Question 1159868
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First, a solution using elementary methods; you should understand it and be able to solve similar problems by this method.<br>
(1) Put the equation of B in slope-intercept form {{{y = mx+b}}} to find the slope.
{{{x+2y = 3}}}
{{{2y = -x+3}}}
{{{y = (-1/2)x+3/2}}}
The slope is (-1/2).<br>
(2) Find the slope of a line perpendicular to B -- it is the negative reciprocal of the slope of B: +(2/1) = 2<br>
(3) Find the equation of line A knowing it has a slope of 2 and passes through (5,-1).
{{{y = mx+b}}}
{{{-1 = 2(5)+b}}}
{{{-1 = 10+b}}}
{{{b = -1-10 = -11}}}
The equation is {{{y = 2x-11}}}<br>
A graph of the two lines -- B red, A green....<br>
{{{graph(400,400,-12,12,-12,12,(-1/2)x+3/2,2x-11)}}}<br>
Alternatively, a shortcut using ideas about vectors.  This is a quick path to the answer you can use if a formal algebraic solution is not required.<br>
Given line B with equation {{{x+2y=3}}}, every line parallel to B will have an equation of the form {{{x+2y=c}}} for some constant c; and every line perpendicular to B will have an equation of the form {{{2x-y=c}}} for some constant c.<br>
Without understanding vectors, what you need to see is that the coefficients of x and y have switched places, and one of them has changed signs.  In this example, the coefficients 1 and 2 become 2 and -1.<br>
So the equation of line A is {{{2x-y=c}}}; and we can determine the constant c knowing that (x,y)=(5,-1) satisfies the equation.<br>
{{{2x-y=c}}}
{{{2(5)-(-1) = c}}}
{{{11 = c}}}<br>
The equation of A is {{{2x-y = 11}}}<br>
That is a different form of the same equation found earlier, which was {{{y = 2x-11}}}<br>