Question

If the origin is the orthocentre of the triangle formed by the points (5, -1), (-2, 3), and (c, d), then find its third vertex.

Answer 1

Step-by-step explanation:

Let the coordinates of triangle be $A(5,-1),\:B(-2,3),\:C(c,d)$

Since, origin $O(0,0)$ is the orthocentre, then we have,

$m_{AO}\times\:m_{BC}=-1$

$\implies \frac{0 - ( - 1)}{0 - 5} . \frac{d - 3}{c - ( - 2)} = - 1$

$\implies \frac{1}{- 5} . \frac{d - 3}{c + 2} = - 1$

$\implies \: d - 3 = 5(c + 2)$

$\implies \: 5c + 10 = d - 3$

$\tt\implies \: \purple{ \bold{ 5c - d + 13 = 0 }} \: \: \: \: ....(i)$

Also,

$m_{BO}\times\:m_{AC}=-1$

$\implies \frac{0 - 3}{0 - ( - 2)} .\frac{d - ( - 1)}{c - 5} = - 1$

$\implies \frac{- 3}{ 2} .\frac{d + 1}{c - 5} = - 1$

$\implies \frac{3}{ 2} .\frac{d + 1}{c - 5} = 1$

$\implies 3(d + 1) = 2(c - 5)$

$\implies 3d + 3= 2c - 10$

$\tt \implies \purple{ \bold{ 2c - 3d - 13 = 0 }} \: \: \: ....(ii)$

From (i) and (ii),

$5c - d + 13 = 0 \\ 2c - 3d - 13 = 0$

By cross multilication, we have,

$\frac{c}{( - 1)( - 13) - ( - 3)(13)} = \frac{d}{(13)(2) - ( - 13)(5)} = \frac{1}{(5)( - 3) - (2)( - 1)}$

$\implies \frac{c}{ 13 + 39} = \frac{d}{26 + 65} = \frac{1}{- 15 + 2}$

$\implies \frac{c}{ 52} = \frac{d}{91} = \frac{1}{- 13}$

$\implies c = \frac{52}{ - 13} \: \: \: and \: \: \: d = \frac{91}{- 13}$

$\implies c = - 4 \: \: \: and \: \: \: d = - 7$

$\sf\:\red{Hence,\: required\:coordinates \: of \:3^{rd}\:vertex\: is\:(-4,-7)}$