Respuesta :
well, first off, let's find what is the slope of BC
[tex]\bf \begin{array}{lllll} &x_1&y_1&x_2&y_2\\ % (a,b) B&({{ 2}}\quad ,&{{ -3}})\quad % (c,d) C&({{ 10}}\quad ,&{{ 1}}) \end{array} \\\\\\ % slope = m slope = {{ m}}= \cfrac{rise}{run} \implies \cfrac{{{ y_2}}-{{ y_1}}}{{{ x_2}}-{{ x_1}}}\implies \cfrac{1-(-3)}{10-2}\implies \cfrac{1+3}{10-2} \\\\\\ \cfrac{4}{8}\implies \cfrac{1}{2}[/tex]
now, a line perpendicular to that one, will have a negative reciprocal slope, thus
[tex]\bf \textit{perpendicular, negative-reciprocal slope for slope}\quad \cfrac{1}{2}\\\\ slope=\cfrac{1}{{{ 2}}}\qquad negative\implies -\cfrac{1}{{{ 2}}}\qquad reciprocal\implies - \cfrac{{{ 2}}}{1}\implies \boxed{-2}[/tex]
now, we know the slope "m" of AB is -2 then, thus
[tex]\bf \begin{array}{lllll} &x_1&y_1&x_2&y_2\\ % (a,b) A&({{ a}}\quad ,&{{ 3}})\quad % (c,d) B&({{ 2}}\quad ,&{{ -3}}) \end{array} \\\\\\ % slope = m slope = {{ m}}= \cfrac{rise}{run} \implies \cfrac{{{ y_2}}-{{ y_1}}}{{{ x_2}}-{{ x_1}}}\implies \cfrac{-3-3}{2-a}=\boxed{-2} \\\\\\ \cfrac{-6}{2-a}=-2\implies -6=-4+2a\implies -2=2a\implies \cfrac{-2}{2}=a \\\\\\ -1=a[/tex]
[tex]\bf \begin{array}{lllll} &x_1&y_1&x_2&y_2\\ % (a,b) B&({{ 2}}\quad ,&{{ -3}})\quad % (c,d) C&({{ 10}}\quad ,&{{ 1}}) \end{array} \\\\\\ % slope = m slope = {{ m}}= \cfrac{rise}{run} \implies \cfrac{{{ y_2}}-{{ y_1}}}{{{ x_2}}-{{ x_1}}}\implies \cfrac{1-(-3)}{10-2}\implies \cfrac{1+3}{10-2} \\\\\\ \cfrac{4}{8}\implies \cfrac{1}{2}[/tex]
now, a line perpendicular to that one, will have a negative reciprocal slope, thus
[tex]\bf \textit{perpendicular, negative-reciprocal slope for slope}\quad \cfrac{1}{2}\\\\ slope=\cfrac{1}{{{ 2}}}\qquad negative\implies -\cfrac{1}{{{ 2}}}\qquad reciprocal\implies - \cfrac{{{ 2}}}{1}\implies \boxed{-2}[/tex]
now, we know the slope "m" of AB is -2 then, thus
[tex]\bf \begin{array}{lllll} &x_1&y_1&x_2&y_2\\ % (a,b) A&({{ a}}\quad ,&{{ 3}})\quad % (c,d) B&({{ 2}}\quad ,&{{ -3}}) \end{array} \\\\\\ % slope = m slope = {{ m}}= \cfrac{rise}{run} \implies \cfrac{{{ y_2}}-{{ y_1}}}{{{ x_2}}-{{ x_1}}}\implies \cfrac{-3-3}{2-a}=\boxed{-2} \\\\\\ \cfrac{-6}{2-a}=-2\implies -6=-4+2a\implies -2=2a\implies \cfrac{-2}{2}=a \\\\\\ -1=a[/tex]
The value of a is -1
What do perpendicular lines have a slope of?
Perpendicular lines have slopes that are negative reciprocals of each other. The given line's slope is 5, which means that the slope of the opposite line must be its negative reciprocal.
Conclusion: The slopes of two perpendicular lines are negative reciprocals of each other. This means that if a line is perpendicular to a line that is the slope of the line is -1/m.
let the slope of line AB is m1 and the slope of line BC be m2
m1= 3+3/a-2 = 6/a-2
m2=1+2/10-2 = 1/2
m1 * m2 = -1
6/a-2 * 1/2 = -1
3 = -a+2
a = -1
Learn more about reciprocal here
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