NCERT Exemplar Solutions
Class 10 - Mathematics - CHAPTER 7: Coordinate Geometry - Exercise 7.3
Question 15

Question. 15

The line segment joining the points \(A(3,2)\) and \(B(5,1)\) is divided at the point \(P\) in the ratio \(1:2\), and \(P\) lies on the line \(3x - 18y + k = 0\). Find the value of \(k\).

Answer:

\(k=19\)

Detailed Answer with Explanation:

Step 1: Recall the section formula.

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

\[ P = \left( \dfrac{mx_2 + nx_1}{m+n}, \; \dfrac{my_2 + ny_1}{m+n} \right) \]

Step 2: Write down what we know.

  • Point \(R = (3, 2)\)
  • Point \(B = (5, 1)\)
  • Ratio \(RP:PB = 1:2\), so \(m=1, n=2\)

Step 3: Find the coordinates of \(P\).

\[ P = \left( \dfrac{1 \cdot 5 + 2 \cdot 3}{1+2}, \; \dfrac{1 \cdot 1 + 2 \cdot 2}{1+2} \right) \]

\[ P = \left( \dfrac{5 + 6}{3}, \; \dfrac{1 + 4}{3} \right) = \left( \dfrac{11}{3}, \; \dfrac{5}{3} \right) \]

Step 4: Use the line equation.

Point \(P\) lies on the line \(3x - 18y + k = 0\). Substituting \(x = 11/3, y = 5/3\):

\[ 3 \left(\dfrac{11}{3}\right) - 18 \left(\dfrac{5}{3}\right) + k = 0 \]

Step 5: Simplify.

\[ 11 - 30 + k = 0 \]

\[ -19 + k = 0 \]

Step 6: Solve for \(k\).

\[ k = 19 \]

Final Answer: \(k = 19\)

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