Question

Find the area of the $△ABC$, coordinates of whose vertices are $A(4,1),B(6,6)$ and $C(8,4)$.

Answer 1

A triangle contains three line segments $AB,BC$ and $CA$, so let's find the equations for these line segment.

Let us first find for $A(4,1)$ and $B(6,6)$

Equation of line $AB$ is given by

$\frac{y-6}{x-6}=\frac{1-6}{4-6}=$

$⟹\frac{y-6}{x-6}=\frac{5}{2}$

$⟹2y-12=5x-30$

$⟹y=\frac{5x-18}{2}$ ........ $\left(i\right)$

Let's find for $B(6,6),C(8,4)$

Equation of line $BC$ is given by

$\frac{y-6}{x-6}=\frac{4-6}{8-6}=$

$⟹\frac{y-6}{x-6}=\frac{-2}{2}=\frac{-1}{1}$

$⟹y-6=-x+6$

$⟹y=-x+12$ ........ $\left(ii\right)$

Let's find for $C(8,4),A(4,1)$

Equation of line $CA$ is given by

$\frac{y-4}{x-8}=\frac{1-4}{4-8}$

$⟹\frac{y-4}{x-8}=\frac{3}{4}$

$⟹4y-16=3x-24$

$⟹y=\frac{3x-8}{4}$ ........ $\left(iii\right)$

Area of triangle $ABC$ is given by

$A\left(△ABC\right)={\int }_4^6\frac{5x-18}{2}dx+{\int }_6^8(-x+12)dx-{\int }_4^8\frac{3x-8}{4}dx$

$=\frac{1}{2}{|\frac{5x^2}{2}-18x|}_4^6+{|-\frac{x^2}{2}+12x|}_6^8-\frac{1}{4}{|\frac{3x^2}{2}-8x|}_4^8$

$=\frac{1}{2}|\frac{5(36-16)}{2}-18(6-4\left)\right|+|-\frac{(64-36)}{2}+12(8-6\left)\right|-\frac{1}{4}{|\frac{3(64-16)}{2}-8(8-4\left)\right|}_4^8$

$=\frac{1}{2}(50-36)+(-14+24)-\frac{1}{4}(72-32)$

$=7+10-10=7$

Hence, $A\left(△ABC\right)=7$ units