Question

Show that the points (-5, 1), (5, 5) and (10, 7) are collinear.

Answer 1

Hint: When we are asked to prove the collinearity of three points, then we have to prove that the triangle formed by these 3 points has an area equal to 0. And, we know that the area of a triangle $=\dfrac{1}{2}\left[ {{x}_{1}}\left( {{y}_{2}}-{{y}_{3}} \right)+{{x}_{2}}\left( {{y}_{3}}-{{y}_{1}} \right)+{{x}_{3}}\left( {{y}_{1}}-{{y}_{2}} \right) \right]$, And whatever value we get, we have to take its modulus.

Complete step-by-step answer:

In this question, we have to prove that the points (-5, 1), (5, 5) and (10, 7) are collinear. So, to prove the collinearity, we have to show that the area of the triangle, formed by these 3 points is 0. To find the area of the triangle, we should know that the area of the triangle $=\dfrac{1}{2}\left[ {{x}_{1}}\left( {{y}_{2}}-{{y}_{3}} \right)+{{x}_{2}}\left( {{y}_{3}}-{{y}_{1}} \right)+{{x}_{3}}\left( {{y}_{1}}-{{y}_{2}} \right) \right]$, where we assume ${{x}_{1}}=-5,{{x}_{2}}=5$ and ${{x}_{3}}=10$. So, we get, ${{y}_{1}}=1,{{y}_{2}}=5$ and ${{y}_{3}}=7$. Therefore, we get the area of the triangle as,
$=\dfrac{1}{2}\left[ \left( -5 \right)\left( 5-7 \right)+\left( 5 \right)\left( 7-1 \right)+\left( 10 \right)\left( 1-5 \right) \right]$
On simplifying the above equation, we get the area of the triangle as,
$\begin{align}
  & =\dfrac{1}{2}\left[ \left( -5 \right)\left( -2 \right)+\left( 5 \right)\left( 6 \right)+\left( 10 \right)\left( -4 \right) \right] \\
 & =\dfrac{1}{2}\left[ 10+30-40 \right] \\
 & =\dfrac{1}{2}\left[ 40-40 \right] \\
 & =\dfrac{1}{2}\left[ 0 \right] \\
 & =0 \\
\end{align}$
So, we get the area of the triangle as 0. We know that if the area of a triangle of three given points is 0, then the points are collinear.
Hence, we have proved that the three points given in the question, (-5, 1), (5, 5) and (10, 7) are collinear.

Note: We can use an alternate method to approach this question, which is as follows: We can prove the collinearity of three points if we can prove the distance between two points is equal to the sum of the distance between another two pair of points. We know that distance between two points can be calculated using distance formula, that is, $\sqrt{{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}+{{\left( {{y}_{2}}-{{y}_{1}} \right)}^{2}}}$, which is the distance between two points, $\left( {{x}_{1}},{{y}_{1}} \right)$ and $\left( {{x}_{2}},{{y}_{2}} \right)$.
In the given question, let us consider (-5, 1) as point A, (5, 5) as point B and (10, 7) as point C. So, by applying the distance formula, we will get the distance AB as,
$\begin{align}
  & AB=\sqrt{{{\left[ 5-\left( -5 \right) \right]}^{2}}+{{\left[ 5-1 \right]}^{2}}} \\
 & \Rightarrow AB=\sqrt{{{\left[ 5+5 \right]}^{2}}+{{\left[ 4 \right]}^{2}}} \\
 & \Rightarrow AB=\sqrt{100+16} \\
 & \Rightarrow AB=\sqrt{116} \\
 & \Rightarrow AB=2\sqrt{29}\ldots \ldots \ldots \left( i \right) \\
\end{align}$
Similarly, we will calculate the distance BC as,
$\begin{align}
  & BC=\sqrt{{{\left( 10-5 \right)}^{2}}+{{\left( 7-5 \right)}^{2}}} \\
 & \Rightarrow BC=\sqrt{{{\left( 5 \right)}^{2}}+{{\left( 2 \right)}^{2}}} \\
 & \Rightarrow BC=\sqrt{25+4} \\
 & \Rightarrow BC=\sqrt{29}\ldots \ldots \ldots \left( ii \right) \\
\end{align}$
And similarly, we will calculate AC as,
$\begin{align}
  & AC=\sqrt{{{\left[ 10-\left( -5 \right) \right]}^{2}}+{{\left[ 7-1 \right]}^{2}}} \\
 & \Rightarrow AC=\sqrt{{{\left( 10+5 \right)}^{2}}+{{\left( 6 \right)}^{2}}} \\
 & \Rightarrow AC=\sqrt{{{\left( 15 \right)}^{2}}+{{\left( 6 \right)}^{2}}} \\
 & \Rightarrow AC=\sqrt{225+36} \\
 & \Rightarrow AC=\sqrt{261} \\
 & \Rightarrow AC=3\sqrt{29}\ldots \ldots \ldots \left( iii \right) \\
\end{align}$
Now, adding equation (i) and equation (iii), we get,
$\begin{align}
  & AB+BC=2\sqrt{29}+\sqrt{29} \\
 & \Rightarrow AB+BC=3\sqrt{29}\ldots \ldots \ldots \left( iv \right) \\
\end{align}$
From equation (iii) and equation (iv), we get,
$AB+BC=AC$
Hence, we have proved that the distance between two points is equal to the sum of the distance between another two pairs of points.