Question ID: #433
Let A(x, y, z) be a point in xy-plane, which is equidistant from three points P(0, 3, 2), Q(2, 0, 3) and R(0, 0, 1). Let B = (1, 4, -1) and C = (2, 0, -2). Then among the statements
(S1): $\Delta ABC$ is an isosceles right angled triangle and
(S2): the area of $\Delta ABC$ is $\frac{9\sqrt{2}}{2}$
- (1) both are true
- (2) only (S1) is true
- (3) only (S2) is true
- (4) both are false
Solution:
Since A lies in the xy-plane, its z-coordinate is 0. Let $A = (x, y, 0)$.
Given $AP = AQ = AR$.
$AP^2 = AR^2$:
$$ x^2 + (y-3)^2 + (0-2)^2 = (x-0)^2 + (y-0)^2 + (0-1)^2 $$
$$ x^2 + y^2 – 6y + 9 + 4 = x^2 + y^2 + 1 $$
$$ -6y + 13 = 1 \Rightarrow 6y = 12 \Rightarrow y = 2 $$
$AQ^2 = AR^2$:
$$ (x-2)^2 + y^2 + (0-3)^2 = x^2 + y^2 + 1 $$
$$ x^2 – 4x + 4 + y^2 + 9 = x^2 + y^2 + 1 $$
$$ -4x + 13 = 1 \Rightarrow 4x = 12 \Rightarrow x = 3 $$
So, $A = (3, 2, 0)$.
Now check $\Delta ABC$ with $A(3, 2, 0)$, $B(1, 4, -1)$, $C(2, 0, -2)$.
$$ AB^2 = (3-1)^2 + (2-4)^2 + (0+1)^2 = 4 + 4 + 1 = 9 \Rightarrow AB = 3 $$
$$ AC^2 = (3-2)^2 + (2-0)^2 + (0+2)^2 = 1 + 4 + 4 = 9 \Rightarrow AC = 3 $$
$$ BC^2 = (2-1)^2 + (0-4)^2 + (-2+1)^2 = 1 + 16 + 1 = 18 \Rightarrow BC = 3\sqrt{2} $$
Since $AB = AC$, it is isosceles.
Since $AB^2 + AC^2 = 9 + 9 = 18 = BC^2$, it is right-angled at A.
Statement (S1) is True.
Area of $\Delta ABC = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times AB \times AC$
$$ \text{Area} = \frac{1}{2} \times 3 \times 3 = \frac{9}{2} $$
Statement (S2) says area is $\frac{9\sqrt{2}}{2}$, which is False.
Ans. (2)
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