Question

Find the orthocenter of a triangle when their vertices are $A(1,2),B(2,6),C(3,-4)$

Answer 1

Given, the vertices of the triangle,

$A=(1,2)$

$B=(2,6)$

$C=(3,-4)$

Slope of $AB$

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

$=\frac{6-2}{2–1}$

$=\frac{4}{1}=4$

Slope of $CF$

= Perpendicular slope of $AB$

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

$=-14$

The equation of $CF$ is given as,

$y–y_1=m(x–x_1)$

$y+4=-14(x–3)$

$4y+16=-x+3$

$x+4y=-13$ ——————————– (1)

Slope of $BC$

$=\frac{y_2–y_1}{x_2–x_1}$

$=\frac{-4–6}{3–2}$

$=\frac{-10}{1}=-10$

Slope of $AD$

=Perpendicular slope of $BC$

$=\frac{-1}{SlopeofBC}$

$=\frac{-1}{-10}$

$=\frac{1}{10}$

The equation of $AD$ is given as,

$y–y_1=m(x–x_1)$

$y+2=\frac{1}{10}(x–1)$

$10y+20=x–1$

$x–10y=21$ ——————————– (2)

Subtracting equation (1) and (2),

$x+4y=-13$

$-x+10y=-21$

——————

$14y=-34$

$y=-2.429$

Substituting the value of $y$ in equation (1),

$x+4y=-13$

$x+4(-2.429)=-13$

$x–9.714=-13$

$x=-3.286$

Orthocenter $=(-3.286,-2.429)$