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Leona [35]
1 year ago
14

Perpendicular to the line y =6 - 3x; passes through (-3, -1)

Mathematics
1 answer:
madam [21]1 year ago
6 0

Answer:

y=\frac{1}{3}x

Step-by-step explanation:

<u>Perpendicular lines</u>

Perpendicular lines have slopes whose product is -1.  In other words, to find a line that is perpendicular to a given line, the new line must have a that is the opposite reciprocal of the original line.

<u>Passing through a point</u>
Additionally, if a line contains a point, then the point must be a solution to the equation (meaning, when you plug in the "x" and "y", the equation must be true).

<u>Finding our perpendicular line</u>

<u>Work on slope first</u>

To find the slope of the original line, rewrite the original line in slope-intercept form: y=mx+b

y=6-3x\\y=6+(-3x)\\y=(-3x)+6\\y=-3x+6\\y=-\frac{3}{1}x+6

In this form, the slope is the number multiplied to the "x", so the original slope is -3/1.

Thus, the slope of the new perpendicular line will be the opposite reciprocal of -3/1 ... which is 1/3.

Writing what we do know about the new perpendicular line, we have y=\frac{1}{3}x+b

<u>Making sure it passes through the point</u>
This new line is also supposed to contain the point (-3,-1), so (-3,-1) must be a solution to the equation, for whatever constant-value the "b" is.  To find out the "b", substitute the known quantities, and solve:

-1=\frac{1}{3} (-3)+b\\-1=-1+b\\(-1)+1=(-1+b)+1\\0=b

Substituting this into our equation:

y=\frac{1}{3}x+(0)\\y=\frac{1}{3}x

<u>Conclusion</u>

So, the equation for a line that is perpendicular to the original line y=6-3x, and that also passes through (-3,-1) is y=\frac{1}{3}x

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