Question

Three vertices of a quadrilateral in order are $(6,1),(7,2)$ and $(-1,0).$ If the area of the quadrilateral is $4$ sq. units, then the locus of the fourth vertex has the equation

• A. $x-7y=1$
• B. $x-7y+15=0$
• C. $x-4y+2=0$
• D. none of these

Answer 1

Answer: (C)

$x-4y+2=0$

Let the coordinates be
$A(6,1),B(7,2),C(x,y)$ and $D(-1,0)$
Area of quadrilateral $ABCD$ $=4$
$\Rightarrow \frac{1}{2}(\left|\begin{array}{cc}6& 1\\ 7& 2\end{array}\right|+\left|\begin{array}{cc}7& 2\\ x& y\end{array}\right|+\left|\begin{array}{cc}x& y\\ -1& 0\end{array}\right|+\left|\begin{array}{cc}-1& 0\\ 6& 1\end{array}\right|)=4\Rightarrow 12-7+7y-2x+y-1=8\Rightarrow -2x+8y-4=0\Rightarrow x-4y+2=0$