Question

If the points $A\left( {x,2} \right),B\left( { - 3, - 4} \right)$ and $C\left( {7, - 5} \right)$ are collinear, then the value of $x$ is
A) $- 63$
B) $63$
C) $60$
D) $- 60$

Answer 1

Here, in order to solve this question, we will use the fact that three points are said to be collinear if the area of triangle formed using these three points is zero. Area of triangle, \vartriangle = \dfrac{1}{2}|\left( {\begin{array}{*{20}{c}} 1&1&1 \\\ {{x_1}}&{{x_2}}&{{x_3}} \\\ {{y_1}}&{{y_2}}&{{y_3}} \end{array}} \right)|
Through this way we can easily find out the value of $x$ . In order to get started, let us first assume the given points from the question as $\left( {{x_1},{y_1}} \right) = \left( {x,2} \right),\left( {{x_2},{y_2}} \right) = \left( { - 3, - 4} \right),\left( {{x_3},{y_3}} \right) = \left( {7, - 5} \right)$ . Now substitute these properly in the determinant and equate it to zero and find the value of $x$ .

Complete step by step solution:
Given that point $\left( {x,2} \right),\left( { - 3, - 4} \right)$ and $\left( {7, - 5} \right)$ are collinear.
We know that when three points are collinear, it means they lie on the same line. Such a set of points cannot form a triangle where all three points must lie on different lines. So, we can say that the area of the triangle formed using these three points will be equal to zero.
Area of triangle, \vartriangle = \dfrac{1}{2}|\left( {\begin{array}{*{20}{c}} 1&1&1 \\\ {{x_1}}&{{x_2}}&{{x_3}} \\\ {{y_1}}&{{y_2}}&{{y_3}} \end{array}} \right)| , where $\left( {{x_1},{y_1}} \right) = \left( {x,2} \right),\left( {{x_2},{y_2}} \right) = \left( { - 3, - 4} \right),\left( {{x_3},{y_3}} \right) = \left( {7, - 5} \right)$ .
Given that points are collinear $\Rightarrow \vartriangle = 0$
Substituting the value in $\vartriangle$ we get,
\Rightarrow \vartriangle = \dfrac{1}{2}|\left( {\begin{array}{*{20}{c}} 1&1&1 \\\ x&{ - 3}&7 \\\ 2&{ - 4}&{ - 5} \end{array}} \right)| = 0
Opening the determinant along first column we get,
\Rightarrow \dfrac{1}{2}|\left( {\begin{array}{*{20}{c}} 1&1&1 \\\ x&{ - 3}&7 \\\ 2&{ - 4}&{ - 5} \end{array}} \right)| = 0 \\\ \Rightarrow \dfrac{1}{2}\left[ {1\left( {\left( { - 3} \right)\left( { - 5} \right) - \left( 7 \right)\left( { - 4} \right)} \right) - 1\left( {\left( { - 5} \right)\left( x \right) - \left( 7 \right)\left( 2 \right)} \right) + 1\left( {\left( { - 4} \right)\left( x \right) - \left( { - 3} \right)\left( 2 \right)} \right)} \right] = 0 \\\ \Rightarrow \dfrac{1}{2}\left[ {1\left( {15 + 28} \right) - 1\left( { - 5x - 14} \right) + 1\left( { - 4x + 6} \right)} \right] = 0 \\\
$\Rightarrow \dfrac{1}{2}\left[ {43 + 5x + 14 - 4x + 6} \right] = 0 \\\ \Rightarrow \dfrac{1}{2}\left[ {x + 63} \right] = 0 \\\ \Rightarrow x + 63 = 0 \\\ \Rightarrow x = - 63 \\\$
The value of $x$ for which the points $\left( {x,2} \right),\left( { - 3, - 4} \right)$ and $\left( {7, - 5} \right)$ are collinear is $- 63$ .

Hence, the correct answer is option (A).

Note: This question can also be solved by opening the determinant directly without using the area of triangle in determinant form. We can directly use, $\dfrac{1}{2}\left[ {{x_1}\left( {{y_2} - {y_3}} \right) + {x_2}\left( {{y_3} - {y_1}} \right) + {x_3}\left( {{y_1} - {y_2}} \right)} \right] = 0$ and substitute value of $\left( {{x_1},{y_1}} \right),\left( {{x_2},{y_2}} \right)$ and $\left( {{x_3},{y_3}} \right)$ directly to get the answer. The answer anyway would be the same. Students can also plot the points and check for themselves whether they are collinear or not.