# Ex.7.1 Q3 Coordinate Geometry Solution - NCERT Maths Class 10

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## Question

Determine if the points $$(1, 5)$$, $$(2, 3)$$ and $$(-2, -11)$$ are collinear.

Video Solution
Coordinate Geometry
Ex 7.1 | Question 3

## Text Solution

Reasoning:

• Three or more points are said to be collinear if they lie on a single straight line.
• The distance between the two points can be measured using the Distance Formula which is given by: \begin{align}&\text{Distance Formula } \\&= \sqrt {{{\left( {{{{x}}_{{1}}} - {{{x}}_{{2}}}} \right)}^2} + {{\left( {{{{y}}_{{1}}} - {{{y}}_{{2}}}} \right)}^2}} .\end{align}

What is Known?

The $$x$$ and $$y$$ co-ordinates of the points between which the distance is to be measured.

What is Unknown?

To determine if the $$3$$ points are collinear.

Steps:

Let the points $$(1, 5)$$, $$(2, 3)$$, and $$(-2, -11)$$ be represented by the $$A$$, $$B$$, and $$C$$ of the given triangle respectively.

We know that the distance between the two points is given by the Distance Formula

\begin{align} = \sqrt {{{\left( {{{{x}}_{{1}}} - {{{x}}_{{2}}}} \right)}^2} + {{\left( {{{{y}}_{{1}}} - {{{y}}_{{2}}}} \right)}^2}} ...\,(1)\end{align}

To find $$AB$$ i.e. the Distance between the Points $$A \;(1, 5)$$ and $$B\; (2, 3)$$

• $$x_1 = 1$$
• $$y_1 = 5$$
• $$x_2 = 2$$
• $$y_2 = 3$$

\begin{align}\therefore \quad {AB} &= \sqrt {{{(1 - 2)}^2} + {{(5 - 3)}^2}} \\&\text{[By substituting in (1)]}\\\\&=\sqrt 5 \end{align}

To find $$BC$$ Distance between Points $$B \;(2, 3)$$ and $$C\; (-2, -11)$$

• $$x_1 = 2$$
• $$y_1 = 3$$
• $$x_2 = -2$$
• $$y_2 = -11$$

\begin{align} \therefore{BC} \!&=\! \sqrt{{{(2 \!-\! (\!-2))}^2}\!+\!{{(3\!+\!(-11))}^2}}\\ &= \sqrt {{4^2} + {{14}^2}} \qquad \\&\text{[By substituting in equation(1)]}\\\\ &= \sqrt {16 + 196} \\&= \sqrt {212} \end{align}

To find $$AC$$ Distance between Points $$A\; (1, 5)$$ and $$C \;(-2, -11)$$

• $$x_1 = 1$$
• $$y_1 = 5$$
• $$x_2 = -2$$
• $$y_2 = -11$$

\begin{align} \therefore {CA} &= \sqrt {{{(1\!-\!(-2))}^2}\!+\!{{(5\!+\!(-11))}^2}}\\&= \sqrt {{3^2} + {{16}^2}} \qquad \qquad \\&\text{ [By substituting in equation (1)]}\\\\&= \sqrt {9 + 256}\\&= \sqrt {265} \end{align}

Since $$AB + AC$$ $$\not=$$ $$BC$$ and $$AB$$ $$\not=$$ $$BC$$ $$+ AC$$ and $$AC$$ $$\not=$$ $$BC$$ Therefore, the points $$(1, 5)$$, $$(2, 3)$$, and $$(-2, -11)$$ are not collinear.

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