Enter a problem...
Algebra Examples
(2,-3)(2,−3) and is perpendicular to the line y=-2x-3y=−2x−3
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 -2−2.
m=-2m=−2
m=-2m=−2
Step 2
The equation of a perpendicular line must have a slope that is the negative reciprocal of the original slope.
mperpendicular=-1-2mperpendicular=−1−2
Step 3
Step 3.1
Move the negative in front of the fraction.
mperpendicular=12mperpendicular=12
Step 3.2
Multiply --12−−12.
Step 3.2.1
Multiply -1−1 by -1−1.
mperpendicular=1(12)mperpendicular=1(12)
Step 3.2.2
Multiply 1212 by 11.
mperpendicular=12mperpendicular=12
mperpendicular=12mperpendicular=12
mperpendicular=12mperpendicular=12
Step 4
Step 4.1
Use the slope 1212 and a given point (2,-3)(2,−3) 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-(-3)=12⋅(x-(2))y−(−3)=12⋅(x−(2))
Step 4.2
Simplify the equation and keep it in point-slope form.
y+3=12⋅(x-2)y+3=12⋅(x−2)
y+3=12⋅(x-2)y+3=12⋅(x−2)
Step 5
Step 5.1
Solve for yy.
Step 5.1.1
Simplify 12⋅(x-2)12⋅(x−2).
Step 5.1.1.1
Rewrite.
y+3=0+0+12⋅(x-2)y+3=0+0+12⋅(x−2)
Step 5.1.1.2
Simplify by adding zeros.
y+3=12⋅(x-2)y+3=12⋅(x−2)
Step 5.1.1.3
Apply the distributive property.
y+3=12x+12⋅-2y+3=12x+12⋅−2
Step 5.1.1.4
Combine 1212 and xx.
y+3=x2+12⋅-2y+3=x2+12⋅−2
Step 5.1.1.5
Cancel the common factor of 22.
Step 5.1.1.5.1
Factor 22 out of -2−2.
y+3=x2+12⋅(2(-1))y+3=x2+12⋅(2(−1))
Step 5.1.1.5.2
Cancel the common factor.
y+3=x2+12⋅(2⋅-1)
Step 5.1.1.5.3
Rewrite the expression.
y+3=x2-1
y+3=x2-1
y+3=x2-1
Step 5.1.2
Move all terms not containing y to the right side of the equation.
Step 5.1.2.1
Subtract 3 from both sides of the equation.
y=x2-1-3
Step 5.1.2.2
Subtract 3 from -1.
y=x2-4
y=x2-4
y=x2-4
Step 5.2
Reorder terms.
y=12x-4
y=12x-4
Step 6