Question

Find the orthocentre of the triangle with sides $x+y=6,2x+y=4$ and $x+2y=5$.

Answer 1

Step 1: Create the diagram

The given sides are,

$x+y=6$ … $\left(1\right)$

$2x+y=4$ … $\left(2\right)$

$x+2y=5$ … $\left(3\right)$

Let $ABC$ be the given triangle.

The coordinates of its vertices $A,B,C$ can be obtained by solving the given equations of its sides in pairs.

Step 2: Calculate coordinates of $A$ and $B$

Upon solving equation $\left(1\right)$ and $\left(3\right)$ we get,

$x+y=6-(x+2y=5)-y=1$

$\Rightarrow$ $y=-1$

Putting this in equation $\left(1\right)$ we get,

$x-1=6$

$\Rightarrow x=6+1=7$

Hence, the coordinates of $A=(7,-1)$.

Upon solving equation $\left(1\right)$ and $\left(2\right)$ we get,

$x+y=6-(2x+y=4)-x=2$

$\Rightarrow$ $x=-2$

Putting this in equation $\left(1\right)$ we get,

$y-2=6$

$\Rightarrow y=6+2=8$

Hence, the coordinates of $B=(-2,8)$.

Step 3: Establish the linear equations of $AD$ and $BE$

Let $AD\perp BC,BE\perp AC$

The point $H$ where $AD$ and $BE$ intersects is called the orthocentre.

Since $AD$ passes through point $A(7,-1)$ and is perpendicular to $BC$, then the equation is

$y+1=\frac{1}{2}\left(x-7\right)$ $\left[\because y-y_1=m\left(x-x_1\right)\right]$

$\Rightarrow 2y+2=x-7$

$\Rightarrow x-2y=9$ …. $\left(4\right)$

Similarly, the equation of $BE$ passing through the point $B(-2,8)$ is

$y–8=2(x+2)$

$\Rightarrow y-8=2x+4$

$\Rightarrow 2x–y=-12$ …. $\left(5\right)$

Step 4: Calculate coordinates of orthocentre

Now by solving equations $\left(4\right)$ and $\left(5\right)$,

$2x-y=-12-2x+4y=-183y=-30$

$\Rightarrow$ $y=-10$

Putting this in equation $\left(4\right)$ we get,

$x-2\left(-10\right)=9$

$\Rightarrow x=9-20=-11$

Therefore, the coordinates of the orthocentre $H$ are $(-11,-10)$.