Question

Find a relation between N and y such that the point in Sy is equidistant from the points (3.6 )and (-3, 4).​

Answer 1

Answer:

Here it is!

Step-by-step explanation:

using distance formula,

d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}d=(x2−x1)2+(y2−y1)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.

Answer 2

Answer:

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.