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

## Question

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

## 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: \[\begin{align}\text{Distance Formula } = \sqrt {{{\left( {{{\text{x}}_{\text{1}}} - {{\text{x}}_{\text{2}}}} \right)}^2} + {{\left( {{{\text{y}}_{\text{1}}} - {{\text{y}}_{\text{2}}}} \right)}^2}} .\end{align}\]

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

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

- \(x_1 = 1\)
- \(y_1 = 5\)
- \(x_2 = 2\)
- \(y_2 = 3\)

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

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

- \(x_1 = 2\)
- \(y_1 = 3\)
- \(x_2 = -2\)
- \(y_2 = -11\)

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

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

- \(x_1 = 1\)
- \(y_1 = 5\)
- \(x_2 = -2\)
- \(y_2 = -11\)

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

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.