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

Let the coordiantes of the point be (a, b)
Now, the distance of the point (a, b) from 4x + 3y + 10 = 0 is given by
$$\begin{gathered} \left|\frac{4a+3b+10}{\sqrt{4^2+3^2}}\right| \\ =\left|\frac{4a+3b+10}{5}\right| \end{gathered}$$
Again, the distance of the point (a, b) from 5x − 12y + 26 = 0 is given by
$$\begin{gathered} \left|\frac{5a-12b+26}{\sqrt{5^2+{\left(-12\right)}^2}}\right| \\ =\left|\frac{5a-12b+26}{13}\right| \end{gathered}$$
Again, the distance of the point (a, b) from 7x + 24y − 50 = 0 is is given by
$$\begin{gathered} \left|\frac{7a+24b-50}{\sqrt{7^2+{\left(24\right)}^2}}\right| \\ =\left|\frac{7a+24b-50}{25}\right| \end{gathered}$$
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).