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

## 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 \(1:1\).

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