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

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 $DC$
$\Rightarrow \sqrt{{\left(h+2\right)}^2+{\left(k-5\right)}^2}=\left|\frac{3h-4k+1}{5}\right|$
$\Rightarrow 25\left({\left(x+2\right)}^2+{\left(y-5\right)}^2\right)={\left(3x-4y+1\right)}^2$