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Coordinate Geometry
EXAMPLES
CBSE_NCERT_TEXTBOOK
Grade 10

Question:

Find the ratio in which the y-axis divides the line segment joining the points (5, – 6) and (–1, – 4). Also find the point of intersection.

Step-by-Step Solution

Key Concept: Use the section formula (internal division) for a point dividing a line segment in a given ratio. For a point on the y‑axis its x‑coordinate is 0, which gives an equation to determine the ratio.
1. Let the two given points be \(A(5,-6)\) and \(B(-1,-4)\).\
2. Suppose the y‑axis (the line \(x=0\)) cuts \(AB\) at point \(P\) which divides \(AB\) internally in the ratio \(AP:PB = m:n\).\
3. By the section formula, the coordinates of \(P\) are\
$$\left(\frac{n\cdot x_A + m\cdot x_B}{m+n},\;\frac{n\cdot y_A + m\cdot y_B}{m+n}\right).$$\
4. Since \(P\) lies on the y‑axis, its x‑coordinate is 0. Hence\
$$\frac{n\cdot 5 + m\cdot (-1)}{m+n}=0 \;\Rightarrow\; 5n - m = 0 \;\Rightarrow\; m = 5n.$$\
5. Therefore the required ratio is\
$$AP:PB = m:n = 5n:n = 5:1.$$\
6. To find the y‑coordinate of \(P\), substitute \(m=5n\) in the y‑coordinate formula:\
$$y_P = \frac{n(-6) + 5n(-4)}{5n + n}=\frac{-6n -20n}{6n}=\frac{-26n}{6n}= -\frac{13}{3}.$$\
7. Hence the point of intersection is\
$$P\;(0,\; -\frac{13}{3}).$$

Correct Answer: Ratio in which the y‑axis divides the segment = 5:1 (from (5,‑6) to (‑1,‑4)). Point of intersection = \((0, -\frac{13}{3})\).
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