# Question #ba70b

Nov 24, 2014

Problem 1

Let $\left\{\begin{matrix}\left({x}_{1} {y}_{1}\right) = \left(3 5\right) \\ \left({x}_{2} {y}_{2}\right) = \left(- 9 5\right)\end{matrix}\right.$

By Slope Formula,

$m = \frac{{y}_{2} - {y}_{1}}{{x}_{2} - {x}_{1}} = \frac{5 - 5}{- 9 - 3} = 0$

By Point-Slope Form $y - {y}_{1} = m \left(x - {x}_{1}\right)$, we have the equation

$y - 5 = 0 \left(x - 3\right) = 0 \implies y = 5$

Problem 2

Let $\left({x}_{1} , {y}_{1}\right) = \left(2 , {\left(2\right)}^{3} - 10\right) = \left(2 , - 2\right)$

By rewriting the equation in Slope-Intercept Form,

$3 x - 2 y + 5 = 0 \implies y = \frac{3}{2} x + \frac{5}{2} \implies$ Slope $= \frac{3}{2}$

By taking the negative reciprocal (since perpendcular),

$m = - \frac{2}{3}$

By Point-Slope Form, we have the equation

$y + 2 = - \frac{2}{3} \left(x - 2\right)$

Problem 3

By setting $y = 0$, we can find the $x$-intercept,

$0 = 5 x - 75 \implies x = 15 \implies$ Let $\left({x}_{1} , {y}_{1}\right) = \left(15 , 0\right)$

By rewriting in Slope-Intercept Form,

$7 y - 5 x = 70 \implies y = \frac{5}{7} x + 10 \implies m = \frac{5}{7}$

By Point-Slope Form, we have the equation

$y = \frac{5}{7} \left(x - 15\right)$

I hope that this was helpful.