# Ex.7.2 Q2 Coordinate Geometry Solution - NCERT Maths Class 10

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## Question

Find the coordinates of the points of trisection of the line segment joining $$(4, -1)$$ and $$(-2, -3)$$.

Video Solution
Coordinate Geometry
Ex 7.2 | Question 2

## 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.

What is Known?

The $$x$$ and $$y$$ co-ordinates of the line segment joining $$(4, -1)$$ and $$(-2, -3)$$.

What is Unknown?

The coordinates of the points of trisection of the line segment joining $$(4, -1)$$ and $$(-2, -3)$$.

Steps:

Given,

• Let line segment joining the points be $$A(4, -1)$$ and $$B(-2, -3)$$.
• Let $$P (x_1, y_1)$$ and $$Q(x_2, y_2)$$ be the points of trisection of the line segment joining the given points i.e., $$AP = PQ = QB$$

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(1) \end{align}

Therefore, by observation point $$P$$ divides $$AB$$ internally in the ratio $$1:2$$.

• Hence $$m: n = 1:2$$

By substituting the values in the Equation (1)

\begin{align}x_1 & \! = \! \frac{{1 \! \times \! ( - 2) \! + \! 2 \! \times \! 4}}{{1 \! + \! 2}}\\x_1 & \! = \! \frac{{ - 2 \! + \! 8}}{3} \\& \! = \! \frac{6}{3} \\& \! = \! 2\\\\ y_1 & \! = \! \frac{{1 \! \times \! ( - 3) \! + \! 2 \! \times \! ( - 1)}}{{1 \! + \! 2}}\\ y_1 & \! = \! \frac{{ - 3 - 2}}{3} \\& \! = \! \frac{{ - 5}}{3}\end{align}

Therefore,

\begin{align}{\text{P}}\left( x_1,y_1 \right) = \left( {2,\frac{{ - 5}}{3}} \right)\end{align}

Therefore, by observation point $$Q$$ divides $$AB$$ internally in the ratio $$2:1$$.

• Hence $$m:n = 2:1$$

By substituting the values in the Equation (1)

\begin{align}x_2 & \! = \! \frac{{2 \! \times \! ( - 2) \! + \! 1 \! \times \! 4}}{{2 \! + \! 1}}\\x_2 & \! = \! \frac{{ - 4 \! + \! 4}}{3} \\& \! = \! 0\\\\y_2 & \! = \! \frac{{2x( - 3) \! + \! 1 \! \times \! ( - 1)}}{{2 \! + \! 1}}\\ y_2 & \! = \! \frac{{ - 6 - 1}}{3} \\& \! = \! \frac{{ - 7}}{3}\end{align}

Therefore,

\begin{align}Q( x_2,y_2) = \left( {0, - \frac{7}{3}} \right)\end{align}

Hence the points of trisection are

\begin{align}P( x_1,y_1) &= \left( {2,\frac{{ - 5}}{3}} \right) \\&\text{and}\;\\ Q(x_2,y_2) &= \left( {0, - \frac{7}{3}} \right)\end{align}

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