Question

The points $(p+1,1)$, $(2p+1,3)$ and $(2p+2,2p)$ are collinear, if $p=$

• A. $-1$, $2$
• B. $\frac{1}{2}$, $2$
• C. $2$, $1$
• D. $-\frac{1}{2}$, $2$

Answer 1

Answer: (D)

Let $A(x_1,y_1)=(p+1,1)$, $B(x_2,y_2)=(2p+1,3)$
and $C(x_3,y_3)=(2p+2,2p)$
Points $A$, $B$ and $C$ are collincar if,
Slope of $AB=$ Slope of $BC$
i.e., $\frac{3-1}{\left(2p+1\right)-\left(p+1\right)}=\frac{2p-3}{\left(2p+2\right)-\left(2p+1\right)}$
$\Rightarrow \frac{2}{p}=\frac{2p-3}{1}$
$\Rightarrow 2p^2-3p-2=0$
$\Rightarrow \left(2p+1\right)\left(p-2\right)=0$
$\Rightarrow p=\frac{-1}{2}$, $2$