Question

the line joining A(a, 3) to B(2 -3) is perpendicular to the line joining C(10,1) to B. The value of a is?

Answer 1

Well, first off, let's find what is the slope of BC

$\bf \begin{array}{lllll} &x_1&y_1&x_2&y_2\\ % (a,b) B&({{ 2}}\quad ,&{{ -3}})\quad % (c,d) C&({{ 10}}\quad ,&{{ 1}}) \end{array} \\\\\\ % slope = m slope = {{ m}}= \cfrac{rise}{run} \implies \cfrac{{{ y_2}}-{{ y_1}}}{{{ x_2}}-{{ x_1}}}\implies \cfrac{1-(-3)}{10-2}\implies \cfrac{1+3}{10-2} \\\\\\ \cfrac{4}{8}\implies \cfrac{1}{2}$

now, a line perpendicular to that one, will have a negative reciprocal slope, thus

$\bf \textit{perpendicular, negative-reciprocal slope for slope}\quad \cfrac{1}{2}\\\\ slope=\cfrac{1}{{{ 2}}}\qquad negative\implies -\cfrac{1}{{{ 2}}}\qquad reciprocal\implies - \cfrac{{{ 2}}}{1}\implies \boxed{-2}$

now, we know the slope "m" of AB is -2 then, thus

$\bf \begin{array}{lllll} &x_1&y_1&x_2&y_2\\ % (a,b) A&({{ a}}\quad ,&{{ 3}})\quad % (c,d) B&({{ 2}}\quad ,&{{ -3}}) \end{array} \\\\\\ % slope = m slope = {{ m}}= \cfrac{rise}{run} \implies \cfrac{{{ y_2}}-{{ y_1}}}{{{ x_2}}-{{ x_1}}}\implies \cfrac{-3-3}{2-a}=\boxed{-2} \\\\\\ \cfrac{-6}{2-a}=-2\implies -6=-4+2a\implies -2=2a\implies \cfrac{-2}{2}=a \\\\\\ -1=a$