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

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

MathematicsStraight LinesAP EAMCETAP EAMCET 2017 (26 Apr Shift 2)

Options:

A $\left(\frac{-7}{5}, \frac{1}{5}\right)$
B $\left(\frac{7}{5}, \frac{-1}{5}\right)$
C $(2,-2)$
D $(-2,2)$

Solution:

1867 Upvotes Verified Answer

The correct answer is: $(-2,2)$

No solution. Refer to answer key.

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