If the perimeter of a triangle is 20 and two | vedclass

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Question

(A) Let the third vertex be $P(x, y)$. The other two vertices are $A(-5, 0)$ and $B(6, 0)$.The perimeter is the sum of the lengths of the sides: $PA +

Vedclass · Accepted Answer

$40 x^2 - 81 y^2 - 40 x - 800 = 0$

Vedclass · Answer

(A) Let the third vertex be $P(x, y)$. The other two vertices are $A(-5, 0)$ and $B(6, 0)$.The perimeter is the sum of the lengths of the sides: $PA + PB + AB = 20$.The distance $AB = \sqrt{(6 - (-5))^2 + (0 - 0)^2} = \sqrt{11^2} = 11$.So,$PA + PB = 20 - 11 = 9$.Substituting the coordinates,$\sqrt{(x + 5)^2 + y^2} + \sqrt{(x - 6)^2 + y^2} = 9$.Rearranging: $\sqrt{(x + 5)^2 + y^2} = 9 - \sqrt{(x - 6)^2 + y^2}$.Squaring both sides: $(x + 5)^2 + y^2 = 81 + (x - 6)^2 + y^2 - 18\sqrt{(x - 6)^2 + y^2}$.$x^2 + 10x + 25 + y^2 = 81 + x^2 - 12x + 36 + y^2 - 18\sqrt{(x - 6)^2 + y^2}$.$22x - 92 = -18\sqrt{(x - 6)^2 + y^2}$.Dividing by $-2$: $46 - 11x = 9\sqrt{(x - 6)^2 + y^2}$.Squaring again: $(46 - 11x)^2 = 81((x - 6)^2 + y^2)$.$2116 - 1012x + 121x^2 = 81(x^2 - 12x + 36 + y^2)$.$2116 - 1012x + 121x^2 = 81x^2 - 972x + 2916 + 81y^2$.$40x^2 - 81y^2 - 40x - 800 = 0$.

If the perimeter of a triangle is $20$ and two of its vertices are $(-5, 0)$ and $(6, 0)$,then the locus of the third vertex is:

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