Question

Find the equations to the sides of an isosceles right angled triangle the equation of whose hypotenuse is 3x + 4y = 4 and the opposite vertex is the point (2, 2)

Answer 1

Given: 

hypotenuse is 3x + 4y = 4 of isosceles right angled triangle the opposite vertex is the point (2, 2). 

To find: 

equation of side of isosceles right angle triangle 

Explanation: 

Here, 

we are given △ABC is an isosceles right angled triangle . 

∠A + ∠B + ∠C = 180° 

⇒ 90° + ∠B + ∠B = 180° 

⇒ ∠B = 45°, ∠C = 45° 

Diagram:

Now, we have to find the equations of the sides AB and AC, where 3x + 4y = 4 is the equation of the hypotenuse BC. 

We know that the equations of two lines passing through a point x₁,y₁ and making an angleα with the given line y = mx + c are

 y - y1 = \(\frac {m±\,tan\ \,\alpha}{1±m\,tan\ \,\alpha}\) (x - x₁)

Here, Equation of the given line is, 

3x + 4y = 4 

⇒ 4y = - 3x + 4

⇒ y = \(-\frac{3}{4}\) x + 1

Comparing this equation with y = mx + c we get,

m = \(-\frac{3}{4}\) 

x₁ = 2, y₁ = 2, α = 45^∘, m = \(-\frac{3}{4}\) 

So, the equations of the required lines are

⇒ x – 7y + 12 = 0 and 7x + y – 16 = 0 

Hence, Equation of given line is x – 7y + 12 = 0 and 7x + y – 16 = 0