keeping in mind that parallel lines have the same exact slope, hmmm what's the slope of 4x - 2y = 3 anyway? well, let's put it in slope-intercept form firstly.
[tex] \bf 4x-2y=3\implies 4x-3=2y\implies \cfrac{4x-3}{2}=y
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\cfrac{4x}{2}-\cfrac{3}{2}=y\implies \stackrel{slope}{2}x-\cfrac{3}{2}=y [/tex]
a)
so we're looking for a line whose slope is 2 and passes through 2,1.
[tex] \bf (\stackrel{x_1}{2}~,~\stackrel{y_1}{1})~\hspace{10em}
slope = m\implies 2
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\stackrel{\textit{point-slope form}}{y- y_1= m(x- x_1)}\implies y-1=2(x-2)
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y-1=2x-4\implies \blacktriangleright y=2x-3 \blacktriangleleft [/tex]
b)
[tex] \bf \stackrel{\textit{perpendicular lines have \underline{negative reciprocal} slopes}}
{\stackrel{slope}{2\implies \cfrac{2}{1}}\qquad \qquad \qquad \stackrel{reciprocal}{\cfrac{1}{2}}\qquad \stackrel{negative~reciprocal}{-\cfrac{1}{2}}}
\\\\[-0.35em]
\rule{34em}{0.25pt} [/tex]
[tex] \bf (\stackrel{x_1}{2}~,~\stackrel{y_1}{1})~\hspace{10em}
slope = m\implies -\cfrac{1}{2}
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\stackrel{\textit{point-slope form}}{y- y_1= m(x- x_1)}\implies y-1=-\cfrac{1}{2}(x-2)
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y-1=-\cfrac{1}{2}x+1\implies \blacktriangleright y=-\cfrac{1}{2}x+2 \blacktriangleleft[/tex]
