Question

Find the value of k for which the points $A\left( -2,3 \right),B\left( 1,2 \right)\ and\ C\left( k,0 \right)$are collinear.

Answer 1

Hint: We will be using the concept of coordinate geometry to solve the problem. We will be using the fact that if three points are collinear then the area formed by these points is zero.

Complete step-by-step answer:
Now, we have three points given to us as,
$A\left( -2,3 \right),B\left( 1,2 \right)\ and\ C\left( k,0 \right)$
Also, we have been given that the three points are collinear. Therefore, the area of the triangle formed by A, B and C is zero.

Now, we know that the area of a triangle whose vertices are $\left( {{x}_{1}},{{y}_{1}} \right),\left( {{x}_{2}},{{y}_{2}} \right)\ and\ \left( {{x}_{3}},{{y}_{3}} \right)$is given by,
$\dfrac{1}{2}\left( {{x}_{1}}\left( {{y}_{2}}-{{y}_{3}} \right)+{{x}_{2}}\left( {{y}_{3}}-{{y}_{1}} \right)+{{x}_{3}}\left( {{y}_{1}}-{{y}_{2}} \right) \right)$.
So, we have the area of $\Delta ABC$as,
$\begin{aligned} & \dfrac{1}{2}\left( -2\left( 2-0 \right)+1\left( 0-3 \right)+k\left( 3-2 \right) \right)=0 \\\ & \dfrac{1}{2}\left( -2\times 2+1\times -3+k\times 1 \right)=0 \\\ & \dfrac{1}{2}\left( -4-3+k \right)=0 \\\ & -7+k=0 \\\ & k=7 \\\ \end{aligned}$
So, we have the value of $k=7$

Note: To solve these types of questions it is important to note that we have used the fact that if three points are collinear then the area formed by these three points is zero. Another approach is that we can compare the slope of the two lines formed through three given points , if they are collinear then the slopes will be equal.