Question

The area of a triangle is 5 and its two vertices are A(2, 1) and B(3, -2). The third vertex lies on: $y=x+3$. Then third vertex is:

• A. $(\frac{7}{2},\frac{13}{2})$
• B. $(\frac{5}{2},\frac{5}{2})$
• C. $(-\frac{3}{3},-\frac{3}{2})$
• D. $\left(0.0\right)$

Answer 1

Answer: (A)

$(\frac{7}{2},\frac{13}{2})$

Let the third vertex be A(x,y).other two vertices of triangle are B(2,1) and C(3,-2).We have

Area of $△ABC=5$ Sq.units

$\Rightarrow \frac{1}{2}\mid (x-4+3y)-(2y+3-2x)\mid =5$

$\Rightarrow \frac{1}{2}\mid x-4+3y-2y-3+2x\mid =5$

$\Rightarrow \frac{1}{2}\mid 3x+y-7\mid =5$

$\Rightarrow 3x+y-7=\pm 10$

$\Rightarrow 3x+y-17=0$ or $3x+y+3=0$

It is given that the third vertex A(x,y) lies on

$y=x+3$

Solving 3x+y-17=0 and y=x+3,we get;

$x=\frac{7}{2},y=\frac{13}{2}$

Solving 3x+y+3=0 and y=x+3, we get;

$x=\frac{-3}{2},y=\frac{3}{2}$

Hence,the coordinate of the third vertex are

$(\frac{7}{2},\frac{13}{2})$ or $(\frac{-3}{2},\frac{3}{2})$