Short answer question :)

Mathematics

1 answer:

alisha [4.7K]3 years ago

3 0

No because the lines are not connecting. The definition of a function is a relationship involving one or more variables. This can't be a function because their is only one variable.

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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?

RideAnS [48]

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

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

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