# 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).

$\begin {array} {1 1} (A)\;x-7y-12=0 & \quad (B)\;x+7y-12=0 \\ (C)\;x-7y+12=0 & \quad (D)\;x+7y+12=0 \end {array}$

Toolbox:
• Slope of a line is $m = -\large\frac{coeff\: of \: x}{coeff\: of \: y}$
• $\tan \theta = \bigg| \large\frac{m_1-m_2}{1+m_1m_2} \bigg|$
Step 1 :
The equation of the hypotenuse is $3x+4y=4$ .
Hence the slope of the line is $-\large\frac{3}{4}$
ASince it is an isoceles right angled triangle the angles should be $45^{\circ}, 45^{\circ}$ and $90^{\circ}$
$\tan 45^{\circ} = \bigg| \large\frac{m- \bigg( -\large\frac{3}{4} \bigg)}{1+m\bigg(-\large\frac{3}{4} \bigg) }\bigg|$
But $\tan 45^{\circ}=1$
$1 = \bigg| \large\frac{m+\large\frac{3}{4}}{1-\large\frac{3m}{4}} \bigg|$
$1-\large\frac{3m}{4} = \pm \bigg( m + \large\frac{3}{4} \bigg)$
Step 2 :
If $1 - \large\frac{3m}{4}=m+\large\frac{3}{4}$
$\Rightarrow m+ \large\frac{3m}{4}$$=1-\large\frac{3}{4} \large\frac{7m}{4}=\large\frac{1}{4} \therefore m = \large\frac{1}{7} The coordinates of B is (2,2) Hence the equation of the line AB is (y-2)=\large\frac{1}{7}$$(x-2)$
$7y-14=x-2$
$\therefore x-7y=-12$
Hence $x-7y+12$ is the required equation.
edited Jul 1, 2014