Question

Find the equation of the line passing through the points $P\left( {5,1} \right)$ and $Q\left( {1, - 1} \right)$. Hence, show that the points $P$, $Q$ and $R\left( {11,4} \right)$ are collinear.

Answer 1

Hint: We are given two passing points of a line segment and we have to find the equation of the line segment. To find the equation of any line segment we need either two passing points of a line segment or slope and one passing point of the line segment. Here, we are given two passing points $P\left( {5,1} \right)$ and $Q\left( {1, - 1} \right)$. From these passing points we can find the slope of a line segment. After this, we will substitute the value of slope and passing point in general equation of line segment to find the equation of the line passing through the points $P\left( {5,1} \right)$ and $Q\left( {1, - 1} \right)$.
For collinearity of three points: $P\left( {5,1} \right)$, $Q\left( {1, - 1} \right)$ and $R\left( {11,4} \right)$: if $R\left( {11,4} \right)$ is collinear to $P\left( {5,1} \right)$ and $Q\left( {1, - 1} \right)$ then it will satisfy the equation of line passing through $P\left( {5,1} \right)$ and $Q\left( {1, - 1} \right)$.
Formula used:
General equation of line = $\left( {y - {y_1}} \right) = m\left( {x - {x_1}} \right)$. Here, $m$ is the slope of the line and $\left( {{x_1},{y_1}} \right)$ is the passing point.
Slope of a line: $m = \dfrac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}$ . Here, $\left( {{x_1},{y_1}} \right)$ and $\left( {{x_2},{y_2}} \right)$ are the passing points of line.

Complete step-by-step solution:
Equation of the line passing through the points $P\left( {5,1} \right)$ and $Q\left( {1, - 1} \right)$:
Let $P\left( {5,1} \right)$ be $\left( {{x_1},{y_1}} \right)$ and $Q\left( {1, - 1} \right)$ be $\left( {{x_2},{y_2}} \right)$. We can take any point as $\left( {{x_1},{y_1}} \right)$ and $\left( {{x_2},{y_2}} \right)$, answer will remain same.
Slope of line $m = \dfrac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}$
Substituting the values of $\left( {{x_1},{y_1}} \right)$ and $\left( {{x_2},{y_2}} \right)$, we get
$ \Rightarrow m = \dfrac{{ - 1 - 1}}{{1 - 5}}$
$ \Rightarrow m = \dfrac{{ - 2}}{{ - 4}}$
Negative- negative cancel each other
$ \Rightarrow m = \dfrac{2}{4}$
$ \Rightarrow m = \dfrac{1}{2}$
General equation of line: $\left( {y - {y_1}} \right) = m\left( {x - {x_1}} \right)$
Now, substitute the value of slope $m = \dfrac{1}{2}$ and passing point, $P\left( {5,1} \right)$ be $\left( {{x_1},{y_1}} \right)$ in the general equation.
$ \Rightarrow \left( {y - 1} \right) = \dfrac{1}{2}\left( {x - 5} \right)$
On cross multiplication, we get
$ \Rightarrow 2\left( {y - 1} \right) = \left( {x - 5} \right)$
On multiplication, we get
 $ \Rightarrow 2y - 2 = x - 5$
Shift all the terms on one side
$ \Rightarrow x - 2y + 2 - 5 = 0$
$ \Rightarrow x - 2y - 3 = 0$
Hence, $x - 2y - 3 = 0$ is the required equation of the line passing through the points $P\left( {5,1} \right)$ and $Q\left( {1, - 1} \right)$.
For collinearity of three points:
Equation of line: $x - 2y - 3 = 0$
Now, if point $R\left( {11,4} \right)$ is collinear to point $P\left( {5,1} \right)$ and $Q\left( {1, - 1} \right)$ then, $R\left( {11,4} \right)$ should satisfy the equation of line.
On substituting point $R\left( {11,4} \right)$ in equation, we get
$ \Rightarrow 11 - 2\left( 4 \right) - 3 = 0$
$ \Rightarrow 11 - 8 - 3 = 0$
On solving, we get
$ \Rightarrow 0 = 0$
Hence, point $P\left( {5,1} \right)$, $Q\left( {1, - 1} \right)$ and $R\left( {11,4} \right)$ are collinear.

Note: Slope is an angle that a line makes with a positive x-axis measured anticlockwise. We should be careful while doing the cross multiplication of the numbers. We should take care of the calculations so as to be sure of our final answer. Another method of finding the collinearity is, the slope of two points should be the same as the other two points. It means the points are on the same line.