Question

plz solve this question for me​

Answer 1

Answer:

we can find this by using distance formula:-

i.e, d^2=(x2-x1)^2 + (y2-y1)^2

first for points (x, y) and (3, 6)

d^2=(3-x)^2 +(6-y)^2.......(1)

now for (x, y) and (-3, 4)

d^2=(-3-x)^2+(4-y)^2 .....(2)

equating eqn 1 and 2 we get

(3-x)^2 + (6-y)^2= (-3-x)^2 +(4-y)^2 =9+x^2- 6x +36+y^2-12y=9+x^2+6x+16+y^2-8y

=-6x+36-12y=6x+16-8y...(after cancellation)

now bringing similiar terms together

-6x-6x-12y+8y= 16-36

-12x-4y=-20

12x+4y=20

3x+y=5

hope it helps ......

Answer 2

Answer:

9y-x-10 = 0

or x-9y+10 = 0

Step-by-step explanation:

distance between two points (a,b) & (c,d)is given as:

$\sqrt{(c - b)^{2} +( d - a)^{2} }$

according to question

point (x,y) is equidistant from (3,6) &(-3,4)

refer attachment...

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