Question

For what values of k, the points (8,1), (3, -2k) and (0, -5) are collinear.

Answer 1

Hint – We will take the help of the area of the triangle to solve this question and find the value of k. Then, by using the appropriate formula for the area of the triangle and by substituting these given points in it, we will get the value of k for which these points are collinear.

Complete step by step answer:

It is given that these (8,1), (3, -2k) and (0, -5) points are collinear, which means
Area of triangle = 0, as no triangle can form if the points are collinear.
Therefore, the formula for area of triangle when three points are given are represented as:
Area of triangle = $\dfrac{1}{2}\left[ {{x_1}\left( {{y_2} - {y_3}} \right) + {x_2}\left( {{y_3} - {y_1}} \right) + {x_3}\left( {{y_1} - {y_2}} \right)} \right]$
Now, we will compare the given points (8,1), (3, -2k) and (0, -5) as $\left( {{x_1},{y_1}} \right),\left( {{x_2},{y_2}} \right)$ and $\left( {{x_3},{y_3}} \right)$
So, we deduce that ${x_1} = 8,{y_1} = 1,{x_2} = 3,{y_2} = - 2k$ and ${x_3} = 0,{y_3} = - 5$
By substituting the above values in the formula, we will get
Area of triangle = $\dfrac{1}{2}\left[ {8\left( { - 2k + 5} \right) + 3\left( { - 5 - 1} \right) + 0\left( {1 + 2k} \right)} \right] = 0$
Simplify the above equation,
$
   \Rightarrow - 16k + 40 - 18 + 0 = 0 \\
   \Rightarrow - 16k + 22 = 0 \\
   \Rightarrow k = \dfrac{{ - 22}}{{ - 16}} \\
 $
Solving further, we will get
$k = \dfrac{{11}}{8}$
So, we can say that there is only one value of k where these points are collinear i.e. $\dfrac{{11}}{8}$

Hence, the answer is $\dfrac{{11}}{8}$.

Note - Collinearity of a set of points is the property of their lying on a single line and a set of points with this property is said to be collinear. As you can see this is the only method to solve this question and one must use the formula of the area of the triangle where its vertices are given. Do remember to equate this formula with 0 as it’s the important step and do write the formula and put the values in it with pure caution.