Question

The line $3x+2y=24$ meets x-axis at A and y-axis at B. The perpendicular bisector of $\overline{AB}$ meets the line through (0, -1) and parallel to x-axis at C. Then the area of $\Delta ABC$ is

• A. 85
• B. 87
• C. 90
• D. 91

Answer 1

The correct option is D 91

$Letthe\Delta beABC.$

$thegivenequationcanbewrittenas$

$\frac{3x}{24}+\frac{2y}{24}=1⟹\frac{x}{8}+\frac{y}{12}=1.$

$Itisoftheform\frac{x}{a}+\frac{y}{b}=1⟹a=8\&b=12$

$SothepointofintersectionwithX-axisisA=(a,0)=(8,0)=(x_1,y_1)$

$andthatwithY-axisisB=(0,b)=(0,1)=(x_2,y_2)$

$.$

$\therefore Mid-pointofAB=(\frac{8+0}{2},\frac{0+12}{2})=(4,6)$

$Nowequationoftheperpendicularbisrctoroftheline3x+2y=24is$

$2x-3y=k........\left(i\right)whenkisaconstant.Thislinepassesthrough(4,6)$

$\therefore substitutingx=4,y=6inequation\left(i\right)weget2\times 4-3\times 6=k⟹k=10$

$\therefore equation\left(i\right)becomes2x-3y=\mathrm{10.........}\left(ii\right).Thislineintersectsy=-\mathrm{1.........}\left(iii\right)$

$soeliminatingyfrom\left(ii\right)\&\left(iii\right)wehave$

$2x-3\times (-1)=10⟹x=\frac{13}{2},y=-1\therefore VertexC=(\frac{13}{2},-1)=(x_3,y_3)$

$\therefore Theareaofthe\Delta is$

$=\frac{1}{2}\{x_1(y_2-y_1)+x_2(y_3-y_1)+x_3(y_1-y_2)\}=\frac{1}{2}\{8(12+1)+0+\frac{13}{2}(0-12)\}=91sq.unit$

$Ans-OptionD$