1. Passing through the point (-2, 3) with slope -4.

Question

Find the equation the following lines satisfying the given conditions. 

1. Passing through the point (-2, 3) with slope -4.

2. Passing through the point (-4, 3) with slope \(\frac{1}{2}\).

3. Line with y-intercept −\(\frac{3}{2}\) and slope \(\frac{1}{2}\)

4. Line with x-intercept – 3 and slope – 2.

5. Line which makes intercepts -3 and 2 on the x- and y-axis respectively.

6. Perpendicular distance from origin is 5 units and the angle the perpendicular makes with the positive direction of x-axis is 30°.

7. Passing through the point (-1, 1) and (2, -4).

Passing through the point (1, -1) and (3, 5).

Answer 1

1. Equation of the line is y – y₁ = m(x – x₁)

⇒ y – 3 = -4(x – (-2))

⇒ y – 3 = -4x – 8 ⇒ 4x + y + 5 = 0.

2. Equation of the line is y – y₁ = m(x – x₁)

⇒ y – 3 = 12(x – (-4))

⇒ 2y – 6 = x + 4 ⇒ x – 2y + 10 = 0.

3. Equation of the line is y = mx + c

⇒ y = \(\frac{1}{2}\) x \(-\frac{3}{2}\) 

⇒ 2y = x – 3

⇒ x – 2y – 3 = 0.

4. Equation of the line is y = m(x – d)

⇒ y = -2(x – (-3)) 

⇒ y = -2x + 6

⇒ 2x + y = 6.

5. Equation of the line is \(\frac{x}{a} + \frac{y}{b}\) = 1

⇒ \(\frac{x}{-3}\) + \(\frac{y}{2}\) = 1 

⇒ 2x – 3y = -6

⇒ 2x – 3y + 6 = 0.

6. Equation of the line is xcosθ + ysinθ = p

⇒ xcos30° + ysin30° = 5

x\(\frac{\sqrt{3}}{2}\) + y\(\frac{1}{2}\) = 5

\(\Rightarrow\) √3x + y = 10

\(\Rightarrow\) √3x + y - 0 = 0

7. Equation of the line is

8. Equation of the line is