Question

A point equidistant from the lines $4x+3y+10=0,5x-12y+26=0$ 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 co-ordinates of the point be $(a,b)$

Now, the distance of the point $(a,b)$ from $4x+3y+10=0$ is given by

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

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

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

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

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

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

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

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

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.