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

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.

Answer 2

Given that,

The point (x,y) is equidistant from the point (3,6) and (-3,4).

To find,

The relation between x and y.

Solution,

here, P = (x,y), A = (3,6) and B = (-3,4)

According to the question,

PA = PB

so, using distance formula

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

Square on both sides,

$\sf {(x - 3)}^{2} + {(y - 6)}^{2} = {(x + 3)}^{2} + {(y - 4)}^{2}$

$\sf {x}^{2} - 2(x)(3) + 9 + {y}^{2} - 2(y)(6) + 36 = {x}^{2} + 2(x)(3) + 9 + {y}^{2} - 2(y)(4) + 16$

$\sf - 6x + 9 - 12y+ 36 =6x + 9 - 8y + 16$

$\sf - 6x - 12y+ 36 =6x - 12y + 16$

$\sf - 6x - 6x - 12y + 8y = 16 - 36$

$\sf - 12x - 4y = - 20$

$\sf - 4(3x + y) = - 20$

$\sf (3x + y) = \dfrac{ - 20}{ - 4}$

$\underline{ \boxed{ \sf 3x + y = 5}}$

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