Question 455112
find the equation of the line that passing through (-2,4) and having the
(a) slope of 7/16.
Use point/slope form: y - y1 = m(x - x1)
y - 4 = {{{7/16}}}(x - (-2))
y - 4 = {{{7/16}}}(x + 2)
y - 4 = {{{7/16}}}x + {{{14/16}}}
y = {{{7/16}}}x + {{{7/8}}} + 4
y = {{{7/16}}}x + {{{7/8}}} + {{{32/8}}}
y = {{{7/16}}}x + {{{39/8}}}
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(b) parallel to the 5x - 3y = 3
Find the slope, put in the point/intercept form
5x - 3y = 3
-3y = -5x + 3
y had to be positive, mult both sides by -1
3y = 5x - 3
Divide both sides by 3
y = {{{5/3}}}x + {{{3/3}}}
y = {{{5/3}}}x + 1
slope = {{{5/3}}}, parallel lines have the same slopes, us the point slope form
y - 4 = {{{5/3}}}(x - (-2))
y - 4 = {{{5/3}}}(x + 2)
y - 4 = {{{5/3}}}x + {{{10/3}}}
y = {{{5/3}}}x + {{{10/3}}} + 4
y = {{{5/3}}}x + {{{10/3}}} + {{{12/3}}}
y = {{{5/3}}}x + {{{22/3}}}
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(c) passing through the origin
slope = {{{4/(-2)}}} = -2
y = -2x is the equation passes thru 0,0 and -2,4
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(d) parallel to the y-axis.
x = -2