Question

The area of a triangle is 5 Two of its vertices are (2, 1) and (3, -2) If the third vertex lies on y = x + 3 then the third vertex is

• A. $(\frac{7}{2},\frac{13}{2})or(\frac{-3}{2},\frac{3}{2})$
• B. $(-\frac{7}{2},\frac{15}{2})or(\frac{3}{2},\frac{3}{2})$
• C. $(-\frac{9}{2},\frac{21}{2})or(\frac{3}{2},\frac{3}{2})$
• D. None of these

Answer 1

Answer: (A)

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

Let the third vertex $=(x,y)$
Given Area of the triangle of the three vertices $=5$
$|\frac{\left(x\right(1+2)+2(-2-y)+3(y-1)}{2}=5$
$\left|\frac{3x-4-2y+3y-3}{2}\right|=5$
$\frac{3x+y-7}{2}=\pm 5$
$3x+y-7=\pm 10$
$3x+y=17$ ........(i)
or $3x+y=-3$ .......(ii)
Given, the point lies on $y=x+3$ ----- (iii)
Solving equation (i) and (iii) $x=\frac{7}{2},y=\frac{13}{2}$ and solving equation (ii) and (iii) $x=-\frac{3}{2},y=\frac{3}{2}$.
So, the co-ordinates of P are $(\frac{7}{2},\frac{13}{2})or(-\frac{3}{2},\frac{3}{2})$