Question

A triangle ABC with vertices A(-1, 0), B(-2, 3/4) & C(-3, -7/6) has its orthocentre H, then the orthocentre of triangle BCH will be

• A. $(-3,-2)$
• B. $(1,3)$
• C. $(-1,2)$
• D. None of these

Answer 1

Answer: (D)

None of these

The orthocenter is the intersecting point or all the altitude of the triangle. The point where the altitude of a triangle meet is known as the orthocenter.

Given the points,

$A(-1,0)\equiv (x_1,y_1),B(-2,\frac{3}{4})\equiv (x_2,y_2),C(-3,\frac{-7}{6})$

To find the orthocenter, H

$SlopeofAB=\frac{y_2-y_1}{x_2-x_1}=\frac{\frac{3}{4}-0}{-2+1}$ $=\frac{-3}{4}$

Slope of $CF$ $=$ perpendicular slope of $AB$

$=\frac{-1}{SlopeofAB}=\frac{4}{3}$

$\therefore$ Equation of $CF\Rightarrow y-y_1=m(x-x_1)$

$y\frac{-3}{4}=\frac{4}{3}(x+3)\Rightarrow 16x-12y+45=0\to \left(1\right)$

$SlopeofBC=\frac{y_2-y_1}{x_2-x_1}=\frac{\frac{-7}{6}\frac{-3}{4}}{-3+2}=\frac{23}{12}$

$\therefore$ Slope of AD=perpendicular slope of $BC=\frac{-12}{23}$

$\therefore$ Equation of $AD=y-y_1=m(x-x_1)$

$y\frac{-23}{12}=\frac{-12}{23}(x+1)\Rightarrow 144x+276y=385⟶\left(2\right)$

Obtaining the value of $x$ and $y$ by solving can (1) and (2)

$16x-12y+45=0⟶\left(1\right)\times 144144x+276y=385⟶\left(2\right)\times 16$
$\frac{2304x+4416y=6160-2304x\pm 1728y=\pm 6480}{6144y=12640}$

$\Rightarrow y=2.057,x=4.36$

To find the orthocenter of $△$BCH

i.e, Let the orthocenter be denoted as 0

$SlopeofBH=\frac{y_2-y_1}{x_2-x_1}$

$=\frac{0.75-2.06}{-2-4.36}=0.206$

$\therefore$ Slope of CR = perpendicular slope of BH

$=\frac{-1}{SlopeofBH}=\frac{-1}{0.206}=-4.85$

Equation of $CR\Rightarrow y-y_1=m(x-x_1)$

$y-0.206=-4.85(x+3)\Rightarrow 4.85x+y=-14.36⟶\left(1\right)$

$SlopeofBC=\frac{y_2-y_1}{x_2-x_1}=\frac{\frac{-7}{6}\frac{-3}{4}}{-3+2}=\frac{23}{12}=1.92$

Slope of HP = perpendicular slope of HP

$\frac{-1}{slopeofHP}=-0.52$

Equation of $HP=y-1.92=-0.52(x-4.36)$

$\Rightarrow 0.52x+y=4.19⟶\left(2\right)$

Obtaining the value of $x$ and $y$ by solving equation (1) and (2)

$\frac{4.85x+y=-14.36-0.52x\mp y=-4.19}{\therefore 4.33x=-18.55}$
$\Rightarrow x=-4.28\text{and}y=6.42$

$\therefore$ The orthocenter of $△BCH=(-4.28,6.42)$