What is the equation in slope-intercept form of the line that passes through the point (2, -2) and is perpendicular to the line represented by y = 2/5 + 2?

Answer 1

You may have left out the x in the line's equation. If so, then you meant y= (2/5)x +2

the line perpendicular to that has slope = negative inverse of 2/5 or -5/2, just change the sign and flip the fraction upside down.

then the perpendicular line is y=-(5/2)x + b where b is the y intercept. Plug in the point to solve for b

-2 = -(5/2)(2) + b

-2 = -5 + b

b =5-2 = 3

the perpendicular line is

y=-5x/2 + 3

Answer 2

If given line is y = ²/₅ x + 2 then gradient = ²/₅ (from y = mx + b slope intercept form).

For perpendicular lines the product of their gradients is -1 (or, each is the negative reciprocal of the other).

Hence our line must have a gradient of -⁵/₂ and an equation of y = -⁵/₂ x + b

We can have infinite perpendicular lines with this equation depending on value of b, but we want one line through point (2, -2).

In which case, substitute point values to get -2 = -⁵/₂ (2) + b

=> -2 = -5 + b

=> b = -2 + 5 = 3

The required equation is y = -⁵/₂ x + 3

You can double-check with Desmos Graphing or similar.