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

Question 8

Step 1: Since point \(A(2,-4)\) is equidistant from \(P(3,8)\) and \(Q(-10,y)\), it means:

\(AP = AQ\)

Step 2: Use the distance formula:

\(AP = \sqrt{(2-3)^2 + (-4-8)^2}\)

\(AQ = \sqrt{(2-(-10))^2 + (-4-y)^2}\)

Step 3: Square both sides to remove the square root:

\((2-3)^2 + (-4-8)^2 = (2+10)^2 + (-4-y)^2\)

Step 4: Simplify each term:

• \((2-3)^2 = (-1)^2 = 1\)
• \((-4-8)^2 = (-12)^2 = 144\)
• \((2+10)^2 = (12)^2 = 144\)
• \((-4-y)^2 = (y+4)^2\) (since squaring removes the sign)

So the equation becomes:

\(1 + 144 = 144 + (y+4)^2\)

\(145 = 144 + (y+4)^2\)

Step 5: Subtract 144 from both sides:

\((y+4)^2 = 1\)

Step 6: Take square root:

\(y+4 = ±1\)

So, \(y = -4+1 = -3\) or \(y = -4-1 = -5\)

Step 7: Now find \(PQ\) using the distance formula.

For \(y = -3\):

\(PQ = \sqrt{(3 - (-10))^2 + (8 - (-3))^2} = \sqrt{13^2 + 11^2} = \sqrt{169 + 121} = \sqrt{290}\)

For \(y = -5\):

\(PQ = \sqrt{(3 - (-10))^2 + (8 - (-5))^2} = \sqrt{13^2 + 13^2} = \sqrt{169 + 169} = \sqrt{338}\)

Final Answer: \(y = -3\) or \(y = -5\). \(PQ = \sqrt{290}\) when \(y = -3\), and \(PQ = \sqrt{338}\) when \(y = -5\).