How could you wirte a slope-intercept equation for a line passing through points (-1,6) f(x) = 2x =9.?

Question

How could you wirte a slope-intercept equation for a line passing through points (-1,6) f(x) = 2x =9.
Also what is an equation perpendicular to that given line?

Answer 1

(-1,6)
f(x) = 2x =9.

x=9/2
y=9

slope=(9-6)/(9/2+1)
=6/11

y-6=6/11(x+1)

and perpendicular line
y-6= - 11/6 (x+1)

Answer 2

2x ? 9y = ?12 9y = 2x + 12 y = (2/9)x + 12/9 comparing with slope intercept style y = mx + c, we come across slope = m = 2/9 Any line perpendicular to this equation has a slope = ?a million/m = ?9/2 A line passing through (x1, y1) and slope M has equation, y - y1 = M(x - x1) Our line passes through, (?2, 3) and has slope ?9/2 y ? 3 = (?9/2)(x ? (?2) y ? 3 = (?9/2)(x + 2) 2y ? 6 = ?9x ? 18 9x + 2y = ?12

Answer 3

the given line 2x=9 can be written can be written as 2x-9=0 which can also written as 2x-0y-9=0 0y=2x-9 this is in the form of y=mx+c here slope is equal to m there fore slope is equal to 2 required line is
perpendicular to the given line . we know that the if two lines are perpendicular there product of their slopes is equal to -1
m1 *m2 = -1
slope of given line m1 is2 substituting
2 * m2 = -1 there fore m2 = -1/2 the equation of line passes through to the point (-1,6) point slope form
(y-y1) = m (x-x1) here x1= -1 y1 = 6 substituting we get
y+6 = -1/2 (x-(-1))
multiplying 2 on both sides we have
2(y+6) = 2 * -1/2 (x+1)
2y+12 = -1(x+1)
2y+12 = -x-1
x+2y+12+1=0
x+2y+13=0 or
x+2y = -13 are the required equations