if the distance of point p(x,y)from the points A(3,6) & B(-3,4)are equal,prove that 3x+y=5
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Answered by
1
A(3,6) & B(-3,4)
P(x,y)
![x = \frac{(3) + ( - 3)}{2} \\ x = 0 \\ y = \frac{6 + 4}{2} \\ y = \frac{10}{2} = 5 x = \frac{(3) + ( - 3)}{2} \\ x = 0 \\ y = \frac{6 + 4}{2} \\ y = \frac{10}{2} = 5](https://tex.z-dn.net/?f=x+%3D++%5Cfrac%7B%283%29+%2B+%28+-+3%29%7D%7B2%7D++%5C%5C+x+%3D+0+%5C%5C+y+%3D++%5Cfrac%7B6+%2B+4%7D%7B2%7D++%5C%5C+y+%3D++%5Cfrac%7B10%7D%7B2%7D++%3D+5)
x=0
y=5
![3x + y = 5 \\ 3(0) + 5 = 5 \\ 5 = 5 3x + y = 5 \\ 3(0) + 5 = 5 \\ 5 = 5](https://tex.z-dn.net/?f=3x+%2B+y+%3D+5+%5C%5C+3%280%29+%2B+5+%3D+5+%5C%5C+5+%3D+5)
LHS=RHS
Proved 3x+y=5 (^_^)
P(x,y)
x=0
y=5
LHS=RHS
Proved 3x+y=5 (^_^)
Answered by
0
here it's proved...this will be fine method I think..
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