Question

find the points on the x axis whose distance from the points (2,3) and (3/2,-1) are in the ratio 2:1

Answer 1

let P(x, 0) is the point on x-axis whose distance from the points A(2,3) and B(3/2,-1) are in the ratio 2:1

distance between joining two points (x1,y1) and (x2,y2) is
√(x2-x1)²+(y2-y1)²

1) PA ² = (2-x)²+(3-0)²
= 4-4x+x²+9
=x²-4x+13----(1)
2)PB² = (3/2 -x)²+(-1-0)²
= 9/4-3x+x²+1
=x²-3x+9/4+1 ----(2)

but
PA/PB =2/1
PA²/PB² = 4/1
(x²-4x+13)/(x²-3x+9/4+1) =4/1

x²-4x+13 = 4(x²-3x+9/4+1)

x²-4x+13 =4x²-12x+9+4
0=4x²-12x+13-x²+4x-13
0 =3x²-8x
3x²-8x=0
x(3x-8)=0
x=0 or 3x-8 =0
x=0 or x= 8/3
therefore required point (x,0) = (0,0)

or
(x,y) = (8/3,0)


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Answer 2

A.t.q=(x-2)2+(0-3)2.
. .......................=2/1
(x-3/2)2+(0+1)2

X2+4-4x+9
.................... =2/1
x2+9/4-3x+1
Cross multiply
x2-4x+13=2(x2)+9/2-6x+2
=> x2+9/2+2-13-6x+4x=0
=> x2-2x-13/2=0
=> 2(x2)-4x-13
. ................... =0
. 2
=> 2(x2)-4x-13=0
=> x=4+_(under root16+104)
...................
4
=> x=4+-2(under root30)
.................................
4
=> x=(2+under root30)
. ..............................
. 2
And
X=(2-under root30)/2