The point on the line $3 x+y+4=0$ which is equidistant from $(-5,6)$ and $(3,2)$ is

The point on the line $3 x+y+4=0$ which is equidistant from $(-5,6)$ and $(3,2)$ is
  1. $\left(\frac{-7}{5}, \frac{1}{5}\right)$
  2. $\left(\frac{7}{5}, \frac{-1}{5}\right)$
  3. $(2,-2)$
  4. $(-2,2)$

Solution

No solution. Refer to answer key.

Asked in: AP EAMCET 2017 (26 Apr Shift 2)

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