Question

A point equidistant from the line $4x+3y+10=0,5x-12y+26=0$ and $7x+24y-50=0$ is

• A. (1, -1)
• B. (1, 1)
• C. (0, 0)
• D. (0, 1)

Answer 1

The correct option is C

(0, 0)

Let the coordinates of the point be (a, b)

Now, the distance of the point (a, b)

from $4x+3y+10=0$ is given by

$\left|\frac{4a+3b+10}{\sqrt{4^2+3^2}}\right|$

$=\left|\frac{4a+3b+10}{5}\right|$

Again, the distance of the point (a, b) from $5x-12y+26=0$ is given by

$\left|\frac{5a-12b+26}{\sqrt{5^2+(-12{)}^2}}\right|$

$=\left|\frac{5a-12b+26}{13}\right|$

Again, the distance of the point (a, b) from $7x+24y-50=0$ is given by

$\left|\frac{7a+24b-50}{\sqrt{7^2+(24{)}^2}}\right|$

$=\left|\frac{7a+24b-50}{25}\right|$

Now $\left|\frac{4a+3b+10}{5}\right|=\left|\frac{5a-12b+26}{13}\right|$

$=\left|\frac{7a+24b-50}{25}\right|$

Only $a=0$ and $b=0$ is satisfying the above equation:

Hence, the correct answer is option (c).