Question

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

पर्याय

• (1, −1)

•  (1, 1)

• (0, 0)

• (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

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

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

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

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

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

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

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

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

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).