Respuesta :
well the line that bisects RS, will cut RS in two equal halves, therefore, that line will cut RS perpendicularly at the midpoint of RS.
now, what the dickens is the midpoint of RS anyway?
[tex]\bf \textit{middle point of 2 points }\\ \quad \\ \begin{array}{lllll} &x_1&y_1&x_2&y_2\\ % (a,b) R&({{ -1}}\quad ,&{{ 6}})\quad % (c,d) S&({{ 5}}\quad ,&{{ 5}}) \end{array}\qquad % coordinates of midpoint \left(\cfrac{{{ x_2}} + {{ x_1}}}{2}\quad ,\quad \cfrac{{{ y_2}} + {{ y_1}}}{2} \right) \\\\\\ \left( \cfrac{5-1}{2}~~,~~\cfrac{5+6}{2} \right)\implies \stackrel{midpoint}{\left(2~~,~~\frac{11}{2} \right)}[/tex]
so, we know that perpendicular line, will have to go through (2, 11/2)
now, a perpendicular line to RS, will have a negative reciprocal slope to it. Well, what is the slope of RS anyway?
[tex]\bf \begin{array}{lllll} &x_1&y_1&x_2&y_2\\ % (a,b) &({{ -1}}\quad ,&{{ 6}})\quad % (c,d) &({{ 5}}\quad ,&{{ 5}}) \end{array} \\\\\\ % slope = m slope = {{ m}}= \cfrac{rise}{run} \implies \cfrac{{{ y_2}}-{{ y_1}}}{{{ x_2}}-{{ x_1}}}\implies \cfrac{5-6}{5-(-1)}\implies \cfrac{5-6}{5+1}\implies -\cfrac{1}{6}[/tex]
and let's check the reciprocal negative of that,
[tex]\bf \textit{perpendicular, negative-reciprocal slope for slope}\quad -\cfrac{1}{6}\\\\ slope=-\cfrac{1}{{{ 6}}}\qquad negative\implies +\cfrac{1}{{{ 6}}}\qquad reciprocal\implies + \cfrac{{{ 6}}}{1}\implies 6[/tex]
so, then, what's is the equation of a line whose slope is 6, and goes through 2, 11/2?
[tex]\bf \begin{array}{lllll} &x_1&y_1\\ % (a,b) &({{ 2}}\quad ,&{{ \frac{11}{2}}}) \end{array} \\\\\\ % slope = m slope = {{ m}}= \cfrac{rise}{run} \implies 6 \\\\\\ % point-slope intercept \stackrel{\textit{point-slope form}}{y-{{ y_1}}={{ m}}(x-{{ x_1}})}\implies y-\cfrac{11}{2}=6(x-2) \\\\\\ y-\cfrac{11}{2}=6x-12\implies -6x+y=-12+\cfrac{11}{2}\implies \stackrel{\textit{standard form}}{-6x+y=-\cfrac{13}{2}}[/tex]
now, what the dickens is the midpoint of RS anyway?
[tex]\bf \textit{middle point of 2 points }\\ \quad \\ \begin{array}{lllll} &x_1&y_1&x_2&y_2\\ % (a,b) R&({{ -1}}\quad ,&{{ 6}})\quad % (c,d) S&({{ 5}}\quad ,&{{ 5}}) \end{array}\qquad % coordinates of midpoint \left(\cfrac{{{ x_2}} + {{ x_1}}}{2}\quad ,\quad \cfrac{{{ y_2}} + {{ y_1}}}{2} \right) \\\\\\ \left( \cfrac{5-1}{2}~~,~~\cfrac{5+6}{2} \right)\implies \stackrel{midpoint}{\left(2~~,~~\frac{11}{2} \right)}[/tex]
so, we know that perpendicular line, will have to go through (2, 11/2)
now, a perpendicular line to RS, will have a negative reciprocal slope to it. Well, what is the slope of RS anyway?
[tex]\bf \begin{array}{lllll} &x_1&y_1&x_2&y_2\\ % (a,b) &({{ -1}}\quad ,&{{ 6}})\quad % (c,d) &({{ 5}}\quad ,&{{ 5}}) \end{array} \\\\\\ % slope = m slope = {{ m}}= \cfrac{rise}{run} \implies \cfrac{{{ y_2}}-{{ y_1}}}{{{ x_2}}-{{ x_1}}}\implies \cfrac{5-6}{5-(-1)}\implies \cfrac{5-6}{5+1}\implies -\cfrac{1}{6}[/tex]
and let's check the reciprocal negative of that,
[tex]\bf \textit{perpendicular, negative-reciprocal slope for slope}\quad -\cfrac{1}{6}\\\\ slope=-\cfrac{1}{{{ 6}}}\qquad negative\implies +\cfrac{1}{{{ 6}}}\qquad reciprocal\implies + \cfrac{{{ 6}}}{1}\implies 6[/tex]
so, then, what's is the equation of a line whose slope is 6, and goes through 2, 11/2?
[tex]\bf \begin{array}{lllll} &x_1&y_1\\ % (a,b) &({{ 2}}\quad ,&{{ \frac{11}{2}}}) \end{array} \\\\\\ % slope = m slope = {{ m}}= \cfrac{rise}{run} \implies 6 \\\\\\ % point-slope intercept \stackrel{\textit{point-slope form}}{y-{{ y_1}}={{ m}}(x-{{ x_1}})}\implies y-\cfrac{11}{2}=6(x-2) \\\\\\ y-\cfrac{11}{2}=6x-12\implies -6x+y=-12+\cfrac{11}{2}\implies \stackrel{\textit{standard form}}{-6x+y=-\cfrac{13}{2}}[/tex]
