Let A(x, y, z) be a point in the xy-plane,whi | vedclass

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(B) Since $A(x, y, z)$ lies in the $xy$-plane,$z = 0$. Thus,$A = (x, y, 0)$.Given $AP^2 = AQ^2 = AR^2$.$AR^2 = x^2 + y^2 + (0 - 1)^2 = x^2 + y^2 + 1$.

Vedclass · Accepted Answer

only $(S1)$ is true

Vedclass · Answer

(B) Since $A(x, y, z)$ lies in the $xy$-plane,$z = 0$. Thus,$A = (x, y, 0)$.Given $AP^2 = AQ^2 = AR^2$.$AR^2 = x^2 + y^2 + (0 - 1)^2 = x^2 + y^2 + 1$.$AP^2 = x^2 + (y - 3)^2 + (0 - 2)^2 = x^2 + y^2 - 6y + 9 + 4 = x^2 + y^2 - 6y + 13$.$AQ^2 = (x - 2)^2 + y^2 + (0 - 3)^2 = x^2 - 4x + 4 + y^2 + 9 = x^2 + y^2 - 4x + 13$.Equating $AP^2 = AR^2 \implies x^2 + y^2 - 6y + 13 = x^2 + y^2 + 1 \implies 6y = 12 \implies y = 2$.Equating $AQ^2 = AR^2 \implies x^2 + y^2 - 4x + 13 = x^2 + y^2 + 1 \implies 4x = 12 \implies x = 3$.So,$A = (3, 2, 0)$.Now,calculate the side lengths of $	riangle ABC$ with $A(3, 2, 0)$,$B(1, 4, -1)$,and $C(2, 0, -2)$:$AB^2 = (3 - 1)^2 + (2 - 4)^2 + (0 - (-1))^2 = 2^2 + (-2)^2 + 1^2 = 4 + 4 + 1 = 9 \implies AB = 3$.$AC^2 = (3 - 2)^2 + (2 - 0)^2 + (0 - (-2))^2 = 1^2 + 2^2 + 2^2 = 1 + 4 + 4 = 9 \implies AC = 3$.$BC^2 = (1 - 2)^2 + (4 - 0)^2 + (-1 - (-2))^2 = (-1)^2 + 4^2 + 1^2 = 1 + 16 + 1 = 18 \implies BC = \sqrt{18} = 3\sqrt{2}$.Since $AB = AC = 3$ and $AB^2 + AC^2 = 9 + 9 = 18 = BC^2$,$	riangle ABC$ is an isosceles right-angled triangle. Thus,$(S1)$ is true.Area of $	riangle ABC = \frac{1}{2} 	imes AB 	imes AC = \frac{1}{2} 	imes 3 	imes 3 = \frac{9}{2}$.Since $\frac{9}{2} 
eq \frac{9\sqrt{2}}{2}$,$(S2)$ is false.Therefore,only $(S1)$ is true.

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