Question

Find the value of $k$ for which points $P\left( {k, - 1} \right)$, $Q\left( {2,1} \right)$and $R\left( {4,5} \right)$ are collinear.

Answer 1

Hint:In order to find the value of the unknown quantity for the points that are collinear, first we must have information about collinear points and non-collinear points.Collinear points are the points that lie on the same line. There can be two or more lines that can lie on the same plane and are called collinear points. And the points that do not lie on the same line are called non-collinear points.

Complete step by step answer:
We are given with the points $P\left( {k, - 1} \right)$, $Q\left( {2,1} \right)$and $R\left( {4,5} \right)$ that are said to be collinear, that means they lie on the same line which gives us the information that if they lie on the same line, their slopes for two points taken at a time would also be equal.

So, we would check the slope for two points at a time and equate them.From the slope formula we know that:
$Slope\left( m \right) = \dfrac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}$ for some points $\left( {{x_1},{y_1}} \right)$, and $\left( {{x_2},{y_2}} \right)$.
For the points $P\left( {k, - 1} \right)$, $Q\left( {2,1} \right)$:
Equating the points with $\left( {{x_1},{y_1}} \right)$, and $\left( {{x_2},{y_2}} \right)$, we get:
$\left( {{x_1},{y_1}} \right) = \left( {k, - 1} \right)$ and $\left( {{x_2},{y_2}} \right) = \left( {2,1} \right)$
Substituting this value in the slope formula, we get the slope 1 as:
$Slope\left( {{m_1}} \right) = \dfrac{{1 - \left( { - 1} \right)}}{{2 - k}} = \dfrac{{1 + 1}}{{2 - k}} = \dfrac{2}{{2 - k}}$ ……(1)

Similarly, for slope 2:
For the points $Q\left( {2,1} \right)$and $R\left( {4,5} \right)$:
Equating the points with $\left( {{x_1},{y_1}} \right)$, and $\left( {{x_2},{y_2}} \right)$, we get:
$\left( {{x_1},{y_1}} \right) = \left( {2,1} \right)$ and $\left( {{x_2},{y_2}} \right) = \left( {4,5} \right)$
Substituting this value in the slope formula, we get the slope 1 as:
$Slope\left( {{m_2}} \right) = \dfrac{{5 - 1}}{{4 - 2}} = \dfrac{4}{2} = 2$ ……(2)

Since, slope 1 and slope 2 are equal, so equating equation 1 and equation 2, we get:
${m_1} = {m_2}$
$ \Rightarrow \dfrac{2}{{2 - k}} = 2$
Dividing both the sides by $2$, we get:
$ \Rightarrow \dfrac{2}{{2\left( {2 - k} \right)}} = \dfrac{2}{2}$
$ \Rightarrow \dfrac{1}{{\left( {2 - k} \right)}} = 1$
Multiplying both sides by $\left( {2 - k} \right)$, we get:
\[ \Rightarrow \dfrac{1}{{\left( {2 - k} \right)}} \times \left( {2 - k} \right) = 1 \times \left( {2 - k} \right)\]
\[ \Rightarrow 1 = 2 - k\]
Subtracting both sides by \[1\]:
\[\Rightarrow 1 - 1 = 2 - k - 1 \\
\Rightarrow 0 = 1 - k \\ \]
Adding k both the sides, we get:
\[\Rightarrow 0 + k = 1 - k + k \\
\therefore k = 1 \\ \]
Therefore, the value of \[k = 1\] for the collinear points $P\left( {k, - 1} \right)$, $Q\left( {2,1} \right)$ and $R\left( {4,5} \right)$.

Note:Collinearity can be checked by three methods that are as follows:
Distance formula: Since they lie on the same line at the same distance, the distance between the two points would also be the same.
Slope formula: As we used the slope formula in the above question, two points' slope will be equal.
Area of Triangle: If the area of the triangle, formed by three points gives zero, that proves that the points are collinear.