How do you write an equation for the line with slope -2/3 that passes through the point (0,4)?

1 Answer
Meave60
Oct 30, 2015

Refer to the explanation.

Explanation:

Use the point-slope form for a linear equation. The general equation is #y-y_1=m(x-x_1)#, where #x_1# and #y_1# are the known point, and,#m# is the slope.

#"Known point"=(0,4)#
#x_1=0#
#y_1=4#
#m=-2/3#

Substitute the known values and solve for #y#.

#y-y_1=m(x-x_1)#

#y-4=-2/3(x-0)=#

Add #4# to both sides.

#y=-2/3(x-0)+4#

Distribute #-2/3#.

#y=-2/3*x-(-2/3*0)+4#

Simplify.

#y=-2/3x+0+4=#

#y=-2/3x+4#

The equation is now in point-intercept form #y=mx+b#, where #m# is the slope and #b# is the y-intercept.

graph{y=-2/3x+4 [-10.15, 9.85, -2.25, 7.75]}

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