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identify the equation of a line in slope- intercept form that is perpendicular to y = 2/3 x - 4 and passes through (6, -2).

Sagot :

Answer:

Step-by-step explanation:

There is a relationship between lines that are perpendicular. Their slopes are negative reciprocals of each other. Example: if one line has a slope of 4/3 then a line with the slope of -3/4 is perpendicular to it (they intersect at right angles).

So to find the slope of a line perpendicular to y = -2/3x + 4 we must know the slope of it. The slope, when the line is in the slope-intercept form (aka y = mx + b), is the multiplier in front of the "x" so in this case it is -2/3. So a line perpendicular to it has a slope of 3/2.

Now that we know the slope (m = 3/2) we can find the equation of the line that has a slope of 3/2 but goes through the point (-2, -2) in a couple of ways.

1) Use the slope-intercept form of a line and plug in the values of x = -2 and y = -2 (from the point (-2, -2) like this:

y = mx + b

y = 3/2x + b

-2 = 3/2(-2) + b

-2 = -3 + b

1 = b

So y = 3/2x + 1

2) We can use the point-slope equation y - y1 = m(x - x1) which works great if you know the value of m (we do, m = 3/2) and some point (x1, y1) on the line (and we do, (-2, -2); so x1 = -2 and y1 = -2):

y - y1 = m(x - x1)

y - y1 = 3/2(x - x1)

y - (-2) = 3/2(x - (-2))

y + 2 = 3/2(x + 2)

Although it looks different than the other equation, it is really the same. Just distribute and combine like-terms and you'll see it's the same. So either is an acceptable answer for this quest