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The locus of a point which moves such that the sum of the square of itsdistance from three vertices of a triangle is constant is a/ancircle (b)straight line(c) ellipse (d)none of theseA. circleB. straight lineC. ellipseD. none of these |
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Answer» Correct Answer - A Let `A(x_(1), y_(1)), B(x_(2), y_(2))` and `C(x_(3), y_(3))` be the vertices of the triangle ABC, and let P (h, k) be any point on the locus. Then, `PA^(2)+PB^(2)+PC^(2)=c " " `(constant) `rArr underset(i=1)overset(3)Sigma(h-x_(i))^(2)+(k-y_(i))^(2)=c` `rArrh^(2)+k^(2)-(2h)/(3)(underset(i=1)overset(3)Sigma x_( i) )-(2k)/(3)(underset(i=1)overset(3)Sigma y_(i))+(1)/(3)underset(i=1)overset(3)Sigma (x_( i)^(2)+y _(i)^(2))-(c)/(3)=0` So, locus of (h, k) is `x^(2)+y^(2)-(2x)/(3)(x_(1)+x_(2)+x_(3))-(2y)/(3)(y_(1)+y_(2)+y_(3))+lambda=0`, where `lambda=underset(i=1)overset(3)Sigma (x_(i)^(2)+y_(i)^(2))-c=0=` constant. Clearly, this locus is a circle with centre at `((x_(1)+x_(2)+x_(3))/(3), (y_(1)+y_(2)+y_(3))/(3))` |
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