Question

If the points $\left( {x + 1,2} \right),\left( {1,x + 2} \right)$ and $\left( {\dfrac{1}{{x + 1}},\dfrac{2}{{x + 1}}} \right)$ are collinear, what is the value of $x$?
A. $4$
B. $0$
C. $ - 4$
D. None of these

Answer 1

Hint: Collinear points are those points which lie on a same straight line. Find the area of the triangle made by the given three points as vertices. Since the points are collinear, so area of the triangle is equal to zero. Using this concept, you’ll obtain an equation and solving the equation you’ll get the value of $x$.

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

Option ‘B’ and ‘C’ 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.