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Coordinate Geometry
EXERCISE 7.2
CBSE_NCERT_TEXTBOOK
Grade 10

Question:

If A and B are (– 2, – 2) and (2, – 4), respectively, find the coordinates of P such that AP = 3 AB 7 and P lies on the line segment AB.

Step-by-Step Solution

Key Concept: Use the section formula for internal division of a line segment. If a point P divides AB internally in the ratio m:n (i.e., AP : PB = m : n), then \(P\bigl(\frac{n x_1 + m x_2}{m+n},\frac{n y_1 + m y_2}{m+n}\bigr)\), where \((x_1,y_1)\) and \((x_2,y_2)\) are the coordinates of A and B respectively.
1. Interpret the given condition\
The statement "AP = 3 AB 7" in NCERT notation means \(AP = \frac{3}{7} \; AB\). Hence \(AP : PB = 3 : 4\) because \(AB = AP + PB\).

2. Identify the ratio\
Let \(m = AP = 3\) and \(n = PB = 4\). So the required ratio is \(m:n = 3:4\).

3. Write the coordinates of A and B\
\[A\equiv (x_1,y_1) = (-2,-2), \quad B\equiv (x_2,y_2) = (2,-4).\]

4. Apply the section formula\
\[P\biggl(\frac{n x_1 + m x_2}{m+n},\;\frac{n y_1 + m y_2}{m+n}\biggr)\]
Substituting \(m=3,\;n=4\):\
\[\begin{aligned}
x_P &= \frac{4(-2) + 3(2)}{3+4} = \frac{-8 + 6}{7} = \frac{-2}{7},\\[4pt]
y_P &= \frac{4(-2) + 3(-4)}{3+4} = \frac{-8 -12}{7} = \frac{-20}{7}.
\end{aligned}\]

5. Result\
Hence the coordinates of point \(P\) are \(\displaystyle \left(-\frac{2}{7},\; -\frac{20}{7}\right)\).

6. Verification (optional)\
- Vector \(\overrightarrow{AB} = (2-(-2),\; -4-(-2)) = (4,-2)\).
- Vector \(\overrightarrow{AP} = \left(-\frac{2}{7}+2,\; -\frac{20}{7}+2\right) = \left(\frac{12}{7},\; -\frac{6}{7}\right)\).
- The ratio \(\frac{\|\overrightarrow{AP}\|}{\|\overrightarrow{AB}\|} = \frac{3}{7}\) confirming the condition.

Thus, \(P\) indeed lies on the line segment \(AB\) and satisfies the required proportion.

Correct Answer: \(\left(-\frac{2}{7},\; -\frac{20}{7}\right)\)
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