question no. 6 ...solution karo bhai ...plzz
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loverayush:
1 u have to apply distance formula to find distance of A to B and then A to C
then u can equate both the eqns
after equating firstly sq both sides and then proceed
u will get the value of x
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If A is equidistant from BandC then A is the mid point of B and C
(X, 2)= (x, y)
( 8,-2)=(x1,y1)
(2,-2)=(x2y2)
Apply the midpoint formulae
X= x1+x2/2
X= 8+2/2 = 5
So, the co-ordinates of Aare (5,2)
Now let A(5,2) (x1y1) and (8,-2)(x2y2)
Distance between AB is = √(x2-x1)^2 + (√y2-y1)^2
√ 3^2 -4^2 = √9-√16 = 3-4 = -1
(X, 2)= (x, y)
( 8,-2)=(x1,y1)
(2,-2)=(x2y2)
Apply the midpoint formulae
X= x1+x2/2
X= 8+2/2 = 5
So, the co-ordinates of Aare (5,2)
Now let A(5,2) (x1y1) and (8,-2)(x2y2)
Distance between AB is = √(x2-x1)^2 + (√y2-y1)^2
√ 3^2 -4^2 = √9-√16 = 3-4 = -1
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