Ex.7.4 Q2 Coordinate Geometry Solution - NCERT Maths Class 10

Find a relation between \(x\) and \(y\) if the points \(\begin{align}(x,y), (1,2)\end{align}\) and \(\begin{align}(7,0)\end{align}\) are collinear.

Text Solution

Reasoning:

Three or more points are said to be collinear if they lie on a single straight line .

What is known?

The \(x\) and \(y\) co-ordinates of the points.

What is unknown?

The relation between \(x\) and \(y\) if the points \(\begin{align}(x,y), (1,2)\end{align}\) and \(\begin{align}(7,0)\end{align}\) are collinear.

Steps:

Given,

• \(\begin{align}\text{Let } A(x_1,y_1) =(x,y)\end{align}\)
• \(\begin{align}\text{Let } B(x_2,y_2) =(1,2)\end{align}\)
• \(\begin{align}\text{Let } C(x_3,y_3) =(7,0)\end{align}\)

If the given points are collinear, then the area of triangle formed by these points will be \(0\).

Area of a triangle

\[\begin{align}=\frac{1}{2} \begin{bmatrix} x_1 \left( y_2 - y_3 \right) + \\  x_2 \left( y_3 - y_1 \right) + \\  x_3 \left( y_1 - y_2 \right) \end{bmatrix} \;\;\dots(1) \end{align}  \]

By substituting the values of vertices, \(A\), \(B\), \(C\) in the Equation \((1)\),

Area of a triangle

\[\begin{align} & \! = \! \frac{1}{2}\begin{bmatrix} x(2 \! - \! 0) \! + \! 1(0 \! - \! y) \! +\\ 7(y \! - \! 2) \end{bmatrix} \\&0 \! = \! \frac{1}{2} \begin{bmatrix} 2x \! - \! y \! + \! 7y \! - \! 14 \end{bmatrix} \\0 & \! = \! \frac{1}{2}\left(2x \! + \! 6y \! - \! 14 \right)\\2x \! + \! 6y \! - \! 14 & \! = \! 0\\ x \! + \! 3 y \! - \! 7 & \! = \! 0\end{align}\]

This is the required relation between \(x\) and \(y\).