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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}

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