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Algebra Examples
y=-0.75xy=−0.75x (8,0)(8,0)
Step 1
Step 1.1
The slope-intercept form is y=mx+by=mx+b, where mm is the slope and bb is the y-intercept.
y=mx+by=mx+b
Step 1.2
Using the slope-intercept form, the slope is -0.75−0.75.
m=-0.75m=−0.75
m=-0.75m=−0.75
Step 2
The equation of a perpendicular line must have a slope that is the negative reciprocal of the original slope.
mperpendicular=-1-0.75mperpendicular=−1−0.75
Step 3
Step 3.1
Divide 11 by -0.75−0.75.
mperpendicular=1.‾3mperpendicular=1.¯3
Step 3.2
Multiply -1−1 by -1.‾3−1.¯3.
mperpendicular=1.‾3mperpendicular=1.¯3
mperpendicular=1.‾3mperpendicular=1.¯3
Step 4
Step 4.1
Use the slope 1.‾31.¯3 and a given point (8,0)(8,0) to substitute for x1x1 and y1y1 in the point-slope form y-y1=m(x-x1)y−y1=m(x−x1), which is derived from the slope equation m=y2-y1x2-x1m=y2−y1x2−x1.
y-(0)=1.‾3⋅(x-(8))y−(0)=1.¯3⋅(x−(8))
Step 4.2
Simplify the equation and keep it in point-slope form.
y+0=1.‾3⋅(x-8)y+0=1.¯3⋅(x−8)
y+0=1.‾3⋅(x-8)y+0=1.¯3⋅(x−8)
Step 5
Step 5.1
Add yy and 00.
y=1.‾3⋅(x-8)y=1.¯3⋅(x−8)
Step 5.2
Simplify 1.‾3⋅(x-8)1.¯3⋅(x−8).
Step 5.2.1
Apply the distributive property.
y=1.‾3x+1.‾3⋅-8y=1.¯3x+1.¯3⋅−8
Step 5.2.2
Multiply 1.‾3 by -8.
y=1.‾3x-10.‾6
y=1.‾3x-10.‾6
y=1.‾3x-10.‾6
Step 6
