Respuesta :
hmmm so hmmm we know AB has the points A(1,2) and B(7,4).
alrite... so the bisector passes through its midpoint... what's the midpoint of AB anyway?
[tex]\bf \textit{middle point of 2 points }\\ \quad \\ \begin{array}{lllll} &x_1&y_1&x_2&y_2\\ % (a,b) A&({{ 1}}\quad ,&{{ 2}})\quad % (c,d) B&({{ 7}}\quad ,&{{ 4}}) \end{array}\qquad % coordinates of midpoint \left(\cfrac{{{ x_2}} + {{ x_1}}}{2}\quad ,\quad \cfrac{{{ y_2}} + {{ y_1}}}{2} \right) \\\\\\ \left(\cfrac{7+1}{2}~~,~~\cfrac{4+2}{2} \right)\implies \left( \cfrac{8}{2}~~,~~\cfrac{6}{2} \right)\implies (4,3)[/tex]
so it goes through 4,3, is a segment perpendicular bisector, that means the slope of it will be the negative reciprocal slope of AB, hmmmmm what's the slope of AB anyway?
[tex]\bf \begin{array}{lllll} &x_1&y_1&x_2&y_2\\ % (a,b) &({{ 1}}\quad ,&{{ 2}})\quad % (c,d) &({{ 7}}\quad ,&{{ 4}}) \end{array} \\\\\\ % slope = m slope = {{ m}}= \cfrac{rise}{run} \implies \cfrac{{{ y_2}}-{{ y_1}}}{{{ x_2}}-{{ x_1}}}\implies \cfrac{4-2}{7-1}\implies \cfrac{2}{6}\implies \cfrac{1}{3}[/tex]
alrite.... so what is the negative reciprocal of that anway?
[tex]\bf \textit{perpendicular, negative-reciprocal slope for slope}\quad \cfrac{1}{3}\\\\ slope=\cfrac{1}{{{3}}}\qquad negative\implies -\cfrac{1}{{{ 3}}}\qquad reciprocal\implies - \cfrac{{{ 3}}}{1}\implies -3[/tex]
ahha, so, we're really looking for the equation of a line that has a slope of -3, and runs through 4,3 then,
[tex]\bf \begin{array}{lllll} &x_1&y_1\\ % (a,b) &({{ 4}}\quad ,&{{ 3}}) \end{array} \\\\\\ % slope = m slope = {{ m}}= \cfrac{rise}{run} \implies -3 \\\\\\ % point-slope intercept \stackrel{\textit{point-slope form}}{y-{{ y_1}}={{ m}}(x-{{ x_1}})}\implies y-3=-3(x-4) \\\\\\ y-3=-3x+12\implies y=-3x+15[/tex]
alrite... so the bisector passes through its midpoint... what's the midpoint of AB anyway?
[tex]\bf \textit{middle point of 2 points }\\ \quad \\ \begin{array}{lllll} &x_1&y_1&x_2&y_2\\ % (a,b) A&({{ 1}}\quad ,&{{ 2}})\quad % (c,d) B&({{ 7}}\quad ,&{{ 4}}) \end{array}\qquad % coordinates of midpoint \left(\cfrac{{{ x_2}} + {{ x_1}}}{2}\quad ,\quad \cfrac{{{ y_2}} + {{ y_1}}}{2} \right) \\\\\\ \left(\cfrac{7+1}{2}~~,~~\cfrac{4+2}{2} \right)\implies \left( \cfrac{8}{2}~~,~~\cfrac{6}{2} \right)\implies (4,3)[/tex]
so it goes through 4,3, is a segment perpendicular bisector, that means the slope of it will be the negative reciprocal slope of AB, hmmmmm what's the slope of AB anyway?
[tex]\bf \begin{array}{lllll} &x_1&y_1&x_2&y_2\\ % (a,b) &({{ 1}}\quad ,&{{ 2}})\quad % (c,d) &({{ 7}}\quad ,&{{ 4}}) \end{array} \\\\\\ % slope = m slope = {{ m}}= \cfrac{rise}{run} \implies \cfrac{{{ y_2}}-{{ y_1}}}{{{ x_2}}-{{ x_1}}}\implies \cfrac{4-2}{7-1}\implies \cfrac{2}{6}\implies \cfrac{1}{3}[/tex]
alrite.... so what is the negative reciprocal of that anway?
[tex]\bf \textit{perpendicular, negative-reciprocal slope for slope}\quad \cfrac{1}{3}\\\\ slope=\cfrac{1}{{{3}}}\qquad negative\implies -\cfrac{1}{{{ 3}}}\qquad reciprocal\implies - \cfrac{{{ 3}}}{1}\implies -3[/tex]
ahha, so, we're really looking for the equation of a line that has a slope of -3, and runs through 4,3 then,
[tex]\bf \begin{array}{lllll} &x_1&y_1\\ % (a,b) &({{ 4}}\quad ,&{{ 3}}) \end{array} \\\\\\ % slope = m slope = {{ m}}= \cfrac{rise}{run} \implies -3 \\\\\\ % point-slope intercept \stackrel{\textit{point-slope form}}{y-{{ y_1}}={{ m}}(x-{{ x_1}})}\implies y-3=-3(x-4) \\\\\\ y-3=-3x+12\implies y=-3x+15[/tex]
