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The equations of the perpendicular bisector of the sides AB and perpendicular bisector of the sides AB and AC of a `triangle ABC` are x-y + 5 = 0 and x + 2y = 0 respectively, if the point Ais (1,-2), then the equation ofthe line BC isA. 14x+23y=40B. 14x-23y=40C. 23x+14y=40D. 23x-14y=40 |
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Answer» Correct Answer - A A is (1, -2) B is image of a w.r.t. x-y+5=0 i.e. (-7, 6) C is image of A w.r.t. x + 2y =0 `" "i.e.((11)/(5),(2)/(5))` Now equation of BC is `y-6= ((2//5)-6)/((11//5)+7)(x+7)` `rArr14x+23y=40` |
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