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Question: Using the distance formula, prove that the points \(A\left( -2,3 \right),B\left( 1,2 \right)\ and\ C...

Using the distance formula, prove that the points A(−2,3),B(1,2) and C(7,0)A\left( -2,3 \right),B\left( 1,2 \right)\ and\ C\left( 7,0 \right) are collinear.

Explanation

Solution

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(−2,3),B(1,2) and C(7,0)A\left( -2,3 \right),B\left( 1,2 \right)\ and\ C\left( 7,0 \right)

Now, if points are collinear then we have to prove that AB + BC = AC.
Now, we know that the distance between two points with coordinate (x1,y1) and (x2,y2)\left( {{x}_{1}},{{y}_{1}} \right)\ and\ \left( {{x}_{2}},{{y}_{2}} \right) is AB=(x1−x2)2+(y1−y2)2AB=\sqrt{{{\left( {{x}_{1}}-{{x}_{2}} \right)}^{2}}+{{\left( {{y}_{1}}-{{y}_{2}} \right)}^{2}}}.
Now, we have the distance AB as,
=(−2−1)2+(3−2)2 =(−3)2+(1)2 =9+1 AB=10units.........(1) \begin{aligned} & =\sqrt{{{\left( -2-1 \right)}^{2}}+{{\left( 3-2 \right)}^{2}}} \\\ & =\sqrt{{{\left( -3 \right)}^{2}}+{{\left( 1 \right)}^{2}}} \\\ & =\sqrt{9+1} \\\ & AB=\sqrt{10}units.........\left( 1 \right) \\\ \end{aligned}
Now, we have the distance BC as,
=(1−7)2+(2−0)2 =(−6)2+(2)2 =36+4 =40 BC=210units.........(2) \begin{aligned} & =\sqrt{{{\left( 1-7 \right)}^{2}}+{{\left( 2-0 \right)}^{2}}} \\\ & =\sqrt{{{\left( -6 \right)}^{2}}+{{\left( 2 \right)}^{2}}} \\\ & =\sqrt{36+4} \\\ & =\sqrt{40} \\\ & BC=2\sqrt{10}units.........\left( 2 \right) \\\ \end{aligned}
Now, we have the distance AC as,
=(−2−7)2+(3−0)2 =(−9)2+(3)2 =81+9 =90 AC=310units.........(3) \begin{aligned} & =\sqrt{{{\left( -2-7 \right)}^{2}}+{{\left( 3-0 \right)}^{2}}} \\\ & =\sqrt{{{\left( -9 \right)}^{2}}+{{\left( 3 \right)}^{2}}} \\\ & =\sqrt{81+9} \\\ & =\sqrt{90} \\\ & AC=3\sqrt{10}units.........\left( 3 \right) \\\ \end{aligned}
Now, on adding (1) and (2) we have,
AB+BC=10+210units =310units \begin{aligned} & AB+BC=\sqrt{10}+2\sqrt{10}units \\\ & =3\sqrt{10}units \\\ \end{aligned}
Now, on equating this with equation (3) we have,
AB + BC = AC
So, we proved that the points A, B, C are collinear. Since, we have proved that,
AB + BC = AC

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.