Question

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

• A. $(-12,-10)$
• B. $(-11,-10)$
• C. $(-1,-10)$
• D. $(1,10)$

Answer 1

The correct option is B $(-11,-10)$

The point where the altitudes of a triangle meet called Orthocentre

We have given a triangle $ABC$ whose vertices are $(1,2),(-2,8),(7,-1)$

In Step 1 : we find slopes Of $AB,BC,CA$

Slope formula:-$\frac{y_2-y_1}{x_2-X_1}$

slope $AB=\frac{(8-2)}{(-2-1)}=-2$

slope $BC=9/-9=-1$

slope $CA=3/-6=-1/2$

In Step 2:

But we know Orthocentre is the point where perpendiculars drawn from vertex to opposite side meet. So

Let's think a triangle ABC and AD, BE, CF are perpendiculars drawn to the vertex.

Slope $AD=-1/slope\left(BC\right)=1$

.......$BE=-1/slope\left(CA\right)=2$

.....$CF=-1/slope\left(AB\right)=-1/-2=1/2$

Step 3:- we have now vertices and slopes of AD,BE,CF we find equations of lines AD,BE and CF

we have $A(1,2)$ and $m=1$ we substitute in the equation $y-y_1=m(x-X_1)$

$y-2=\left(1\right)(x-1)$

$x-y=-1$ ............ -[ eq 1]

$B(-2,8)$ and slope $BE=2$

$y-8=2(x+2)$

$2x-y=-12$........... -[eq 2]

solving eq (1) and eq (2) ,we get $x=-11,y=-10$

So, Orthotcentre is $(-11,-10)$.