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.