Write an equation in point-slope form, slope intercept form, and standard form for a line that passes through (-2,8) with slope 8/5 please help : don’t give answers if you aren’t sure

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Step-by-step explanation:

Since you're given a point and slope, it is convenient to start with that form.

Point-slope form

  y -k = m(x -h) . . . . . line with slope m through point (h, k)

  y -8 = 8/5(x +2) . . . point-slope equation

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Slope-intercept form

  y = mx + b . . . . . line with slope m and y-intercept b

The above equation can be rearranged to this form.

  y = 8/5x +16/5 +8

  y = 8/5x +56/5 . . . . . slope-intercept form

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Standard form

  ax +by = c

Multiplying by 5 and subtracting the y-term gives the general form equation ...

  5y = 8x +56

  8x -5y +56 = 0

  8x -5y = -56 . . . . . . add -56 to put into standard form

check the picture below.

well, the angle at the vertex d is clearly not a right-angle, let's check for the one at the vertex f which does look like,   however if it's indeed, that means ef is perpendicular to df, in which case their slopes will be negative reciprocal of each other.

so the negative reciprocal of the slope of ef is 1, however the slope of df is not 1, is 1/2, so nope, they're not perpendicular lines.

the correct answer is the second option