Question

Question 10 Find a relation between x and y such that the point (x, y) is equidistant from the point (3, 6) and (− 3, 4).

Class 10 - Math - Coordinate Geometry Page 162

Answer 1

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.

Answer 2

Answer:

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