Question

Find equation of the line that contains the point (4,-2) and is perpendicular to the line y= _2x+8

Answer 1

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Answer:

y = 1/2x -4

Explanation:

We presume the given line is ...

y = -2x +8

This is in slope-intercept form, which allows us to determine easily that the slope of this line is -2.

A perpendicular line will have a slope that is the opposite reciprocal of -2:

m = -1/(-2) = 1/2

The y-intercept of the desired line can be found from the point (x, y) = (4, -2) using the equation ...

b = y - mx

b = -2 -(1/2)(4) = -4

Now, we know the slope and y-intercept of the desired perpendicular line through (4, -2), so we can write its equation as ...

y = 1/2x -4

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Additional comment

"Slope-intercept form" is ...

y = mx + b . . . . . . where m is the slope and b is the y-intercept

Answer 2

y = 1/2x - 4

Explanation:

If two lines are perpendicular to each other, they have opposite slopes.

The first line is y = -2x + 8. Its slope is -2. A line perpendicular to this one will have a slope of 1/2.

Plug this value (1/2) into your standard point-slope equation of y = mx + b.

y = 1/2x + b

To find b, we want to plug in a value that we know is on this line: in this case, it is (4, -2). Plug in the x and y values into the x and y of the standard equation.

-2 = 1/2(4) + b

To find b, multiply the slope and the input of x (4)

-2 = 2 + b

Now, subtract 2 from both sides to isolate b.

-4 = b

Plug this into your standard equation.

y = 1/2x - 4

This equation is perpendicular to your given equation (y = -2x + 8) and contains point (4, -2)

Hope this helps!