Question

Find a relation between x and y such that the point (x, y) is equidistant from the point (3,6) and (-3,4).​

Answer 1

By Using distance formula,

d =√(x2 -x1)2 + (y2 – y1)2

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

distance between P and R =√(x + 3)2+ (y-4)2

Similarly, distance between P and Q =√(x- 3)2+ (y – 6)2

(x, y) is the equidistant from the points (3,6) and (-3,4) i.e.,PQ = PR

√(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

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

${ \tt{ \large{ \color{palegreen}{hope \: it \: helps}}}}$

Answer 2

Step-by-step explanation:

By Using distance formula,

d =√(x2 -x1)2 + (y2 – y1)2

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

distance between P and R =√(x + 3)2+ (y-4)2

Similarly, distance between P and Q =√(x- 3)2+ (y – 6)2

(x, y) is the equidistant from the points (3,6) and (-3,4) i.e.,PQ = PR

√(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

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

${ \tt{ \large{ \color{palegreen}{hope \: it \: helps}}}}$