NCERT Exemplar Solutions
Class 10 - Mathematics - CHAPTER 7: Coordinate Geometry - Exercise 7.3
Question 13

Question. 13

If \((a,b)\) is the midpoint of the segment joining \(A(10,-6)\) and \(B(k,4)\) and \(a-2b=18\), find \(k\) and \(|AB|\).

Answer:

\(k=22\), \(|AB|=2\sqrt{61}\)

Detailed Answer with Explanation:

Step 1: Recall the midpoint formula

The midpoint of a line joining points \((x_1, y_1)\) and \((x_2, y_2)\) is:

\[ \left( \dfrac{x_1 + x_2}{2}, \dfrac{y_1 + y_2}{2} \right) \]

Step 2: Apply to given points

Point \(A(10, -6)\), point \(B(k, 4)\).

So, midpoint \((a,b)\) is:

\[ a = \dfrac{10 + k}{2}, \quad b = \dfrac{-6 + 4}{2} \]

Step 3: Simplify the \(y\)-coordinate of midpoint

\[ b = \dfrac{-6 + 4}{2} = \dfrac{-2}{2} = -1 \]

Step 4: Use the condition \(a - 2b = 18\)

Substitute values:

\[ a - 2b = 18 \]

\[ \dfrac{10 + k}{2} - 2(-1) = 18 \]

\[ \dfrac{10 + k}{2} + 2 = 18 \]

Step 5: Solve for \(k\)

Subtract 2 from both sides:

\[ \dfrac{10 + k}{2} = 16 \]

Multiply both sides by 2:

\[ 10 + k = 32 \]

Subtract 10:

\[ k = 22 \]

Step 6: Find length of \(|AB|\)

Formula for distance between two points \((x_1,y_1)\) and \((x_2,y_2)\) is:

\[ |AB| = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]

Substitute \(A(10,-6)\), \(B(22,4)\):

\[ |AB| = \sqrt{(22 - 10)^2 + (4 - (-6))^2} \]

\[ |AB| = \sqrt{(12)^2 + (10)^2} \]

\[ |AB| = \sqrt{144 + 100} = \sqrt{244} \]

\[ |AB| = 2\sqrt{61} \]

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