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Ex.7.2 Q8 Coordinate Geometry Solution - NCERT Maths Class 10

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Question

If \(A\) and \(B\) are \((-2, -2)\) and \((2, -4)\), respectively, find the coordinates of \(P\) such that \(\begin{align}AP= \frac{3}{7}AB\end{align}\) and \(P\) lies on the line segment \(AB\).

 Video Solution
Coordinate Geometry
Ex 7.2 | Question 8

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.

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

What is the known?

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

The ratio in which \(P\) divides \(AB\).

What is the unknown?

Co-ordinates of \(P\)

Steps:

From the Figure,

Given,

  • The coordinates of point \(A\) and \(B\) are \((-2, -2)\) and \((2, -4)\) respectively.
  • \(\begin{align}{{AP}} = \frac{3}{7}{{AB}}\end{align}\)

Hence \(\begin{align}\frac{{{{AB}}}}{{{{AP}}}} = \frac{7}{3}\end{align}\)

We know that \(\begin{align}AB = AP + PB\end{align}\)

from figure,

\[\begin{align}\frac{{{{AP}} + {{PB}}}}{{{{AP}}}} &= \frac{{3 + 4}}{3}\\1 + \frac{{{{PB}}}}{{{{AP}}}}& = 1 + \frac{4}{3}\\\frac{{{{PB}}}}{{{{AP}}}}& = \frac{4}{3}\end{align}\]

Therefore, \(AP:PB = 3:4\)

Point \(P(x, y)\) divides the line segment \(AB\) in the ratio \(3:4.\) Using Section Formula,

Coordinates of

\[\begin{align}P({{x}},{{y}}) & \! = \! \begin{bmatrix}\left({\frac{{3 \times 2 + 4 \times ( - 2)}}{{3 + 4}}}\right), \\ \left(\frac{{3 \times ( - 4) + 4 \times ( - 2)}}{{3 + 4}}\right) \end{bmatrix}\\ & \! = \! \left[ {\frac{{6 - 8}}{7},\;\frac{{ - 12 - 8}}{7}} \right]\\ & \! = \! \left[ { - \frac{2}{7},\; - \frac{{20}}{7}} \right]\end{align}\]