Question

Orthocentre of the triangle whose vertices are A(3,4), B(0,0) & C(4,0) is

• A. $(3,\frac{3}{4})$
• B. $(3,\frac{5}{4})$
• C. $(3,12)$
• D. $(2,0)$

Answer 1

The correct option is A $(3,\frac{3}{4})$

$Letthevertices\text{of triangle ABC be}$
$\text{A}\left(\text{3}\text{.4}\right),B(0,0),C(4,0)$
$LetAD\text{be the perpendicular on BCand BE be the perpendicular on AC}$
$\text{Equation of BC}$
$\text{y=0}.......\left(i\right)$
$EquationofAD$
$x=3....\left(ii\right)$
$slopeofBC=0$
$SlopeofAC=\frac{y_2-y_1}{x_2-x_1}=\frac{4-0}{3-4}=-4$
$slopeofBE=\frac{1}{4}$
$EquationofBEpas\sin gthrough(0,0)$
$y=mx+c$
$y=\frac{x}{4}+c$
$itpassesthrough(0,0)$
$0=0+c$
$hencetheequationofBE$
$y=\frac{x}{4}......\left(iii\right)$
$pluggingx=3from\left(ii\right)$
$y=\frac{3}{4}$
$Henceorthocenter(3,\frac{3}{4})$