – 3y = 6 and –3x + 2y = 6.

56. What are the points on the y-axis whose distance from the line \(\frac x3+\frac y4=1\) is 4 units.

Answer:

Let (0, b) be the point on the y-axis whose distance from line \(\frac x3+\frac y4=1\) is 4 units. 

The given line can be written as

4x + 3y – 12 = 0 ………………….… (1) 

On comparing equation (1) to the general equation of line Ax + By + C = 0, we obtain 

A = 4, B = 3, and C = –12.

It is known that the perpendicular distance (d) of a line Ax + By + C = 0 from a point (x₁, y₁) is given by

Therefore, if (0, b) is the point on the y-axis whose distance from line \(\frac x3+\frac y4=1\) is 4 units, then:

Thus, the required points are \((0,\frac{32}3)\) and \((0,-\frac 83)\).

57. Find the perpendicular distance from the origin to the line joining the points

(cos θ, sin θ) and (cos \(\varnothing\), sin \(\varnothing\))

Answer:

The equation of the line joining the points (cos θ, sin θ) and (cos \(\varnothing\), sin \(\varnothing\)) is given by

It is known that the perpendicular distance (d) of a line Ax + By + C = 0 from a point (x₁, y₁) is given by

Therefore, the perpendicular distance (d) of the given line from point (x₁, y₁) = (0, 0) is

58. Find the equation of the line parallel to y-axis and drawn through the point of intersection of the lines x – 7y + 5 = 0 and 3x + y = 0.

Answer:

The equation of any line parallel to the y-axis is of the form 

x = a ……………….… (1) 

The two given lines are

 x – 7y + 5 = 0 ……………….(2) 

3x + y = 0 …………………… (3) 

On solving equations (2) and (3), we obtain

x = \(-\frac5{22}\) and y = \(\frac{15}{22}\) .

Therefore, \(\left(-\frac5{22},\frac{15}{22}\right)\) is the point of intersection of lines (2) and (3).

Since line x = a passes through point \(\left(-\frac5{22},\frac{15}{22}\right)\) So, a = \(-\frac5{22}\) 

Thus, the required equation of the line is x = \(-\frac5{22}\) 

59. Find the equation of a line drawn perpendicular to the line \(\frac x4 + \frac y6=1\) through the point, where it meets the y-axis.

Answer:

The equation of the given line is \(\frac x4 + \frac y6=1\)

This equation can also be written as 3x + 2y – 12 = 0 

\(y = \frac{-3}2x+6\), which is of the form y = mx + c

\(\therefore\) Slope of the = \(-\frac32\) given line 

\(\therefore\) Slope of line perpendicular to the given line = \(-\cfrac1{(-\frac{3}{2})}=\frac23\)

Let the given line intersect the y-axis at (0, y). 

On substituting x with 0 in the equation of the given line, we obtain 

\(\frac y6=1\)

⇒ y = 6

\(\therefore\) The given line intersects the y-axis at (0, 6). 

The equation of the line that has a slope of 2/3 and passes through point (0, 6) is

Thus, the required equation of the line is 2x – 3y + 18 = 0.

60. Find the area of the triangle formed by the lines y – x = 0, x + y = 0 and x – k = 0.

Answer:

The equations of the given lines are 

y – x = 0 ………………… (1) 

x + y = 0 ……………….. (2) 

x – k = 0 ……………… (3) 

The point of intersection of lines (1) and (2) is given by 

x = 0 and y = 0 

The point of intersection of lines (2) and (3) is given by 

x = k and y = –k 

The point of intersection of lines (3) and (1) is given by 

x = k and y = k 

Thus, the vertices of the triangle formed by the three given lines are (0, 0), (k, –k), and (k, k). 

We know that the area of a triangle whose vertices are (x₁, y₁), (x₂, y₂), and (x₃, y₃) is

Therefore, area of the triangle formed by the three given lines

61. Find the value of p so that the three lines 3x + y – 2 = 0, px + 2y – 3 = 0 and 2x – y – 3 = 0 may intersect at one point.

