Question

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 $(2,2).$

Answer 1

Toolbox:

• Slope of a line is $m=-\frac{coeffofx}{coeffofy}$
• $\tan \theta =|\frac{m_1-m_2}{1+m_1{m}_2}|$

Step 1 :

The equation of the hypotenuse is $3x+4y=4$ .

Hence the slope of the line is $-\frac{3}{4}$

Since it is an isoceles right angled triangle the angles should be ${45}^{\circ },{45}^{\circ }$ and ${90}^{\circ }$

$\tan {45}^{\circ }=|\frac{m-(-\frac{3}{4})}{1+m(-\frac{3}{4})}|$

But $\tan {45}^{\circ }=1$

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

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

Step 2 :

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

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

$\frac{7m}{4}=\frac{1}{4}$

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

The coordinates of B is (2,2)

Hence the equation of the line AB is

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

$7y-14=x-2$

$\therefore x-7y=-12$

Hence $x-7y+12$ is the required equation.