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# help coordinates

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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