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

- 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.

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