# Ex.7.2 Q4 Coordinate Geometry Solution - NCERT Maths Class 10

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## Question

Find the ratio in which the line segment joining the points (-$$3, 10$$) and $$(6, -8)$$ is divided by $$(-1, 6)$$.

## 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 $$m1 : m2$$ 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 line segment which is divided by the point $$(-1, 6)$$.

What is Unknown?

The ratio in which the line segment joining the points $$(-3, 10)$$ and $$(6, -8)$$ is divided by $$(-1, 6).$$

Steps:

From the figure,

Given,

• Let the ratio in which the line segment joining $$A(-3, 10)$$ and $$B(6, -8)$$ is divided by point $$P(-1, 6)$$ be $$k:1$$.

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 (2)

Therefore,

\begin{align} - 1 &= \frac{{6{\text{k}} - 3}}{{{\text{k}} + 1}}\\ - {\text{k}} - 1 &= 6{\text{k}} - 3\\7{\text{k}} &= 2 \qquad ({\text{By Cross Multiplying & Transposing}})\\{\text{k}} &= \frac{2}{7}\end{align}

Hence the point $$P$$ divides $$AB$$ in the ratio $$2:7$$

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