How do you find the possible values for a if the points (6,a), (5,0) has a distance of $sqrt17$ ?

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Answer 1 · 2017-04-03T12:07:57

$a=-4$ or $a=4$

The distance between two points $(x_1,y_1)$ and $(x_2,y_2)$ is

$sqrt((x_2-x_1)^2+(y_2-y_1)^2)$

Hence, distance between two points $(6,a)$ and $(5,0)$ is

$sqrt((5-6)^2+(0-a)^2)$ and this should be $sqrt17$

Therefore $sqrt((5-6)^2+(0-a)^2)=sqrt17$

or $(5-6)^2+(0-a)^2=17$

i.e. $1+a^2=17$

or $a^2-16=0$

or $(a+4)(a-4)=0$

i.e. $a=-4$ or $a=4$

How do you find the possible values for a if the points (6,a), (5,0) has a distance of sqrt17sqrt17?