What is the equation of the line that passes through the points ?

(4/5,1/5)(1/2,3/2)
slope = (3/2 - 1/5) / (1/2 - 4/5) = (15/10 - 2/10) / (5/10 - 8/10) = (13/10) / (- 3/10) = 13/10 * - 10/3 = - 130/30 reduces to -13/3

y = mx + b
slope(m) = -13/3
use either of ur points (4/5,1/5)....x = 4/5 and y = 1/5
now we sub and find b, the y int
1/5 = -13/3(4/5) + b
1/5 = - 52/15 + b
1/5 + 52/15 = b
3/15 + 52/15 = b
55/15 (reduces to 11/3) = b

so ur equation is : y = -13/3x + 11/3 <==

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apologiabiology apologiabiology · Answer 2 · 2017-02-16T23:59:49+01:00

y-y1=m(x-x1)
that is the equation of a line that passes through the point (x1,y1) and has a slope of m

slope between (x1,y1) and (x2,y2) is (y2-y1)/(x2-x1)

so first convert them to common denomenator which is 10

[tex]( \frac{8}{10} , \frac{2}{10} )[/tex] and [tex]( \frac{5}{10} , \frac{15}{10} )[/tex]
slope=[tex] \frac{\frac{15}{10}-\frac{2}{10}}{ \frac{5}{10}-\frac{8}{10} } = \frac{\frac{13}{10}}{ \frac{-3}{10} } = \frac{13}{-3}= \frac{-13}{3} [/tex]

a point is [tex]( \frac{1}{2} , \frac{3}{2} )[/tex]

so
[tex]y-\frac{3}{2}=\frac{-13}{3}(x-\frac{1}{2})[/tex]
[tex]y-\frac{3}{2}=\frac{-13}{3}x+\frac{13}{6}[/tex]
times both sides by 6
[tex]6y-9=-26x+13[/tex]
[tex]6y=-26x+22[/tex]
[tex]y= \frac{-13}{3}x+ \frac{11}{3} [/tex] or in standard form