# 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 \(A(x_1, y_1)\) and \(B(x_2, y_2)\), internally, in the ratio \(m_1 : m_2\) is given by the Section Formula.

\[\begin{align} P(x,y)& \!=\! \left[ {\frac{{mx_2 \! + \! nx_1}}{{m \! + \! n}}\!,\!\frac{{my_2 \! + \! ny_1}}{{m \! + \! n}}} \!\right]\end{align}\]

**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

\[\begin{align} P(x,y)& \!=\! \left[ {\frac{{mx_2 \! + \! nx_1}}{{m \! + \! n}}\!,\!\frac{{my_2 \! + \! ny_1}}{{m \! + \! n}}} \!\right]\;\;\dots(2) \; \end{align}\]

Therefore,

\[\begin{align} - 1 &= \frac{{6k - 3}}{{k + 1}}\\ - k - 1 &= 6k - 3\\7k &= 2\end{align}\]

By Cross Multiplying & Transposing

\[\begin{align}k &= \frac{2}{7}\end{align}\]

Hence the point \(P\) divides \(AB\) in the ratio \(2:7\)