Question

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

Answer 1

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


`( (8)/(10) , (2)/(10) )` and
`( (5)/(10) , (15)/(10) )`
slope=
`((15)/(10)-(2)/(10))/( (5)/(10)-(8)/(10) ) = ((13)/(10))/( (-3)/(10) ) = (13)/(-3)= (-13)/(3)`

a point is
`( (1)/(2) , (3)/(2) )`

so

`y-(3)/(2)=(-13)/(3)(x-(1)/(2))`

`y-(3)/(2)=(-13)/(3)x+(13)/(6)`
times both sides by 6

`6y-9=-26x+13`

`6y=-26x+22`

`y= (-13)/(3)x+ (11)/(3)` or in standard form

Answer 2

(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 <==