# Ex.7.2 Q1 Coordinate Geometry Solution - NCERT Maths Class 10

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## Question

Find the coordinates of the point which divides the join of $$(-1, 7)$$ and $$(4, -3)$$ in the ratio $$2:3$$.

## 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

\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 points which is to be divided in the ratio $$2:3$$

What is Unknown?

The coordinates of the point which divides the join of $$(-1, 7)$$ and $$(4, -3)$$ in the ratio $$2:3$$

Steps:

Given,

• Let $$P(x, y)$$ be the required point. • Let $$A(-1, 7)$$ and $$B(4, -3)$$
• $$\rm m: n = 2:3$$
• Hence
• $$x_1 = -1$$
• $$y_1 = 7$$
• $$x_2 = 4$$
• $$y_2 = -3$$

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 (1)

By substituting the values in the Equation (1)

\begin{align}{\text{x}} &= \frac{{2 \times 4 + 3{\text{x}}( - 1)}}{{2 + 3}} \qquad \qquad {\text{y}} = \frac{{2 \times ( - 3) + 3 \times 7}}{{2 + 3}}\\{\text{x}} &= \frac{{8 - 3}}{5} \qquad \qquad \qquad \qquad \; {\text{y}} = \frac{{ - 6 + 21}}{5}\\{\text{x}} &= \frac{5}{5} = 1 \qquad \qquad \qquad \qquad\; {\text{y}} = \frac{{15}}{5} = 3\end{align}

Therefore, the co-ordinates of point $$P$$ are $$(1, 3)$$.

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