Question

The area of a traingle is 5 square units, two of its verices are (2,1) and (3,-2).The third vertex lies on y=x+3.The third vertex is

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

Answer 1

The correct option is B $(-\frac{3}{2},\frac{3}{2})$

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

Area of ABC =$5$ sq. Units

Area of triangle $=\frac{1}{2}|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)|$

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

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

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

$3x+y-17=0\left(i\right)$

or $3x+y+3=0\left(ii\right)$

Given that, $A(x,y$) lies on $y=x+3\left(iii\right)$

On solving $\left(i\right)$ and$\left(iii\right)$ we get, $x=\frac{7}{2}$ and $y=\frac{13}{2}$ and,

On solving (ii) and (iii) we get, $x=-\frac{3}{2}$ and $y=\frac{3}{2}$