Question

"Is the line through points P(-8, -10) and Q(-5, -12) perpendicular to the line through points R(9, -6) and S(17, -5)? Explain." Any help would be welcomed. I don't want the answer, I'd like to know how this is solved. Thanks y'all!

Answer 1

find their slopes first \(\bf \begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
P&({\color{red}{ -8}}\quad ,&{\color{blue}{ -10}})\quad Q&({\color{red}{ -5}}\quad ,&{\color{blue}{ -12}})\\
R&({\color{red}{ 9}}\quad ,&{\color{blue}{ -6}})\quad S&({\color{red}{ 17}}\quad ,&{\color{blue}{ -5}})
\end{array}
\\\quad \\

slope = {\color{green}{ m}}= \cfrac{rise}{run} \implies \cfrac{{\color{blue}{ y_2}}-{\color{blue}{ y_1}}}{{\color{red}{ x_2}}-{\color{red}{ x_1}}}
\)

if the multiplication of their slopes gives " -1 ", they're perpendicular

that is... is the slope of one, is the NEGATIVE RECIPROCAL of the other, then they're perpendicular

say for example if one is a/b or \(\bf \cfrac{a}{{\color{blue}{ b}}}\qquad negative\implies -\cfrac{a}{{\color{blue}{ b}}}\qquad reciprocal\implies -\cfrac{{\color{blue}{ b}}}{a}\)

Answer 2

Okay, this is really well written and that part actually makes sense to me now, but I have a slightly dumb question: how exactly would you multiply them? Is it cross multiplication or am I just completely missing something? Thanks for your patience. :)

Answer 3

@jdoe0001

Answer 4

yes... plain fraction multiplication \(\bf \cfrac{a}{b}\cdot \cfrac{c}{d}\implies \cfrac{a\cdot c}{b\cdot d}
\\ \quad \\
thus\implies \cfrac{\cancel{ a }}{{\color{blue}{ \cancel{ b }}}}\cdot -\cfrac{{\color{blue}{ \cancel{ b }}}}{\cancel{ a }}\implies -1\)

Answer 5

Thank you. @jdoe0001

Answer 6

yw