Question

find the relation between x and y when x y is equal distance from 36 and - 34​

Answer 1

Step-by-step explanation:

using distance formula,

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

(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}

(x+3)

2

+(y−4)

2

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

(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}

(x−3)

2

+(y−6)

2

=

(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.