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

## 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 \(2:7\)