Question

$A(p,0),B(4,0),C(5,6)$ and $D(1,4)$ are the vertices of a quadrilateral $ABCD$.

The value of $p$ for which $\angle ADC$ is obtuse and $\angle BAD$ is a right angle is :

• A. $1$
• B. $2$
• C. $-3$
• D. $-4$

Answer 1

The correct option is A $1$

$\angle BAD$ is a right angle, so

$(AB{)}^2+(AD{)}^2=(BD{)}^2$

$\Rightarrow \left[\right(p-4{)}^2]+[(p-1{)}^2+4^2]=3^2+4^2$
$\Rightarrow 2p^2-10p+8=0\Rightarrow p=1,4$

If $p=4$, then $A\equiv B$

$\therefore p=1$ (only possible solution)