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Coordinate Geometry
EXERCISE 7.2
CBSE_NCERT_TEXTBOOK
Grade 10

Question:

Find the ratio in which the line segment joining A(1, – 5) and B(– 4, 5) is divided by the x-axis. Also find the coordinates of the point of division.

Step-by-Step Solution

Key Concept: Use the section formula (internal division) for a point dividing a line segment in the ratio \(m:n\). Since the required point lies on the x‑axis, its y‑coordinate is zero. Set the y‑coordinate from the section formula to zero and solve for the ratio \(m:n\). Then substitute the ratio back into the formula to obtain the x‑coordinate.
1. Let the required point be \(P(x,0)\), which divides the segment \(AB\) internally in the ratio \(m:n\) (i.e., \(AP:PB = m:n\)).

2. Section formula (internal division):
$$\begin{aligned}
x &= \frac{m\,x_B + n\,x_A}{m+n},\\[4pt]
y &= \frac{m\,y_B + n\,y_A}{m+n}.
\end{aligned}$$
Here \(A(1,-5)\) and \(B(-4,5)\).

3. Use the condition \(y=0\) (since \(P\) lies on the x‑axis):
$$0 = \frac{m\cdot 5 + n\cdot (-5)}{m+n} \;\Rightarrow\; 5m - 5n = 0 \;\Rightarrow\; m = n.$$
Hence the line segment is divided in the ratio \(1:1\) (the midpoint).

4. Find the x‑coordinate using \(m=n\):
$$x = \frac{m\,(-4) + n\,(1)}{m+n}
= \frac{-4m + n}{2m}
= \frac{-4m + m}{2m}
= \frac{-3m}{2m}
= -\frac{3}{2}.$$

5. Coordinates of the point of division:
$$P\left(-\frac{3}{2},\;0\right).$$

6. Answer: The x‑axis divides the segment \(AB\) in the ratio \(1:1\) and the point of division is \\((-\frac{3}{2},\;0)\).

Correct Answer: Ratio = 1:1 (midpoint). Point of division = \((-\frac{3}{2},\;0)\).
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