Ex.7.2 Q1 Coordinate Geometry Solution - NCERT Maths Class 10

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Question

Find the coordinates of the point which divides the join of \((-1, 7)\) and \((4, -3)\) in the ratio \(2:3\).

 

Text Solution

Reasoning:

The coordinates of the point \(P(x, y)\) which divides the line segment joining the points \(A(x1, y1)\) and \(B(x2, y2)\), internally, in the ratio \(\rm m1 : m2\) is given by the Section Formula

\(\begin{align}{\text{P}}({\text{x}},{\text{y}}) = \left[ {\frac{{{\text{m}}{{\text{x}}_2} + {\text{n}}{{\text{x}}_1}}}{{{\text{m}} + {\text{n}}}},\frac{{{\text{m}}{{\text{y}}_2} + {\text{n}}{{\text{y}}_1}}}{{{\text{m}} + {\text{n}}}}} \right] & & \end{align}\)

What is Known?

The \(x\) and \(y\) co-ordinates of the points which is to be divided in the ratio \(2:3\)

What is Unknown?

The coordinates of the point which divides the join of \((-1, 7)\) and \((4, -3)\) in the ratio \(2:3\)

Steps:

Given,

  • Let \(P(x, y)\) be the required point.

  • Let \(A(-1, 7)\) and \(B(4, -3)\)
  • \(\rm m: n = 2:3\)
  • Hence
    • \(x_1 = -1\)
    • \(y_1 = 7\)
    • \(x_2 = 4\)
    • \(y_2 = -3\)

By Section formula

\(\begin{align}{\text{P}}({\text{x}},{\text{y}}) = \left[ {\frac{{{\text{m}}{{\text{x}}_2} + {\text{n}}{{\text{x}}_1}}}{{{\text{m}} + {\text{n}}}},\frac{{{\text{m}}{{\text{y}}_2} + {\text{n}}{{\text{y}}_1}}}{{{\text{m}} + {\text{n}}}}} \right] & & ...\,\end{align}\) Equation (1)

By substituting the values in the Equation (1)

\(\begin{align}{\text{x}} &= \frac{{2 \times 4 + 3{\text{x}}( - 1)}}{{2 + 3}} \qquad \qquad {\text{y}} = \frac{{2 \times ( - 3) + 3 \times 7}}{{2 + 3}}\\{\text{x}} &= \frac{{8 - 3}}{5} \qquad \qquad \qquad \qquad \; {\text{y}} = \frac{{ - 6 + 21}}{5}\\{\text{x}} &= \frac{5}{5} = 1 \qquad \qquad \qquad \qquad\; {\text{y}} = \frac{{15}}{5} = 3\end{align}\)

Therefore, the co-ordinates of point \(P\) are \((1, 3)\).