Given: Points are A(-2, 2, 3) and B(13, -3, 13)
To find: the locus of point P which moves in such a way that 3PA = 2PB
Let the required point P(x, y, z)
According to the question:
3PA = 2PB
⇒ 9PA2 = 4PB2
Formula used:
The distance between any two points (a, b, c) and (m, n, o) is given by
\(\sqrt{(a−m)^2+(b−n)^2+(c−0)^2}\)
Therefore,
The distance between P(x, y, z) and A(-2, 2, 3) is PA,
= \(\sqrt{(x-(-2))^2+(y-2)^2+(z-3)^2}\)
= \(\sqrt{(x+2)^2+(y-2)^2+(z-3)^2}\)
The distance between P(x, y, z) and B(13, -3, 13) is PB,
= \(\sqrt{(x-13)^2+(y-(-3))^2+(z-13)^2}\)
= \(\sqrt{(x-13)^2+(y+3)^2+(z-13)^2}\)
As 9PA2 = 4PB2
9{(x + 2)2+ (y – 2)2 + (z – 3)2} = 4{(x – 13)2 + (y + 3)2 + (z – 13)2}
⇒ 9{x2+ 4 + 4x + y2 + 4 – 4y + z2 + 9 – 6z} = 4{x 2+ 169 – 26x + y2 + 9 + 6y + z 2 + 169 – 26z}
⇒ 9{x2 + 4x + y2 – 4y + z2 – 6z + 17} = 4{x2 – 26x + y2 + 6y + z2 – 26z + 347}
⇒ 9x2 + 36x + 9y2 – 36y + 9z2 – 54z + 153 = 4x2 – 104x + 4y 2 + 24y + 4z 2 – 104z + 1388
⇒ 9x2 + 36x + 9y2 – 36y + 9z2 – 54z + 153 – 4x2 + 104x – 4y2 – 24y – 4z2 + 104z – 1388 = 0
⇒ 5x2 + 5y2 + 5z2 + 140x – 60y + 50z – 1235 = 0
Hence locus of point P is 5x2 + 5y2 + 5z2 + 140x – 60y + 50z – 1235 = 0
