Write an equation for a line through P. That is parallel to line L. Write another equation for a line through P that is perpendicular to line L. P(2,4) L: 2x+3y=4
Please help! I know I need to find slope and plug it in, but I am lost...
hmmm so.. can you solve "2x+3y=4" for "y"? what would that give you?
Y=-2/3 +4/3
yeap... any ideas on what the slope might be?
Sorry -2/3x+4/3
-2/3
well... so a line parallel to that one, will have the same slope of -2/3 and we also know it passes through P(2,4) so \(\bf \begin{array}{lllll} &x_1&y_1\\ P&({\color{red}{ 2}}\quad ,&{\color{blue}{ 4}})\quad \end{array} \\\quad \\ slope = {\color{green}{ m}}= -\cfrac{2}{3} \\ \quad \\ y-{\color{blue}{ y_1}}={\color{green}{ m}}(x-{\color{red}{ x_1}})\qquad \textit{plug in the values and solve for "y"}\\ \qquad \uparrow\\ \textit{point-slope form} \)
and you'd do, more or less the same for the perpendicular one, with the exception that a perpendicular has a NEGATIVE RECIPROCAL slope to that one, thus \(\bf -\cfrac{2}{{\color{purple}{ 3}}}\qquad negative\to +\cfrac{2}{{\color{purple}{ 3}}}\qquad reciprocal\to \cfrac{{\color{purple}{ 3}}}{2}\) so for the reciprocal we know the slope is 3/2 and passes through P(2,4) thus \(\bf \begin{array}{lllll} &x_1&y_1\\ P&({\color{red}{ 2}}\quad ,&{\color{blue}{ 4}})\quad \end{array} \\\quad \\ slope = {\color{green}{ m}}= \cfrac{3}{2} \\ \quad \\ y-{\color{blue}{ y_1}}={\color{green}{ m}}(x-{\color{red}{ x_1}})\qquad \textit{plug in the values and solve for "y"}\\ \qquad \uparrow\\ \textit{point-slope form}\)
3/2(x-2) +4
for the perpendicular, yes I gather you don't have to solve for "y", unless you need it in slope-intercept form.... which seems you don't have to
What would the simplified version look like? 3/2x+4?
@jdoe0001 can you help me finish this problem really quickly! I'm sorry for keeping you busy
ok \(\bf y-4=\cfrac{3}{2}(x-2)\implies y-4=\cfrac{3}{2}x-3\implies y=\cfrac{3}{2}x+1\)
Thanks a ton!,
yw
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