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| 1. |
find the values of y for which the distance between the point p(2, -3) and q(10,y) is 10 untis |
| Answer» Using Distance formula, we have{tex}10 = \\sqrt {{{(2 - 10)}^2} + {{( - 3 - y)}^2}} {/tex}{tex} \\Rightarrow 10 = \\sqrt {{{( - 8)}^2} + 9 + {y^2} + 6y} {/tex}\xa0{tex}{/tex}{tex} \\Rightarrow 10 = \\sqrt {64 + 9 + {y^2} + 6y} {/tex}Squaring both sides, we get100 = 73 + y2 + 6y⇒ y2 + 6y - 27 = 0Solving this Quadratic equation by factorization, we can write⇒ y2 + 9y − 3y - 27 = 0⇒ y(y + 9) - 3(y + 9) = 0⇒ (y + 9)(y − 3) = 0⇒ y = 3, −9 | |