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Q.
The orthocentre of the triangle formed by joining the points $(1, - 3)$, $(6,1)$ and $(4, - 1)$ is
Straight Lines
Solution:
Given points are $A(1, - 3),B(6, 1)$ and $C(4, - 1)$ Orthocentre is the point of intersection of the altitudes drawn from vertex to opposite side. We need the equations of any two sides and equations of corresponding altitudes.
Now, equation of $BC$ is $y + 1 = 1 (x - 4)$
$\left(\because m_{BC}=\frac{-1-1}{4-6}=1\right)$
$\Rightarrow x - y - 5 = 0$
$\therefore $ Equation of $AD$ is $y+3 = -1 \left(x- 1\right)\,\left[AD\,\bot\,BC \therefore m_{AD}=-1\right]$
$\Rightarrow x + y + 2 = 0\quad\left(i\right)$
Now equation of $AB$ is $4x - 5y - 19 = 0$
$\therefore $ Equation of $CF$ is $5x + Ay - 16 = 0\quad\ldots\left(ii\right)$
Now solving $(i)$ and $(ii)$, we get
$x = 24$,
$y = -26$
$\therefore $ Coordinate of orthocentre is $(24, -26)$.
