Question

The locus of the point which moves such that its distance from the point $(4,5)$ is equal to its distance from the line $7x-3y-13=0$ is

• A. A straight line
• B. A circle
• C. A parabola
• D. An ellipse

Answer 1

Answer: (A)

A straight line

$(4,5)$ lies on $7x-3y-13=0$

Hence, if a point $(h,k)$ is equidistant from line and $(4,5)$, it must lie on a line perpendicular to $7x-3y-13=0$ and passing through $(4,5)$ as shown in figure.

Using the concept: line perpendicular to $ax+by+c=0$ is $bx-ay+c^{\prime }=0$

Line perpendicular to $7x-3y-13=0$ will be $3x+7y=c$

Now $(4,5)$ lies on it

$\therefore 12+35=c=47$

The equation is $3x+7y=47$.