Write an equation of the line that passes through (6, 1) and is perpendicular to x +2y-12=0.
2 answers:
Answer:
y = 2x - 11
Step-by-step explanation:
<u>Perpendicular lines</u> have negative reciprocal slopes—multiplying the slopes of two lines results in a product of -1.
Given the linear equation, x + 2y - 12 = 0, and the other point, (6, 1):
Convert the given linear equation into its slope-intercept form:
x + 2y - 12 = 0
Subtract x from both sides:
x - x + 2y - 12 = - x 0
2y - 12 = -x
Next, add 12 to both sides of the equation:
2y - 12 + 12 = -x + 12
2y = -x + 12
Divide both sides by 2 in order to isolate y:
y = - ½x + 6
Since the slope of the given equation is - ½, then the other line perpendicular to it must have a slope of 2:
Let m₁ = - ½
m₂ = 2
m₁ × m₂ = -1
-½ × 2 = -1
Hence, the slope of the other line, m₂ = 2. Next, using the slope of the other line, m₂ = 2, and the other given point, (6, 1), substitute these values into the slope-intercept form to solve for the y-intercept, (b):
y = mx + b
1 = 2(6) + b
1 = 12 + b
1 - 12 = 12 - 12 + b
-11 = b
Thus, the <u>y-intercept</u> of the other line is b = -11.
Therefore, the linear equation of the line that is perpendicular to x + 2y - 12 = 0 is:
<h3>y = 2x - 11.</h3>As a proof, attached is the graph of both equations, where it shows that they are perpendicular from each other.
You might be interested in