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Question: If the algebraic sum of distances of points (2,1), (3,2) and (–4, 7) from the line y = mx + c is zer...

If the algebraic sum of distances of points (2,1), (3,2) and (–4, 7) from the line y = mx + c is zero, then this line will always pass through a fixed point whose co-ordinate is–

A

(1, 3)

B

(1, 10)

C

(1, 6)

D

(13,103)\left( \frac { 1 } { 3 } , \frac { 10 } { 3 } \right)

Answer

(13,103)\left( \frac { 1 } { 3 } , \frac { 10 } { 3 } \right)

Explanation

Solution

P1 + P2 + P3 = 0

+ + = 0

2m + 3m – 4m – 1 – 2 – 7 + 3c = 0

m – 10 + 3c = 0 mx – y + c = 0

10y\frac { - 10 } { - y } = 31\frac { 3 } { 1 } x = 1/3, y = 10/3