Question

If the vertices of a triangles are $A(7,-1),B(-2,8)$ and $C(1,2)$ then which of following is/are correct?

• A. Altitude through $A$ is $x-2y-9=0$
• B. Altitude through $B$ is $2x-y+12=0$
• C. Altitude through $C$ is $x-y+1=0$
• D. Orthocentre is $(2,4)$

Answer 1

Answer: (A), (B), (C)

Altitude through $C$ is $x-y+1=0$


Let $AD,BE$ and $CF$ be three altitudes of triangle $ABC$.
Let $m_1,m_2$ and $m_3$ be the slopes of $AD,BE$ and $CF$ respectively.
Then, $AD\perp BC$
$\Rightarrow$ Slope of $AD\times$ Slope of $BC=-1$
$\Rightarrow m_1\times \left(\frac{2-8}{1+2}\right)=-1$
$\Rightarrow m_1=\frac{1}{2}$
$BE\perp AC$
$\Rightarrow$ Slope of $BE\times$ Slope of $AC=-1$
$\Rightarrow m_2\times \left(\frac{2+1}{1-7}\right)=-1$
$\Rightarrow m_2=2$
And. $CF\perp AB$
$\Rightarrow$ Slope of $CF\times$ Slope of $AB=-1$
$\Rightarrow m_3\times \left(\frac{8+1}{-2-7}\right)=-1$
$\Rightarrow m_3=1$
Since $AD$ passes through $A(7,-1)$ and has slope of
$m_1=\frac{1}{2}$
So, its equation is :
$y+1=\frac{1}{2}(x-7)$
$\Rightarrow x-2y-9=0$
Similarly, equation of $BE$ is :
$y-8=2(x+2)$
$\Rightarrow 2x-y+12=0$
Equation of $CF$ is :
$y-2=1(x-1)$
$\Rightarrow x-y+1=0$

Orthocentre is the intersection of the altitudes.
$x-2y-9=\mathrm{0...}\left(i\right)$
$2x-y+12=\mathrm{0...}\left(ii\right)$
Solving $\left(i\right)$ and $\left(ii\right)$ we get
$x=-11,y=-10$