Question

The sides of a triangle are $y=x,y=2x$ and $y=3x+4$. Find the coordinates of its orthocentre.

• A. $(-1,6)$
• B. $(-8,5)$
• C. $(-6,7)$
• D. $(-18,6)$

Answer 1

The correct option is D $(-18,6)$

The point where the altitudes of a triangle meet called Orthocentre

We have given a triangle $ABC$ whose vertices are $(0,0),(-2,-2),(-4,-8)$

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

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

slope $AB=\frac{(0+2)}{(0+2)}=1$

slope $BC=6/2=3$

slope $CA=8/4=2$

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 $AD=-1/slope\left(BC\right)=-1/3$

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

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

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

we have $A(0,0)$ and $m=-1/3$ we substitute in the equation $y-y_1=m(x-X_1)$

$y-0=(-1/3)(x-0)$

$3y+x=0$ ............ -[ eq 1]

$B(-2,-2)$ and slope $BE(-1/2)$

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

$2y+x+6=0$........... -[eq 2]

solving eq.(1) and eq (2) ,we get $x=-18,y=6$

So, Orthotcentre is $(-18,6)$.