Question

If $x+8y-22=0,5x+2y-34=0,2x-3y+13=0$ are the three sides of a triangle. Then the area of the triangle is

• A. $36$ square unit
• B. $19$ square unit
• C. $42$ square unit
• D. $72$ square unit

Answer 1

The correct option is B $19$ square unit

Three sides of a triangle have been given. We find the intersection of these lines, taking two at a time to obtain the vertices and hence the area.

$x+8y-22=0...\left(1\right)$

$5x+2y-34=0...\left(2\right)$

$2x-3y+13=0...\left(3\right)$

Multiplying equation $\left(2\right)$ by $4$ and subtracting that from equation $\left(1\right)$, we have

$x+8y-22-(20x+8y-136)=0$

$\Rightarrow -19x+114=0$

$\Rightarrow x=6$ and correspondingly $y=2$

Next, multiplying equation $\left(2\right)$ by $3$ and equation $\left(3\right)$ by $2$ and adding the two, we have

$15x+6y-102+(4x-6y+26)=0$

$\Rightarrow 19x-76=0$ or $x=4$

$\Rightarrow 20+2y-34=0$

$\Rightarrow y=7$

Third, multiplying equation $\left(1\right)$ by $2$ and subtracting equation $\left(3\right)$ from this, we have

$2x+16y-44-(2x-3y+13)=0$

$\Rightarrow 2x+16y-44-2x+3y-13=0$

$\Rightarrow 19y-57=0$ or $y=3$

$\Rightarrow 2x-9+13=0$ or $x=-2$

The three vertices thus become $(6,2),(4,7),(-2,3)$

The area of the triangle thus becomes $\frac{1}{2}\times [6\times 7+4\times 3+(-2)\times 2-(2\times 4+7\times (-2)+6\times 3\left)\right]$

$=\frac{1}{2}\times [42+12-4-(8-14+18\left)\right]$

$=\frac{1}{2}\times [50-12]$

$=19$ sq unit