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What is the value of k if point P(k, k sqrt(3)) is 4 units from the line 5x+8y=17?

 Jun 7, 2021
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5x+8y-17=0

Formula for distance between point (p, q) and line ax+by+c=0 is:  

distance=\(\frac{\left|ap+bq+c\right|}{\sqrt{a^2+b^2}}\) 

since in this case, p=k and q=\(k\sqrt{3}\) , a=5, b=8, and c=-17, we can substitute:

\(\frac{\left|ap+bq+c\right|}{\sqrt{a^2+b^2}}=\frac{\left|5k+8k\sqrt{3}-17\right|}{\sqrt{5^2+8^2}}=\frac{\left|5k+8k\sqrt{3}-17\right|}{\sqrt{89}}\)

Since the distance is 4, \(\frac{\left|5k+8k\sqrt{3}-17\right|}{\sqrt{89}}=4\)

                                      solving this, we get \(k=-\frac{\left(-4\sqrt{89}+17\right)\left(5-8\sqrt{3}\right)}{167}\) or \(\:k=-\frac{\left(4\sqrt{89}+17\right)\left(5-8\sqrt{3}\right)}{167}\)

                                      tell me if I'm wrong, because this is a pretty complicated solution. 

 

JP

 Jun 7, 2021

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