Question

Straight line $3x+4y=5$ and $4x-3y=15$ intersect at the point $A.$ Points $B$ and $C$ are chosen on these two lines such that $AB=AC$. Possible equation of $BC$ passing through $(1,2)$ can be

• A. $5x+2y-9=0$
• B. $7x+y-9=0$
• C. $x-7y+13=0$
• D. $7x+3y-13=0$

Answer 1

Answer: (B), (C)

The correct options are
B $7x+y-9=0$
C $x-7y+13=0$

Given Lines

$3x+4y=5$----(1) and $4x-3y=15$------(2)

On comparing eq (1) and (2) with $y=mx+c$

$m_1=\frac{-3}{4},m_2=\frac{4}{3}$

$m_1{m}_2=\frac{4}{3}\times -\frac{3}{4}=-1$

Hence given lines are perpendicular

So, line $AB=AC$ SO a right angled isosceles $\Delta$

Hence, the line $BC$ through $(1,2)$ will make angle of ${45}^0$ with given lines

So, possible equation of $BC$ are

$(y-2)=\frac{m\pm \tan {45}^0}{1\mp m\tan {45}^0}(x-1)$

where $m=$ slope of $AB-\frac{-3}{4}$

$\Rightarrow (y-2)=\frac{-\frac{3}{4}\pm 1}{1\mp (-\frac{3}{4})}(x-1)$

$\Rightarrow (y-2)=\frac{-3\pm 4}{4\pm 3}(x-1)$

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

$\Rightarrow x-7y+13=0$ and $\Rightarrow 7x+y-9=0$