Answer:

The equations of the given lines are 

3x + y – 2 = 0 …………………. (1)

px + 2y – 3 = 0 ……………….. (2)

2x – y – 3 = 0 ………………… (3)

On solving equations (1) and (3), we obtain 

x = 1 and y = –1

Since these three lines may intersect at one point, the point of intersection of lines (1) and (3) will also satisfy line (2). 

p (1) + 2 (–1) – 3 = 0 

p – 2 – 3 = 0 p = 5 

Thus, the required value of p is 5.

Answer 9

62. If three lines whose equations are y = m₁x + c₁, y = m₂x + c₂ and y = m₃x + c₃ are concurrent, then show that  m₁(c₂ - c₃) + m₂(c₃ - c₁) + m₃ (c₁ - c₂) = 0.

Answer:

The equations of the given lines are 

y = m₁x + c₁ ………………. (1) 

y = m₂x + c₂ …………….… (2) 

y = m₃x + c₃ ………………. (3) 

On subtracting equation (1) from (2), we obtain

On substituting this value of x in (1), we obtain

is the point of intersection of lines (1) and (2). 

It is given that lines (1), (2), and (3) are concurrent. Hence, the point of intersection of lines (1) and (2) will also satisfy equation (3).

63. Find the equation of the lines through the point (3, 2) which make an angle of 45° with the line x –2y = 3.

Answer:

Let the slope of the required line be m₁. 

The given line can be written as \(y =\frac12x-\frac32\), which is of the form y = mx + c

\(\therefore \) Slope of the given line = m₂ = \(\frac12\) 

It is given that the angle between the required line and line x – 2y = 3 is 45°. 

We know that if θ is the acute angle between lines l₁ and l₂ with slopes m₁ and m₂ , then

Case I: m₁ = 3 

The equation of the line passing through (3, 2) and having a slope of 3 is: 

y – 2 = 3 (x – 3) 

y – 2 = 3x – 9 

3x – y = 7

Case II: m₁ = \(-\frac13\)

The equation of the line passing through (3, 2) and having a slope of \(-\frac13\) is :

Thus, the equations of the lines are 3x – y = 7 and x + 3y = 9.

64. Find the equation of the line passing through the point of intersection of the lines 4x + 7y – 3 = 0 and 2x – 3y + 1 = 0 that has equal intercepts on the axes.

Answer:

Let the equation of the line having equal intercepts on the axes be

On solving equations 4x + 7y – 3 = 0 and 2x – 3y + 1 = 0, we obtain

\(\therefore\left(\frac1{13},\frac5{13}\right)\)is the point of intersection of the two given lines. 

Since equation (1) passes through \(\left(\frac1{13},\frac5{13}\right)\) point,

\(\therefore \) Equation (1) becomes 

Thus, the required equation of the line is 13x + 13y = 6.

65. Show that the equation of the line passing through the origin and making an angle θ with the line

Answer:

Let the equation of the line passing through the origin be y = m₁x. If this line makes an angle of θ with line y = mx + c, then angle θ is given by

Case I:

Case II:

Therefore, the required line is given by 

66. In what ratio, the line joining (–1, 1) and (5, 7) is divided by the line x + y = 4?

Answer:

The equation of the line joining the points (–1, 1) and (5, 7) is given by

The equation of the given line is 

x + y – 4 = 0 ……… (2) 

The point of intersection of lines (1) and (2) is given by 

x = 1 and y = 3 

Let point (1, 3) divide the line segment joining (–1, 1) and (5, 7) in the ratio 1:k. 

Accordingly, by section formula,

Thus, the line joining the points (–1, 1) and (5, 7) is divided by line x + y = 4 in the ratio 1:2.

67. Find the distance of the line 4x + 7y + 5 = 0 from the point (1, 2) along the line 2x – y = 0.

Answer:

The given lines are

2x – y = 0 …………………… (1) 

4x + 7y + 5 = 0 ……………… (2) 

A (1, 2) is a point on line (1). 

Let B be the point of intersection of lines (1) and (2).

On solving equations (1) and (2), we obtain 

\(\therefore \) Coordinates of point B are \(\left(\frac{-5}{18},\frac{-5}{9}\right).\)

By using distance formula, the distance between points A and B can be obtained as

Thus, the required distance is \(\frac{23\sqrt5}{18}\) units.

