Question

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

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

Answer 1

The correct option is A $x-7y+13=0$ and $7x+y-9=0$

Let $m_1$ and $m_2$ be the slope of lines $3x+4y=5$ and $4x-3y=15$ respectively
Then, $m_1=-\frac{3}{4}$ and $m_2=\frac{4}{3}$.
Clearly, $m_1{m}_2=-1$
So, lines $AB$ and $AC$ are at right angle. thus the triangle ABC is a right angles isosceles $△$
Hence, the line $BC$ through $(1,2)$ will make an angle of ${45}^O$ with the given lines.
So, the possible equation of $BC$ are
$(y-2)=\frac{m\pm \tan {45}^O}{1\mp m\tan {45}^O}(x-1)$

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 $(y-2)=-7(x-1)$

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