Question

For what value of $k$ are the two straight lines $3x+4y=1$ and $4x+3y+2k=0$ equidistant from the point $(1,1)$?

• A. $\frac{1}{2}$
• B. $2$
• C. $-2$
• D. $-\frac{1}{2}$

Answer 1

Answer: (D)

$-\frac{1}{2}$

Distance of a point $(m,n)$ from line $Ax+By+C=0$ is,

$\frac{|Am+Bn+C|}{\sqrt{A^2+B^2}}$

Since, the lines $3x+4y=1$ and $4x+3y+2k=0$ are equidistant from the point $(1,1)$

We have,

$\frac{|3+4-1|}{\sqrt{3^2+4^2}}=\frac{|3+4+2k|}{\sqrt{3^2+4^2}}$

$⟹2k+7=6⟹k=-\frac{1}{2}$ or $2k+7=-6⟹k=-\frac{13}{2}$

Hence, option D is correct.