Question

Using integration, find the area of the triangle $PQR$, whose vertices are $P(2,5),Q(4,7)$ and $R(6,2)$.

Answer 1

Using integration finding area of triangle $P(2,5)Q(4,7)$ and $R(6,2)$

Equation of line $PQ$

$(y-5)=\frac{7-5}{4-2}(x-2)\Rightarrow y=x+3$

Equation of line $QR$

$(y-7)=\frac{2-7}{6-4}(x-4)\Rightarrow y=\frac{-5}{2}x+17$

Equation of line $PR$

$(y-5)=\frac{2-5}{6-2}(x-2)\Rightarrow y=\frac{-3}{4}x+\frac{13}{2}$

Area of $△PQR$

$=\left(areaunderPQ\right)+\left(areaunderQR\right)-\left(areaunderPR\right)$

$={\int }_2^4(x+3)dx+{\int }_4^6(-\frac{5x}{2}+17)dx-{\int }_2^6(\frac{-3}{4}x+\frac{13}{2})dx$

$={[\frac{x^2}{2}+3x]}_2^4+{[\frac{-5x^2}{4}+17x]}_4^6-{[\frac{3x^2}{8}+\frac{13x}{2}]}_2^6$

$=6+3(4-2)+\frac{-5}{4}(36-16)+17(6-4)-\frac{3}{8}\times 32-\frac{13}{2}\times 4$

$=6+6-\frac{5}{4}\times 20+34-12-26$

$=12-25+34-14=7sq.unit$