Two straight lines $3 x+4 y=5$ and $4 x-3 y=15$ intersect at the point $A$. The equations of the lines passing through $(1,2)$ and intersecting the given lines at $B$ and $C$ such that $A B=A C$ are

• A

  $x+4 y=9,4 x-y=2$
• B

  $9 x-2 y=5,2 x+9 y=20$
• C

  $6 x-y=4, x+6 y=13$
• Correct answer
  D

  $7 x+y=9, x-7 y+13=0$

Solution

Given line $3 x+4 y=5 \text { and } 4 x-3 y=15$ Slope of line $3 x+4 y=5$ is i.e., $m_1=\frac{-3}{4}$ Slope of line $4 x-3 y=15$ i.e., $m_2=\frac{4}{3}$
Let the slope of line $B C=m$ Given, $A B=A C$ $\because \quad \tan \alpha=\tan \beta$ $\frac{m-\frac{4}{3}}{1+\frac{4 m}{3}}=\frac{-\frac{3}{4}-m}{1-\frac{3 m}{4}}$ $\frac{3 m-4}{3+4 m}=\frac{-3-4 m}{4-3 m}$ $12 m^2-9 m^2-16+12 m=-9-12 m-12 m-16 m^2$ $\Rightarrow \quad 7 m^2+48 m-7=0$ $(7 m-1)(m+7)=0$ $m=\frac{1}{7}, m=-7$ Equation of line passing through $(1,2)$ and slope $\frac{1}{7}$ is, $=y-2=\frac{1}{7}(x-1)=x-7 y+13=0$ Equation of line passing through $(1,2)$ and slope $=-7$ is $y-2=-7(x-1)$ $7 x+y=9$

Hence, the correct answer is option: 'D'.