Let A(x, y, z) be a point in xy-plane, which is equidistant from three points (0, 3, 2), (2, 0, 3) and (0, 0, 1). Let B = (1, 4, –1) and C = (2, 0, –2). Then among the statements (S1) : ABC is an isosceles right angled triangle, and (S2) : the area of A

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Solution

Q.
Let A(x, y, z) be a point in xy-plane, which is equidistant from three points (0, 3, 2), (2, 0, 3) and (0, 0, 1). Let B = (1, 4, –1) and C = (2, 0, –2). Then among the statements

(S1) : $△$ ABC is an isosceles right angled triangle, and

(S2) : the area of $△$ ABC is $\frac{9 \sqrt{2}}{2}$ .

1 only (S1) is true

2 both are false

3 both are true

4 only (S2) is true

Ans.
(1)

Given, A(x, y, z) be a point in xy-plane. Let the point P(0, 3, 2), Q(2, 0, 3) and R(0, 0, 1)

The distance of the point AP = AQ = AR

$\Rightarrow {(x - 0)}^{2} + {(y - 3)}^{2} + {(z - 2)}^{2}$

$= {(x - 2)}^{2} + {(y - 0)}^{2} + {(z - 3)}^{2}$

$= {(x - 0)}^{2} + {(y - 0)}^{2} + {(z - 1)}^{2}$

In xy-plane, z = 0

So, $x^{2} - 4 x + y^{2} + 9 + 4 = x^{2} + y^{2} + 1 \Rightarrow - 4 x + 13 = 1 \Rightarrow x = 3$

And $x^{2} + y^{2} - 6 y + 9 + 4 = x^{2} + y^{2} + 1 \Rightarrow y = 2$

So, A(3, 2, 0), B(1, 4, –1) and C(2, 0, –2).

In $△$ ABC

$A B = \sqrt{1 + 4 + 4} = 3, B C = \sqrt{1 + 16 + 1} = \sqrt{18}, C A = \sqrt{1 + 4 + 4} = 3$

So, AB = AC and $A B^{2} + A C^{2} = {(B C)}^{2}$

$∴$ $△$ ABC is an isosceles right angled triangle.

So, (S1) is true.

Also, Area of $△ A B C = \frac{1}{2} \times (Base) (Height) = \frac{1}{2} \times 3 \times 3 = \frac{9}{2}$

So, (S2) is false.

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