What is the value of k if point P(k, k sqrt(3)) is 4 units from the line 5x+8y=17?
+208 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
