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P is a point on the circle `x^2+y^2=9` Q is a point on the line `7x+y+3=0`. The perpendicular bisector of PQ is `x-y+1=0`. Then the coordinates of P are:A. `(0,-3)`B. `(0,3)`C. `(72//25,21//25)`D. `(-72//25,21//25)` |
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Answer» Correct Answer - 3 Any point on the line `7x+y+3=0` is `Q(t ,-3-7t),t inR`. Now, `P(h,k)` is the image of point Q in the line `x-y+1=0`. Then, `(h-t)/(1)=(k-(-3-7t))/(-1)` `=-(2{t-(-3-7t)+1})/(1+1)` `=-8t-4` or `(h,k) -= (-7t-4,t+1)` This point lies on the circle `x^(2)+y^(2)=9`. Then, `(-7t-4)^(2)+(t+1)^(2)=9` or `50t^(2)+58t+8=0` or `25t^(2)+29t+4=0` or `(25t+4)(t+1)=0` or `t=(-4)/(25),t= -1` i.e., `(h,k) -= (-(72)/(25),(21)/(25))` or `(3,0)` |
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