1.

The point P is equidistant from A(1, 3), B(–3, 5) and C(5, –1). Then PB is equal to :(a) \(5\sqrt2\)(b) 5 (c) \(5\sqrt5\)(d) \(5\sqrt{10}\)

Answer»

(d) \(5\sqrt{10}\)

. Let P ≡ (x, y). Then, PA = PB and PB = PC. 

∴ PA2 = PB2 ⇒ (x – 1)2 + (y – 3)2 = (x + 3)2 + (y – 5)2 

⇒ x2 – 2x + 1 + y2 – 6y + 9 = x2 + 6x + 9 + y2 – 10y + 25 

⇒ –8x + 4y – 24 = 0 ⇒ 2x – y + 6 = 0            ...(i) 

PB2 = PC2 ⇒ (x + 3)2 + (y – 5)2 = (x – 5)2 + (y + 1)2 

⇒ x2 + 6x + 9 + y2 – 10y + 25 = x2 – 10x + 25 + y2 + 2y + 1 

⇒ 16x – 12y + 8 = 0 ⇒ 4x – 3y + 2 = 0           ...(ii) 

From (i), y = 2x + 6. Putting in (ii), we have 

4x – 3(2x + 6) + 2 = 0 

⇒ 4x – 6x – 18 + 2 = 0 ⇒ –2x –16 = 0 ⇒ x = –8 

∴ y = 2 × (– 8) + 6 = –10. 

∴ PB = \(\sqrt{(-8+3)^2+(-10-5)^2}\) = \(\sqrt{(-5)^2+(-15)^2}\)

\(\sqrt{250}\) = \(5\sqrt{10}\).



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