asimmon42 asimmon42
  • 07-11-2018
  • Mathematics
contestada

how do you write the equation of a line with a slope of -2/15 that passes through the point ( - 5,5 )

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wegnerkolmp2741o
wegnerkolmp2741o wegnerkolmp2741o
  • 07-11-2018

point slope form

y-y1 = m(x-x1)

y-5 = -2/15(x--5)

y-5 = -2/15 (x+5)

change to slope intercept form

distribute

y-5 =-2/15x -10/15

y-5 = -2/15x -2/3

add 5 to each side

y = -2/15x -2/3 + 5

y = -2/15x -2/3 + 15/3

y = -2/15x +13/3

Answer Link
sammisam8
sammisam8 sammisam8
  • 08-11-2018
a lines equation is generally

mx+b

where m is the slope (-2/15)
and b is the y intercept

point: (-5,5)

y = mx +/- b

5 = -2/15 - 5 (plugging in)

multiply b by the numerator

5 = 10/15

multiply y by the denomenator

75/15 - 10/15

subtract ^

65/15 (4.33 when calculated)
Answer Link

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good job