NCERT Class X Chapter 7: Coordinate Geometry Exercise 7.2 Question 2
NCERT Class X Chapter 7: Coordinate Geometry
Question:
Find the coordinates of the points of trisection of the line segment joining (4, –1) and (–2, –3).
Given:
The endpoints of the line segment are:
A(4, -1) and B(-2, -3)
To Find:
The coordinates of the points that divide the line segment AB into three equal parts (i.e., the points of trisection).
Formula:
If a point divides the line joining \((x_1, y_1)\) and \((x_2, y_2)\) in the ratio \(m:n\), then its coordinates are:
$$ \left( \frac{mx_2 + nx_1}{m+n},\ \frac{my_2 + ny_1}{m+n} \right) $$Solution:
Step 1: Let the points of trisection be \(P\) and \(Q\).
\(P\) divides AB in the ratio \(1:2\) (i.e., \(AP:PB = 1:2\)).
\(Q\) divides AB in the ratio \(2:1\) (i.e., \(AQ:QB = 2:1\)).
Step 2: Find the coordinates of \(P\) (dividing AB in \(1:2\)).
$$ P = \left( \frac{1 \times (-2) + 2 \times 4}{1+2},\ \frac{1 \times (-3) + 2 \times (-1)}{1+2} \right) $$Step 3: Calculate the x-coordinate and y-coordinate of \(P\).
\[ \begin{align*} x_P &= \frac{1 \times (-2) + 2 \times 4}{3} = \frac{-2 + 8}{3} = \frac{6}{3} = 2 \\ y_P &= \frac{1 \times (-3) + 2 \times (-1)}{3} = \frac{-3 + (-2)}{3} = \frac{-5}{3} \end{align*} \]
Step 4: Find the coordinates of \(Q\) (dividing AB in \(2:1\)).
$$ Q = \left( \frac{2 \times (-2) + 1 \times 4}{2+1},\ \frac{2 \times (-3) + 1 \times (-1)}{2+1} \right) $$Step 5: Calculate the x-coordinate and y-coordinate of \(Q\).
\[ \begin{align*} x_Q &= \frac{2 \times (-2) + 1 \times 4}{3} = \frac{-4 + 4}{3} = \frac{0}{3} = 0 \\ y_Q &= \frac{2 \times (-3) + 1 \times (-1)}{3} = \frac{-6 + (-1)}{3} = \frac{-7}{3} \end{align*} \]
Result:
The coordinates of the points of trisection of the line segment joining (4, –1) and (–2, –3) are:
\[ P \left(2,\ -\frac{5}{3}\right) \quad \text{and} \quad Q \left(0,\ -\frac{7}{3}\right) \]
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