Question

Two straight lines $3x+4y=5$ and $4x-3y=15$ intersect at point $A.$ Two points $B$ and $C$ are chosen on these lines such that $AB=AC.$ If the two possible equations of $BC$ passing through $(1,2)$ are $a_{1}x+b_{1}y+13=0$ and $a_{2}x+b_{2}y+9=0$, then the value of $a_2{b}_1-a_1{b}_2$ is equal to

Answer 1

$L_1:3x+4y=5,m_1=\frac{-3}{4}$ and
$L_2:4x-3y=15,m_2=\frac{4}{3}$
Solving both, we get point of intersection $A\equiv (3,-1)$
Also, $m_1\cdot m_2=-1$
$\Rightarrow \angle A={90}^{\circ }$

Now, if we choose points $B$ and $C$ on line $L_1$ and $L_2$ respectively such that $AB=AC,$ then $△ABC$ will be isosceles triangle.
$\Rightarrow \angle ABC=\angle BCA={45}^{\circ }$
$BC$ is passing through $(1,2)$
So possible equation of $BC$ are :
$(y-2)=\tan (\theta \pm {45}^{\circ })(x-1)$
$($ Here, $\tan \theta =-\frac{3}{4}$ $)$
$\Rightarrow (y-2)=\frac{\tan \theta \pm \tan {45}^{\circ }}{1\mp \tan \theta \cdot \tan {45}^{\circ }}(x-1)$
$\Rightarrow x-7y+13=0\cdots \left(1\right)$
$\Rightarrow a_1=1,b_1=-7$
or, $-7x-y+9=0\cdots \left(2\right)$
$\Rightarrow a_2=-7,b_2=-1$
$\therefore a_2{b}_1-a_1{b}_2=49+1=50$