find the value of x for which the distance between the points p(4,-5) and q(12,x) is 10 units
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p(4,-5) and q(12,x)
100=(12-4)^2+(x+5)^2
100=8^2+x^2+5^+2*x*5
100=64+x^2+25+10x
100-89=x^2+10x
11=x^2+10x
x^2+10x-11=0
x^2+11x-1x-11=0
x(x+11)-1(x+11)
(x-1)(x+11)
x=1,-11
THANKS
100=(12-4)^2+(x+5)^2
100=8^2+x^2+5^+2*x*5
100=64+x^2+25+10x
100-89=x^2+10x
11=x^2+10x
x^2+10x-11=0
x^2+11x-1x-11=0
x(x+11)-1(x+11)
(x-1)(x+11)
x=1,-11
THANKS
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