Question

Find the equation of locus of a point which is at a distance of 2d from (1,0) and at a distance of d from (2,0).

Answer 1

Answer:

y =0

No need to show the whole steps i just did to make u understand...

Most of the things are understood

Step-by-step explanation:

Take a point to be P(a,0)

which will have co ordinates as

Where as A(1,0) and B(2,0)

So AP=2d

and BP=d

therfore the ratio is

AP:BP=2:1

By SECTION FORMULA

x=[(m1)(x2)+(m2)(x1)] ÷ (m1 + m2)

therefore

x= (2×2+1×1)÷3

x=1

y=0

m = (y2-y1) ÷ (x2-x1)

m= (0-0) ÷ (2-1)

m= 0

therefore the slope or gradient is 0

General eq of a straight line is :

y= mx+c

putting values to find y- intercept or c.

0=0×1+c

c = 0

therefore eq of the straight line is

y= 0+0

y=0;

Hope it helps..

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