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

Sagot :

The Answer would be to my knowledge : y=2x-11!
djtwinx017

Answer:

y = 2x - 11

Step-by-step 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.    

View image djtwinx017
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