Q. 65.0( 3 Votes )
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).
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
we are given △ABC is an isosceles right angled triangle .
∠A + ∠B + ∠C = 180°
⇒ 90° + ∠B + ∠B = 180°
⇒ ∠B = 45°, ∠C = 45°
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
Equation of the given line is,
3x + 4y = 4
⇒ 4y = - 3x + 4
Comparing this equation with y = mx + c
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
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