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 \(A(x_1, y_1)\) and \(B(x_2, y_2)\), internally, in the ratio \(m_1 : m_2\) is given by the Section Formula

\[\begin{align}\!P(x,y) \!=\! \left[ {\frac{{mx_2} \!+\! {nx_1}}{{m \!+ n}},\frac{{my_2} \!+\! {ny_1}}{{m} +\! {n}}}\! \right]\! \end{align}\]

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

By Section formula

\[\begin{align}\!P(x,y) \!=\! \left[ {\frac{{mx_2} \!+\! {nx_1}}{{m \!+ n}},\frac{{my_2} \!+\! {ny_1}}{{m} +\! {n}}}\! \right]\!\;\;\dots(1) \end{align}\] 

By substituting the values in the Equation (1)

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

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