this is from 10th class ssc 7th chapter exercise 7.1 15th one 15 find a relation between x and y such that the point x , y is equidistant from the points (- 2,-8) and (- 3 ,-5)
Answers
Step-by-step explanation:
Given :-
The points (- 2,-8) and (- 3 ,-5)
To find :-
Find a relation between x and y such that the point (x , y) is equidistant from the points (- 2,-8) and (- 3 ,-5) ?
Solution :-
Given points are (-2,-8) and (-3,-5)
Let A = (-2,-8)
Let B = (-3,-5)
Let the required point be P(x,y)
P us the equidistant from the points A and B
A___________P___________B
If P is the equidistant from the points A and B then AP = PB .
Finding AP:-
Let (x1, y1) = (-2,-8) => x1 = -2 and y1 = -8
Let (x2, y2) = (x,y) => x2 = x and y2 = y
We know that
The distance between two points (x1, y1) and
(x2, y2) is √[(x2-x1)²+(y2-y1)²] units
=> AP = √[(x-(-2))²+(y-(-8))²]
=> AP = √[(x+2)²+(y+8)²]
=> AP = √(x²+4x+4+y²+16y+64)
=> AP = √(x²+4x+16y+y²+68) units------(1)
Finding PB :-
Let (x1, y1) = (x,y) => x1 = x and y1 = y
Let (x2, y2) = (-3,-5) => x2 = -3 and y2 = -5
We know that
The distance between two points (x1, y1) and
(x2, y2) is √[(x2-x1)²+(y2-y1)²] units
=> PB = √[(-3+x)²+(-5+y)²]
=> PB = √[(x-3)²+(y-5)²]
=> PB = √(x²-6x+9+y²-10y+25)
=> PB = √(x²-6x-10y+y²+34) units------(2)
We have ,
AP = PB
From (1) & (2)
=>√(x²+4x+16y+y²+68)=√(x²-6x-10y+y²+34)
On squaring both sides then
=>[√(x²+4x+16y+y²+68)]²=[√(x²-6x-10y+y²+34)]²
=> x²+4x+16y+y²+68 = x²-6x-10y+y²+34
=> x²+4x+16y+y²-x²+6x+10y-y² = 34-68
=> 4x+16y+6x+10y = -34
=> 10x+26y = -34
=> 2(5x+13y) = -34
=> 5x+13y = -34/2
=> 5x+13y = -17 or
=> 5x+13y+17 = 0
Answer:-
The relation between x and y for the given problem is 5x+13y+17 = 0
Used formulae:-
The distance between two points (x1, y1) and
(x2, y2) is √[(x2-x1)²+(y2-y1)²] units