# Ex.7.2 Q5 Coordinate Geometry Solution - NCERT Maths Class 10

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## Question

Find the ratio in which the line segment joining $$A\; (1, -5)$$ and $$B\; (-4, 5)$$ is divided by the $$x$$-axis. Also find the coordinates of the point of division.

## 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.

What is Known?

The $$x$$ and $$y$$ co-ordinates of the line segment which is divided by the $$x$$-axis.

What is Unknown?

The ratio in which the line segment joining $$A\;(1, -5)$$ and $$B\; (-4, 5)$$ is divided by the $$x$$-axis and the coordinates of the point of division

Steps:

From the Figure, Given,

• Let the ratio be $$k : 1$$.
• Let the line segment joining $$A\;(1, -5)$$ and $$B\; (-4, 5)$$

By 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] & & ...\,\rm{Equation} \,(1)\end{align}

By substituting the values in Equation (1)

Therefore, the coordinates of the point of division is \begin{align}\left( {\frac{{ - 4{\text{k}} + 1}}{{{\text{k}} + 1}},\;\frac{{5{\text{k}} - 5}}{{{\text{k}} + 1}}} \right)\end{align}

We know that $$y$$-coordinate of any point on $$x$$-axis is $$0$$.

\begin{align}∴\; \frac{{5{{k}} - 5}}{{{{k}} + 1}} &= 0\\\;\;\;5{{k}} - 5 &= 0\\\;\;\;\, \to 5{{k}}& = 5 \qquad (\text{By cross multiplying Transposing)}\\{{k}} &= 1\end{align}

Therefore, $$x$$-axis divides it in the ratio $$1:1$$.

\begin{align}{\text{Division point}} &= \left( {\frac{{ - 4(1) + 1}}{{1 + 1}},\frac{{5(1) + 5}}{{1 + 1}}} \right)\\ &= \left( {\frac{{ - 4 + 1}}{2},\frac{{5 + 5}}{2}} \right)\\ &= \left( {\frac{{ - 3}}{2},0} \right)\end{align}

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