Question

which of the fallowing functions has a slope 3/2 and contains the midpoint segment between (6, 3) and (-2, 11)?

Answer 1

Well, we know the slope is 3/2, what's the midpoint of those anyway?

$\bf \textit{middle point of 2 points }\\ \quad \\ \begin{array}{lllll} &x_1&y_1&x_2&y_2\\ % (a,b) &({{ 6}}\quad ,&{{ 3}})\quad % (c,d) &({{ -2}}\quad ,&{{ 11}}) \end{array}\qquad % coordinates of midpoint \left(\cfrac{{{ x_2}} + {{ x_1}}}{2}\quad ,\quad \cfrac{{{ y_2}} + {{ y_1}}}{2} \right) \\\\\\ \left( \cfrac{-2+6}{2}~~,~~\cfrac{11+3}{2} \right)\implies (2,7)$

so, what's the equation of a line whose slope is 3/2 and runs through 2,7?

$\bf \begin{array}{lllll} &x_1&y_1\\ % (a,b) &({{ 2}}\quad ,&{{ 7}}) \end{array} \\\\\\ % slope = m slope = {{ m}}= \cfrac{rise}{run} \implies \cfrac{3}{2} \\\\\\ % point-slope intercept \stackrel{\textit{point-slope form}}{y-{{ y_1}}={{ m}}(x-{{ x_1}})}\implies y-7=\cfrac{3}{2}(x-2) \\\\\\ y-7=\cfrac{3}{2}x-3\implies y=\cfrac{3}{2}x+4$