Question

What is the value of point P if the points $\left( {p + 1,1} \right),\left( {2p + 1,3} \right),\left( {2p + 2,2p} \right)$ are collinear?
A. $p = - 1$
B. $p = \dfrac{1}{2}$
C. $p = 2$
D. $p = - \dfrac{1}{2}$

Answer 1

Hint: If three points are collinear then we will get no triangle made by the points and hence the area of the triangle made by the points is equal to zero. So, find the area of the triangle made by the given three points and equate it to zero and then solve the resulting equation to get the value of $p$.

Formula Used:
Area of a triangle having coordinates of the vertices $A\left( {{x_1},{y_1}} \right),B\left( {{x_2},{y_2}} \right),C\left( {{x_3},{y_3}} \right)$ is given by $\dfrac{1}{2}\left| {{x_1}\left( {{y_2} - {y_3}} \right) + {x_2}\left( {{y_3} - {y_1}} \right) + {x_3}\left( {{y_1} - {y_2}} \right)} \right|$ square units.

Complete step by step solution:
Here the coordinates of the vertices are $\left( {p + 1,1} \right),\left( {2p + 1,3} \right)$ and $\left( {2p + 2,2p} \right)$
Let be the triangle in which $A = \left( {p + 1,1} \right),B = \left( {2p + 1,3} \right),C = \left( {2p + 2,2p} \right)$
So, ${x_1} = p + 1,{y_1} = 1,{x_2} = 2p + 1,{y_2} = 3,{x_3} = 2p + 2,{y_3} = 2p$
Substitute these values in the formula.
Area of is
$\begin{array}{l}\dfrac{1}{2}\left| {\left( {p + 1} \right)\left\{ {\left( 3 \right) - \left( {2p} \right)} \right\} + \left( {2p + 1} \right)\left\{ {\left( {2p} \right) - \left( 1 \right)} \right\} + \left( {2p + 2} \right)\left\{ {\left( 1 \right) - \left( 3 \right)} \right\}} \right|\\ = \dfrac{1}{2}\left| {\left( {p + 1} \right)\left( {3 - 2p} \right) + \left( {2p + 1} \right)\left( {2p - 1} \right) + \left( {2p + 2} \right)\left( { - 2} \right)} \right|\end{array}$
$\begin{array}{l} = \dfrac{1}{2}\left| {3p + 3 - 2{p^2} - 2p + 4{p^2} - 1 - 4p - 4} \right|\\ = \dfrac{1}{2}\left| {2{p^2} - 3p - 2} \right|\end{array}$
The given points being collinear, the area of the triangle is equal to zero.
So, $\dfrac{1}{2}\left| {2{p^2} - 3p - 2} \right| = 0$
$ \Rightarrow \left| {2{p^2} - 3p - 2} \right| = 0$
Modulus of an expression is equal to zero if the expression itself is equal to zero.
$ \Rightarrow 2{p^2} - 3p - 2 = 0$
Factorize the expression on the left hand side of the equation.
$\begin{array}{l}2{p^2} - 3p - 2\\ = 2{p^2} - 4p + p - 2\\ = 2p\left( {p - 2} \right) + 1\left( {p - 2} \right)\\ = \left( {p - 2} \right)\left( {2p + 1} \right)\end{array}$
So, $\left( {p - 2} \right)\left( {2p + 1} \right) = 0$
Product of two factors is equal to zero if any of the factors is equal to zero.
$ \Rightarrow p - 2 = 0$ or $2p + 1 = 0$
Solve these two equations for $p$.
$p - 2 = 0 \Rightarrow p = 2$
or,
$2p + 1 = 0 \Rightarrow 2p = - 1 \Rightarrow p = - \dfrac{1}{2}$
Finally, we get $p = 2, - \dfrac{1}{2}$

Option ‘C’ and ‘D’ is correct

Note: Area of a triangle is equal to zero. It means that actually no triangle can’t be made by the three points as the vertices. Area of an actual triangle is always positive. Modulus of any expression is equal to zero if the expression itself is equal to zero.