Question

Using the distance formula, show that the points $A\left( 3,-2 \right),B\left( 5,2 \right)\ and\ C\left( 8,8 \right)$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 A, B, C are collinear i.e. they lie in a line then,
AB + BC = AC.

Complete step-by-step answer:

Now, we have been given three points as,
$A\left( 3,-2 \right),B\left( 5,2 \right)\ and\ C\left( 8,8 \right)$

Now, if points A, B, C are collinear then we have to prove that AB + BC = AC.
Now, we know that the distance between two points with coordinate $\left( {{x}_{1}},{{y}_{1}} \right)\ and\ \left( {{x}_{2}},{{y}_{2}} \right)$ is $AB=\sqrt{{{\left( {{x}_{1}}-{{x}_{2}} \right)}^{2}}+{{\left( {{y}_{1}}-{{y}_{2}} \right)}^{2}}}$.
Now, we have the distance AB as,
$\begin{aligned} & =\sqrt{{{\left( 3-5 \right)}^{2}}+{{\left( -2-2 \right)}^{2}}} \\\ & =\sqrt{{{\left( -2 \right)}^{2}}+{{\left( -4 \right)}^{2}}} \\\ & =\sqrt{4+16} \\\ & =\sqrt{20} \\\ & AB=2\sqrt{5}units.........\left( 1 \right) \\\ \end{aligned}$
Now, we have the distance AC as,
$\begin{aligned} & =\sqrt{{{\left( 3-8 \right)}^{2}}+{{\left( -2-8 \right)}^{2}}} \\\ & =\sqrt{{{\left( -5 \right)}^{2}}+{{\left( 10 \right)}^{2}}} \\\ & =\sqrt{25+100} \\\ & =\sqrt{125} \\\ & AC=5\sqrt{5}units.........\left( 2 \right) \\\ \end{aligned}$
Now, we have the distance BC as,
$\begin{aligned} & =\sqrt{{{\left( 5-8 \right)}^{2}}+{{\left( 2-8 \right)}^{2}}} \\\ & =\sqrt{{{\left( -3 \right)}^{2}}+{{\left( -6 \right)}^{2}}} \\\ & =\sqrt{9+36} \\\ & =\sqrt{45} \\\ & BC=3\sqrt{5}units.........\left( 3 \right) \\\ \end{aligned}$
Now, on adding (1) and (3) we have,
$AB+BC=5\sqrt{5}units$
Now, on equating this with equation (2) we have,
AB + BC = AC
So, the points A, B, C are collinear.

Note: To solve these type of question it is important to note that we have used a fact that if A, B, C are collinear then,
AB + BC = AC
Also, it has to be noted that we have taken A, B, C as their order on graph.