Question

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) : △ABCs an isosceles right angled triangle, and

(S2) : the area of △ABC is \(9\sqrt2\over2\)

a.) Both S1 and S2 are true

b.)Only S1

c.)Only S2

d.) Both are False

Answer 1

Answer: (b) only (S1)

\(\mathrm{A}(\mathrm{x}, \mathrm{y}, \mathrm{z})\) Let \(\mathrm{P}(0,3,2), \mathrm{Q}(2,0,3), \mathrm{R}(0,0,1)\)

\(\mathrm{AP}=\mathrm{AQ}=\mathrm{AR}\)

\(x^{2}+(y-3)^{2}+(z-2)^{2}=(x-2)^{2}+y^{2}+(z-3)^{2}=x^{2}+y^{2}+(z-1)^{2}\)

In xy plane z = 0

So, \(x^{2}-4 x+4+y^{2}+9=x^{2}+y^{2}+1\)

\(\Rightarrow \mathrm{y}=2\)

\(\mathrm{x}=3\)

\(9+y^{2}-6 y+9+4=x^{2}+y^{2}+1\)

So, \(\mathrm{A}(3,2,0)\ \text{also}\ \mathrm{B}(1,4,-1)\ \& \ \mathrm{C}(2,0,-2)\)

Now \(A B=\sqrt{4+4+1}=3\)   

\(\mathrm{AC}=\sqrt{1+4+4}=3 \)

\( \mathrm{BC}=\sqrt{1+16+1}=\sqrt{18}\)     

\(\mathrm{AB}=\mathrm{AC}\)

isosceles \(\Delta\ \& \ \mathrm{AB}^{2}+\mathrm{AC}^{2}=\mathrm{BC}^{2}\)

right angle \(\Delta\)

Area of \(\triangle \mathrm{ABC}=\frac{1}{2} \times\) base.height

\(\frac{1}{2} \times 3 \times 3=\frac{9}{2}\)

So only \(\mathrm{S}_{1}\) is true