9.Straight Line
hard

Two vertices of a triangle are $(5, - 1)$ and $( - 2,3)$. If orthocentre is the origin then coordinates of the third vertex are

A

$(7, 4)$

B

$(-4, 7)$

C

$(4, -7)$

D

$(-4, -7)$

(IIT-1979) (IIT-1983) (AIEEE-2012)

Solution

(d) Let the third vertex be $(h, k)$. Obviously the equation of $AB$ is $\sqrt {\frac{5}{2}} $$ \Rightarrow 4x + 7y – 13 = 0$.
Thus equation of $CE$ is $7x – 4y + \lambda = 0$, but it passes through origin, so $\lambda = 0$. Hence $7x – 4y = 0$ ….(i)
Also equation of $CE$ is $y – 0 = \frac{k}{h}(x – 0)$
==> $ – kx + hy = 0$……(ii)
Here (i) and (ii) are same lines, therefore $(h, k)$ is $( – 4, – 7)$.

Standard 11
Mathematics

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