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If A (3, y) is equidistant from points P (8, -3) and Q (7,6) , find the value of y and find the distance AQ. |
Answer» Coordinates are P(8, -3) and Q(7,6) The point A (3, y) is equidistant. Using distance formula = \(\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\) ⇒ PA = QA \(\sqrt{(3 - 8)^2 + (y + 3)^2}\) = \(\sqrt{(3 - 7)^2 + (y - 6)^2}\) On squaring both sides, we get (3 - 8)2 + (y + 3)2 = (3 - 7)2 + (y - 6)2 25 + y2 + 6y + 9 = 16 + y2 - 12y + 36 y = 1 AQ = \(\sqrt{(3 - 7)^2 + (1 - 6)^2}\) = \(\sqrt{16 + 25}\) = \(\sqrt{41}\) units |
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