The distance between the points (3,2) and (0,x) is 5 .Find x
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Here,
let (3,2)be =(x₁,y₁)
(0,x)be=(x₂,y₂)
Distance between them (d)=√(x₂-x₁)²+(y₂-y₁)²
or, 5 =√(0-3)²+(x-2)²
or, 5 =√(-3)²+x²-4x+4
or, 5 =√9+x²-4x+4
or, 5 =√x²-4x+13
or, 5²=x²-4x+13
or,25=x²-4x+13
or,x²-4x-12=0
or,x²+2x-6x-12=0
or,x(x+2)-6(x+2)=0
∵(x+2)(x-6)=0
Now,
Either x+2=0 Or x-6=0
∵x= -2 ∵x=6
So,x= -2 or 6
let (3,2)be =(x₁,y₁)
(0,x)be=(x₂,y₂)
Distance between them (d)=√(x₂-x₁)²+(y₂-y₁)²
or, 5 =√(0-3)²+(x-2)²
or, 5 =√(-3)²+x²-4x+4
or, 5 =√9+x²-4x+4
or, 5 =√x²-4x+13
or, 5²=x²-4x+13
or,25=x²-4x+13
or,x²-4x-12=0
or,x²+2x-6x-12=0
or,x(x+2)-6(x+2)=0
∵(x+2)(x-6)=0
Now,
Either x+2=0 Or x-6=0
∵x= -2 ∵x=6
So,x= -2 or 6
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