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2 years ago

10

Given line y-1/6x+9,find a perpendicular line to the given line passes through points (6,-3)?

1 answer:

DerKrebs [107]2 years ago

7 0

Answer:

y + 3 = (6)(x - 6)

Step-by-step explanation:

The proper format for this equation y-1/6x+9 is y = (-1/6)x+9.

Any line perpendicular to y = (-1/6)x+9 has the slope which is the negative reciprocal of that of y = (-1/6)x+9, or the negative reciprocal of -1/6. That is +6.

Thus, the desired line, which passes through (6, -3), is found using the point-slope formula for the equation of a straight line:

y + 3 = (6)(x - 6)

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soldier1979 [14.2K]

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Step-by-step explanation:

So this sounds like a problem for the Law of Cosines. The largest angle is always opposite the largest side in a triangle.

So 11 is the largest side so the angle opposite to it is what we are trying to find. Let's call that angle, X.

My math is case sensitive.

X is the angle opposite to the side x.

Law of cosines formula is:

$x^2=a^2+b^2-2ab \cos(X)$

So we are looking for X.

We know x=11, a=4, and b=8 (it didn't matter if you called b=4 and a=8).

$11^2=4^2+8^2-2(4)(8)\cos(X)$

$121=16+64-64\cos(X)$

$121=80-64\cos(X)$

Subtract 80 on both sides:

$121-80=-64\cos(X)$

$41=-64\cos(X)$

Divide both sides by -64:

$\frac{41}{-64}=\cos(X)$

Now do the inverse of cosine of both sides or just arccos( )

[these are same thing]

$\arccos(\frac{-41}{64})=X$

Time for the calculator:

X=129.8 approximately

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