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

Question 4

Step 1: Recall the distance formula between two points \((x_1, y_1)\) and \((x_2, y_2)\):

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

Step 2: Here, the coordinates are:

• Point A: \((-3, -14)\)
• Point B: \((a, -5)\)
• Distance: \(d = 9\)

Step 3: Substitute values into the formula:

\[ 9 = \sqrt{(a - (-3))^2 + (-5 - (-14))^2} \]

Step 4: Simplify the terms:

• \(a - (-3) = a + 3\)
• \(-5 - (-14) = -5 + 14 = 9\)

So:

\[ 9 = \sqrt{(a+3)^2 + (9)^2} \]

Step 5: Square both sides to remove the square root:

\[ 9^2 = (a+3)^2 + 9^2 \]

\[ 81 = (a+3)^2 + 81 \]

Step 6: Subtract 81 from both sides:

\[ 81 - 81 = (a+3)^2 \]

\[ 0 = (a+3)^2 \]

Step 7: Take square root of both sides:

\[ a + 3 = 0 \]

Step 8: Solve for \(a\):

\[ a = -3 \]

Final Answer: \(a = -3\).