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 angle α with the given line y = mx + c are

Here,
Equation of the given line is,

3x + 4y = 4

⇒ 4y = - 3x + 4

Comparing this equation with y = mx + c

we get,

x1 = 2, y1 = 2, α = 45∘, m

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