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Coordinate Geometry
EXERCISE 7.2
CBSE_NCERT_TEXTBOOK
Grade 10

Question:

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

Step-by-Step Solution

Key Concept: Use the section formula (internal division) for coordinates. If a point P(x, y) divides the line segment joining A(x₁, y₁) and B(x₂, y₂) in the ratio m : n, then \(x = \frac{nx_1 + mx_2}{m+n}\) and \(y = \frac{ny_1 + my_2}{m+n}\).
1. Identify the given points:
\[A \equiv (-1,\,7)\quad\text{and}\quad B \equiv (4,\,-3)\]
2. The required point P divides AB internally in the ratio \(m:n = 2:3\) (i.e., \(AP : PB = 2 : 3\)).
3. Apply the section formula:
\[x_P = \frac{n\,x_A + m\,x_B}{m+n},\qquad y_P = \frac{n\,y_A + m\,y_B}{m+n}\]
Here, \(x_A = -1,\; y_A = 7,\; x_B = 4,\; y_B = -3,\; m = 2,\; n = 3\).
4. Compute the x‑coordinate:
\[x_P = \frac{3(-1) + 2(4)}{2+3} = \frac{-3 + 8}{5} = \frac{5}{5} = 1\]
5. Compute the y‑coordinate:
\[y_P = \frac{3(7) + 2(-3)}{2+3} = \frac{21 - 6}{5} = \frac{15}{5} = 3\]
6. Hence the required point P is \((1,\,3)\).

Correct Answer: (1, 3)
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