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Write an equation of the line that passes through (6, 1) and is perpendicular to x +2y-12=0.

User Berriel
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The Answer would be to my knowledge : y=2x-11!
User KCK
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Answer:

y = 2x - 11

Explanation:

Perpendicular lines 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 y-intercept of the other line is b = -11.

Therefore, the linear equation of the line that is perpendicular to x + 2y - 12 = 0 is:

y = 2x - 11.

As a proof, attached is the graph of both equations, where it shows that they are perpendicular from each other.

Write an equation of the line that passes through (6, 1) and is perpendicular to x-example-1
User Sonjia
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