Question

Find the equation of a line which passes through point (-4, 15) and is perpendicular to the line formed by the equation y = 2x + 6. Express it in slope-intercept form.

Answer 1

one method is to rearrange the equation into standard form ax + by = c:

y - 2x = 6 ; now we can ignore the 6 altogether, but swap the coeffs of x and y then change the operator like this:

2y+x = n ; insert our point (-4,15) to adjust for "n"

2(15) -4 = n

30-4 = n

26 = n ; now put it altogether...

2y +x = 26 ; and put it into slope intercept form

y = (-1/2)x + 13 if i see it right

Answer 2

where do you get that equation: 2y+x=n?... i think you mean y-2x=n

Answer 3

it is the perpendicular form of the equation; think of it this way:

ax + by = c1 is standard form; such that slope = -a/b

a perp line has a flipped slope that changes its sign: b/a

the perp equation of the line is then:

bx - ay = c2