Question

(3,6) और (-3,4) से
4
5. सभी खण्ड कीजिए।
(क) x और y में एक ऐसा संबंध ज्ञात कीजिए कि बिन्दु (x, y)
बिन्दु
समदूरस्थ हो।
(ख) यदि 3cos+A4हो तो जाँचिए कि-
1-tan? A
1+tan' A
= cos-A-sin'A है या नहीं।
2​

Answer 1

Given : point (x,y) is equidistant from (3,6) and (-3,4).

To find : relation between x and y

Solution:

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

=> √(x - 3)² + (y - 6)² = √(x -(-3))² + (y - 4)²

=> (x - 3)² + (y - 6)² =  (x + 3)² + (y - 4)²

=> x² -6x + 9+ y² -12y + 36  = x² + 6x + 9  + y² - 8y + 16

=>  -12x - 4y  + 20 = 0

=> -3x - y + 5 = 0

=> 3x + y = 5

Another way :

Slope of (3,6) and (-3,4).

= 1/3

sloe of perpendicular line = - 3

Mid point of  (3,6) and (-3,4). = (0 , 5)

Equation of line = y - 5 = (-3)(x -0)

=> y - 5 = -3x  

=> 3x + y = 5

Learn more:

Find the value of x if p(x 2) is equidistant from the point a(2 -1)and b ...

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The point which is equidistant to the points (9, 3),(7, -1) ,(-1,3)

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Answer 2

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3x + y = 5