Question

10. Find a relation between x and y such that the point (x,y) is equidistant from the point
(3.0) and (-3,4)

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help me to do this​

Answer 1

Since they are equidistant

Therefore,

√(x-3)^2+(y-0)^2 = √(x-(-3))^2+(y-4)^2

Squaring both sides,

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

x^2 - 6x +9 +y^2= x^2 +6x +9+ y^2 -8y+16

-6x-6x= -8y+16 ( all others will cancel out)

8y -12x= 16

Dividing by 4,

2y-3x=4

Answer 2

Step-by-step explanation:

we have to find the relation between x and y such that the point (x, y) is the equidistant from the points (3, 6) and (-3,4).

solution : using distance formula,

d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}

let P = (x, y) , Q = (3, 6) and R = (-3, 4).

distance between P and R = \sqrt{(x+3)^2+(y-4)^2}

similarly, distance between P and Q = \sqrt{(x-3)^2+(y-6)^2}

a/c to question, (x, y) is the equidistant from the points (3,6) and (-3,4)

i.e.,PQ = PR

⇒\sqrt{(x-3)^2+(y-6)^2}=\sqrt{(x+3)^2+(y-4)^2}

squaring both sides we get,

⇒(x - 3)² + (y - 6)² = (x + 3)² + (y - 4)²

⇒x² + 9 - 6x + y² + 36 - 12y = x² + 9 + 6x + y² + 16 - 8y

⇒-6x + 45 - 12y = 6x - 8y + 25

⇒-6x - 6x - 12y + 8y = 25- 45

⇒-12x - 4y = -20

⇒3x + y = 5

Therefore the relation between x and y is 3x + y = 5.