Question

Equation of one of the sides of an isosceles right-angled triangle whose hypotenuse is $3x+4y=4$ and the opposite vertex of the hypotenuse is $(2,2)$, will be

• A. $x–7y+12=0$
• B. $7x+y–12=0$
• C. $x–7y+16=0$
• D. $7x+y+16=0$

Answer 1

The correct option is A

$x–7y+12=0$

Step 1. Draw a triangle $ABC$ right angled at $B$

As shown in the figure,

hypotenuse is along the line $3x+4y-4=0$

$\therefore$Slope of $AC=\frac{-3}{4}$

$\because ABC$ is an isosceles right triangle,

$\Rightarrow$$\angle BAC=\angle ACB$

$=45°$

Step 2. Let the slope of the line making an angle $45°$ with $AC$ be $m$

As we know, angle between two lines is $\mathit{t}\mathit{a}\mathit{n}\mathbf{}\mathit{\theta }\mathbf{=}\left|\frac{m_1-m_2}{1+m_1{m}_2}\right|$

$\therefore$ $\tan 45°=\left|\frac{m-\left(-{\displaystyle \frac{3}{4}}\right)}{1+m\left(-{\displaystyle \frac{3}{4}}\right)}\right|$

$\Rightarrow$$\frac{4m+3}{4-3m}=\pm 1$

$\Rightarrow$ $4m+3=4-3m$

Or, $4m+3=3m-4$

$\Rightarrow$ $m=\frac{1}{7}$

Or, $m=-7$

Now, if the slope of $BC=\frac{1}{7}$

then the slope of $AB=-7$

Step 3. Find the equation lines:

So, the equation of $BC$ will be

$y-2=\frac{1}{7}\left(x-2\right)$

$\Rightarrow$$x-7y+12=0$

and equation of $AB$ will be

$y-2=-7\left(x-2\right)$

$\Rightarrow$$7x+y-16=0$

Hence, Option ‘A’ is Correct.