Question

The base of vertices of an isosceles triangle PQR are Q(1, 3) and R(-2, 7). The vertex P can be

• A. (1, 6)
• B. (1/2, 5)
• C. (5/6, 6)
• D. None of these

Hint:

Equal sides condition gives equation of perpendicular bisector of QR.

Answer 1

The Correct Option is C

To determine the coordinates of vertex P in the isosceles triangle PQR, we need to satisfy the property that two sides of the triangle are equal in length. Given that Q(1, 3) and R(-2, 7) are the base vertices of the triangle, we need to find the vertex P such that the distance from P to Q equals the distance from P to R.

Let's denote the coordinates of point P as (x, y). The equation for distance between two points \((x_1, y_1)\) and \((x_2, y_2)\) is:

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

For PQR to be isosceles, d(PQ) = d(PR), which gives:

1. Distance PQ: \(d(PQ) = \sqrt{(x - 1)^2 + (y - 3)^2}\)

2. Distance PR: \(d(PR) = \sqrt{(x + 2)^2 + (y - 7)^2}\)

For equality, set \(d(PQ) = d(PR)\):

\(\sqrt{(x - 1)^2 + (y - 3)^2} = \sqrt{(x + 2)^2 + (y - 7)^2}\)

Squaring both sides to eliminate the square root gives:

\((x - 1)^2 + (y - 3)^2 = (x + 2)^2 + (y - 7)^2\)

Expanding both sides, we have:

\(x^2 - 2x + 1 + y^2 - 6y + 9 = x^2 + 4x + 4 + y^2 - 14y + 49\)

After simplifying, canceling common terms and combining like terms, this simplifies to:

\(-6x + 2y = 43\)

Now, substitute each option to identify the point which satisfies the equation:

• For option (1, 6), substituting x = 1, y = 6: \(-6(1) + 2(6) \neq 43\)
• For option (1/2, 5), substituting x = 1/2, y = 5: \(-6(1/2) + 2(5) \neq 43\)
• For option (5/6, 6), substituting x = 5/6, y = 6: \(-6(5/6) + 2(6) = 43\) ✔️

Hence, the correct vertex P satisfying the isosceles triangle condition is:
(5/6, 6)