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| 1. |
Find the value of y for which the distance between the points p (2, - 3 )and q(10,y)is 10 units |
| Answer» Given that\xa0distance between the points P (2, -3) and Q (10, y) is 10\xa0Therefore using distance formula√(2 -10)2 + (-3-y)2 = 10√ (-8)2 + ( 3 + y)2 = 10squaring on both side64 + (y +3)2 = 100(y + 3)2 = 36y + 3 = ± 6y+3=6or y+3= -6Therefore, y=3 or y= -9 | |