keeping in mind that perpendicular lines have <u>negative reciprocal</u> slopes, let's find the slope of 3x + 4y = 9, by simply putting it in slope-intercept form.
![\bf 3x+4y=9\implies 4y=-3x+9\implies y=-\cfrac{3x+9}{4}\implies y=\stackrel{slope}{-\cfrac{3}{4}}x+\cfrac{9}{4} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ \stackrel{\textit{perpendicular lines have \underline{negative reciprocal} slopes}} {\stackrel{slope}{-\cfrac{3}{4}}\qquad \qquad \qquad \stackrel{reciprocal}{-\cfrac{4}{3}}\qquad \stackrel{negative~reciprocal}{+\cfrac{4}{3}}\implies \cfrac{4}{3}}](https://tex.z-dn.net/?f=%20%5Cbf%203x%2B4y%3D9%5Cimplies%204y%3D-3x%2B9%5Cimplies%20y%3D-%5Ccfrac%7B3x%2B9%7D%7B4%7D%5Cimplies%20y%3D%5Cstackrel%7Bslope%7D%7B-%5Ccfrac%7B3%7D%7B4%7D%7Dx%2B%5Ccfrac%7B9%7D%7B4%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20%5Crule%7B34em%7D%7B0.25pt%7D%5C%5C%5C%5C%20%5Cstackrel%7B%5Ctextit%7Bperpendicular%20lines%20have%20%5Cunderline%7Bnegative%20reciprocal%7D%20slopes%7D%7D%20%7B%5Cstackrel%7Bslope%7D%7B-%5Ccfrac%7B3%7D%7B4%7D%7D%5Cqquad%20%5Cqquad%20%5Cqquad%20%5Cstackrel%7Breciprocal%7D%7B-%5Ccfrac%7B4%7D%7B3%7D%7D%5Cqquad%20%5Cstackrel%7Bnegative~reciprocal%7D%7B%2B%5Ccfrac%7B4%7D%7B3%7D%7D%5Cimplies%20%5Ccfrac%7B4%7D%7B3%7D%7D%20)
so we're really looking for the equation of a line whose slope is 4/3 and runs through 8, -4.
Whenever you multiply by a negative number.
Hopefully that helps you a little.
X-4y=-12
4x-y=12
eliminate y's
multily 2nd equaiton by -4 and add to top one
x-4y=-12
<u>-16x+4y=-48 +
</u>-15x+0y=-60
-15x=-60
divide both sides by -15
x=4
sub back
x-4y=-12
4-4y=-12
minus 4 both sides
-4y=-16
divide both sides by -4
y=4
x=4
y=4
(x,y)
(4,4)<u>
</u>
(60mi / 1hr) * 12hr =
(12hr * 60mi) / 1hr because the hr and mi are being multiplied we can cancel the hrs =
12 * 60mi = 720mi
distance = speed * time
speed = distance / time
time = distance / speed
the equation of the line is
y = 6/5xy + 7/5x - 2/5