68. Find the direction in which a straight line must be drawn through the point (–1, 2) so that its point of intersection with the line x + y = 4 may be at a distance of 3 units from this point.

Answer:

Let y = mx + c be the line through point (–1, 2). 

Accordingly, 2 = m (–1) + c. 

⇒ 2 = –m + c 

⇒ c = m + 2 

∴ y = mx + m + 2 ……………… (1) 

The given line is 

x + y = 4 ……………….. (2) 

On solving equations (1) and (2), we obtain

 

is the point of intersection of lines (1) and (2).

Since this point is at a distance of 3 units from point (– 1, 2), according to distance formula,

Thus, the slope of the required line must be zero i.e., the line must be parallel to the x-axis.

Answer 10

69. Find the image of the point (3, 8) with respect to the line x + 3y = 7 assuming the line to be a plane mirror.

Answer:

The equation of the given line is

 x + 3y = 7 … (1) 

Let point B (a, b) be the image of point A (3, 8). 

Accordingly, line (1) is the perpendicular bisector of AB.

Slope of AB = \(\frac{b-8}{a-3}\), while the slope of line (1) = \(-\frac13\)

Since line (1) is perpendicular to AB,

The mid-point of line segment AB will also satisfy line (1). 

Hence, from equation (1), we have

On solving equations (2) and (3), we obtain a = –1 and b = –4. 

Thus, the image of the given point with respect to the given line is (–1, –4).

70. If the lines y = 3x + 1 and 2y = x + 3 are equally inclined to the line y = mx + 4, find the value of m.

Answer:

The equations of the given lines are 

y = 3x + 1 ………………… (1) 

2y = x + 3 ……………..… (2) 

y = mx + 4 ……………….. (3) 

Slope of line (1), m₁ = 3 

Slope of line (2), m₂ = 1/2 

Slope of line (3), m₃ = m

It is given that lines (1) and (2) are equally inclined to line (3). This means that the angle between lines (1) and (3) equals the angle between lines (2) and (3).

Hence, this case is not posible.

Thus, the required value of m is \(\frac{1\pm5\sqrt2}{7}\)

71. If sum of the perpendicular distances of a variable point P (x, y) from the lines x + y – 5 = 0 and 3x – 2y + 7 = 0 is always 10. Show that P must move on a line.

Answer:

The equations of the given lines are x + y – 5 = 0 … (1) 3x – 2y + 7 = 0 … (2) 

The perpendicular distances of P (x, y) from lines (1) and (2) are respectively given by

It is given that d₁ + d₂ = 10

which is the equation of a line. 

Similarly, we can obtain the equation of line for any signs of (x + y - 5) and (3x - 2y + 7)

Thus, point P must move on a line.

72. Find equation of the line which is equidistant from parallel lines 9x + 6y – 7 = 0 and 3x + 2y + 6 = 0.

Answer:

The equations of the given lines are 

9x + 6y – 7 = 0 … (1)

3x + 2y + 6 = 0 … (2) 

Let P (h, k) be the arbitrary point that is equidistant from lines (1) and (2). 

The perpendicular distance of P (h, k) from line (1) is given by

The perpendicular distance of P (h, k) from line (2) is given by  

                         

Since P (h, k) is equidistant from lines (1) and (2), d₁ = d₂  

            

      

∴ 9h + 6k – 7 = – 9h – 6k – 18 

⇒ 18h + 12k + 11 = 0 

Thus, the required equation of the line is 18x + 12y + 11 = 0. 

73. A ray of light passing through the point (1, 2) reflects on the x-axis at point A and the reflected ray passes through the point (5, 3). Find the coordinates of A.

Answer:

Let the coordinates of point A be (a, 0). 

Draw a line (AL) perpendicular to the x-axis. 

We know that angle of incidence is equal to angle of reflection. 

Hence, let 

∠BAL = ∠CAL = Φ 

Let ∠CAX = θ 

∴∠OAB = 180° – (θ + 2Φ) = 180° – [θ + 2(90° – θ)] 

= 180° – θ – 180° + 2θ = θ 

∴ ∠BAX = 180° – θ

From equations (1) and (2), we obtain

Thus, the coordinates of point A are \(\left(\frac{13}{15},0\right). \)