The orthocentre of triangle formed by lines 4 x-7 y+10=0, x+y=5 and 7 x+4 y=15 is

Question

The orthocentre of triangle formed by lines 4 x-7 y+10=0, x+y=5 and 7 x+4 y=15 is (A) (1, 2) (B) (1, -2) (C) (-1, -2) (D) (-1, 2)

Tardigrade Admin · Accepted Answer

Orthocentre of right angled triangle is same as the vertex of right angle. Therefore, point of intersection of 7 x +4 y-15=0 and 4 x-7 y+10=0 is (1,2)

Q. The orthocentre of triangle formed by lines $4 x - 7 y + 10 = 0$ , $x + y = 5$ and $7 x + 4 y = 15$ is

(1, 2)

(1, -2)

(-1, -2)

(-1, 2)

Orthocentre of right angled triangle is same as the vertex of right angle. Therefore, point of intersection of $7 x$ $+ 4 y - 15 = 0$ and $4 x - 7 y + 10 = 0$ is $(1, 2)$

Q. The orthocentre of triangle formed by lines 4x−7y+10=0, x+y=5 and 7x+4y=15 is