The vertices of triangle ABC are A (2,0,0), B (0,1,0), C (0,0,2). Its orthocentre is H and circumcentre is S. P is a point equidistant from A , B , C and the origin O. PA =

Question

The vertices of triangle ABC are A (2,0,0), B (0,1,0), C (0,0,2). Its orthocentre is H and circumcentre is S. P is a point equidistant from A , B , C and the origin O. PA = (A) 1 (B) √2 (C) √(3/2) (D) (3/2)

Tardigrade Admin · Accepted Answer

P ( x , y , z ) ⇒ x 2+ y 2+ z 2=( x -2)2+ y 2+ z 2 = x 2+( y -1)2+ z 2= x 2+ y 2+( z -2)2 ⇒ x =1, y =(1/2), z =1 P (1, (1/2), 1), A (2,0,0) ⇒ AP =(3/2)

Q. The vertices of $△ A BC$ are $A (2, 0, 0), B (0, 1, 0), C (0, 0, 2)$ . Its orthocentre is $H$ and circumcentre is $S$ . $P$ is a point equidistant from $A, B, C$ and the origin $O$ .
$P A =$

1

$2$

$\frac{3}{2}$

$\frac{3}{2}$

$P (x, y, z) \Rightarrow x^{2} + y^{2} + z^{2} = (x - 2)^{2} + y^{2} + z^{2} < b r / >= x^{2} + (y - 1)^{2} + z^{2} = x^{2} + y^{2} + (z - 2)^{2}$
$\Rightarrow x = 1, y = \frac{1}{2}, z = 1$
$P (1, \frac{1}{2}, 1), A (2, 0, 0) \Rightarrow A P = \frac{3}{2}$

Q. The vertices of △ABC are A(2,0,0),B(0,1,0),C(0,0,2). Its orthocentre is H and circumcentre is S. P is a point equidistant from A,B,C and the origin O. PA=

1

2​

23​​

23​

P(x,y,z)⇒x2+y2+z2=(x−2)2+y2+z2<br/>=x2+(y−1)2+z2=x2+y2+(z−2)2 ⇒x=1,y=21​,z=1 P(1,21​,1),A(2,0,0)⇒AP=23​

Q. The vertices of $△ A BC$ are $A (2, 0, 0), B (0, 1, 0), C (0, 0, 2)$ . Its orthocentre is $H$ and circumcentre is $S$ . $P$ is a point equidistant from $A, B, C$ and the origin $O$ .
$P A =$

$2$

$\frac{3}{2}$

$\frac{3}{2}$

$P (x, y, z) \Rightarrow x^{2} + y^{2} + z^{2} = (x - 2)^{2} + y^{2} + z^{2} < b r / >= x^{2} + (y - 1)^{2} + z^{2} = x^{2} + y^{2} + (z - 2)^{2}$
$\Rightarrow x = 1, y = \frac{1}{2}, z = 1$
$P (1, \frac{1}{2}, 1), A (2, 0, 0) \Rightarrow A P = \frac{3}{2}$