Question

Let the point $A$ lies on $3x-4y+1=0$ , the point $B$ lies on $4x+3y-7=0$ and the point $C$ is $\left(- 2,5\right)$ . If $ABCD$ is a rhombus, then the locus of $D$ is

• A. $25\left(\left(x + 2\right)^{2} + \left(y - 5\right)^{2}\right)=\left(3 x - 4 y + 1\right)^{2}$
• B. $\left(3 x - 4 y + 1\right)^{2}+\left(4 x + 3 y - 7\right)^{2}=1$
• C. $\left(3 x - 4 y + 1\right)^{2}-\left(4 x + 3 y - 7\right)^{2}=1$
• D. $\left(4 x + 3 y - 7\right)^{2}-\left(3 x - 4 y + 1\right)^{2}=1$

Answer 1

Answer: (A)

$C\left(- 2,5\right)$ lies on $4x+3y-7=0$
$\Rightarrow $ $BC$ is perpendicular to $3x-4y+1=0$
$\Rightarrow $ $DA$ is perpendicular to $3x-4y+1=0$
$\Rightarrow $ Because $DA=DC$
so, the distance of $D$ from $3x-4y+1=0$ is equal to $DC$
$\Rightarrow \sqrt{\left(h + 2\right)^{2} + \left(k - 5\right)^{2}}=\left|\frac{3 h - 4 k + 1}{5}\right|$
$\Rightarrow 25\left(\left(x + 2\right)^{2} + \left(y - 5\right)^{2}\right)=\left(3 x - 4 y + 1\right)^{2}$