NCERT Exemplar Solutions
Class 10 - Mathematics - CHAPTER 7: Coordinate Geometry - Exercise 7.3
Question 11

Question. 11

Find the ratio in which the point \(P\big(\dfrac{3}{4},\dfrac{5}{12}\big)\) divides the segment joining \(A\big(\dfrac{1}{2},\dfrac{3}{2}\big)\) and \(B(2,-5)\).

Answer:

\(1:5\) (\(AP:PB\)).

Detailed Answer with Explanation:

Step 1: Recall the section formula.

If a point \(P(x,y)\) divides a line joining two points \(A(x_1,y_1)\) and \(B(x_2,y_2)\) in the ratio \(m:n\), then

\[ x = \dfrac{mx_2 + nx_1}{m+n}, \quad y = \dfrac{my_2 + ny_1}{m+n} \]

Step 2: Write down the given values.

  • \(A(x_1,y_1) = \big(\tfrac{1}{2}, \tfrac{3}{2}\big)\)
  • \(B(x_2,y_2) = (2, -5)\)
  • \(P(x,y) = \big(\tfrac{3}{4}, \tfrac{5}{12}\big)\)

Step 3: Apply the section formula for the \(x\)-coordinate.

\[ \tfrac{3}{4} = \dfrac{m(2) + n(\tfrac{1}{2})}{m+n} \]

Step 4: Simplify the numerator.

\[ \tfrac{3}{4} = \dfrac{2m + \tfrac{n}{2}}{m+n} \]

Step 5: Remove the fraction in the numerator.

Multiply numerator and denominator by 2: \[ \tfrac{3}{4} = \dfrac{4m + n}{2m + 2n} \]

Step 6: Cross multiply.

\[ 3(2m + 2n) = 4(4m + n) \]

\[ 6m + 6n = 16m + 4n \]

Step 7: Rearrange terms.

\[ 6n - 4n = 16m - 6m \] \[ 2n = 10m \]

Step 8: Simplify the ratio.

\[ n = 5m \quad \Rightarrow \quad m:n = 1:5 \]

Step 9: Verify with the \(y\)-coordinate (optional check).

Using \(y = \tfrac{my_2 + ny_1}{m+n}\): Substitute \(m:n = 1:5\), You will get \(y = \tfrac{5}{12}\), which matches the given \(P\).

Final Answer: The point \(P\) divides \(AB\) in the ratio \(1:5\) (that is, \(AP:PB = 1:5\)).

NCERT Exemplar Solutions Class 10 – Mathematics – CHAPTER 7: Coordinate Geometry – Exercise 7.3 | Detailed Answers