Find the equation of one of the sides of an isosceles right angled triangle whose hypotenuse is given by 3x + 4y = 4 and the opposite vertex of the hypotenuse is (2, 2).

Given that equation of the hypotenuse is 3x + 4y = 4

and opposite vertex of the hypotenuse is (2, 2)

Firstly, we find the slope of the given equation

3x + 4y = 4

It can be re-written as 4y = 4 – 3x

Since, the above equation is in y = mx + b form

So,

Now, let the slope of AC be m

Now, we find the value of m, by using the formula

Putting the values of m₁ and m₂ in above eq., we get

[∵ tan 45° = 1]

Taking (+) sign, we get

⇒ 4m + 3 = 4 – 3m

⇒ 4m + 3m = 4 – 3

⇒ 7m = 1

Taking (-) sign, we get

⇒ 4m + 3 = - (4 – 3m)

⇒ 4m + 3 = - 4 + 3m

⇒ 4m – 3m = - 4 – 3

⇒ m = -7

If , then equation of AC is

y – y₁ = m(x – x₁)

⇒ 7y – 14 = x – 2

⇒ x – 7y – 2 + 14 = 0

⇒ x – 7y + 12 = 0

If m = -7, then equation of AC is

y – 2 = (-7)(x – 2)

⇒ y – 2 = -7x + 14

⇒ 7x + y = 16

Hence, the required equations are x – 7y + 12 = 0 and 7x + y = 16