What is an equation of the line that is perpendicular to y=-3/4x+6 and passes through the point (3,9)

Mathematics

1 answer:

goldenfox [79]3 years ago

5 0

Answer:

$y=\frac{4}{3}x+5$

Step-by-step explanation:

The line y= -3/4+6 in the comments has slope -3/4. A perpendicular line will have the negative reciprocal of -3/4. It's 4/3.

To write the equation, substitute 4/3 and the point (3,9) into

$y-y_1 = m(x-x_1)$

$y-9=\frac{4}{3}(x-3)\\\\y-9=\frac{4}{3}x -\frac{4*3}{3}\\\\y-9=\frac{4}{3}x-4\\\\y=\frac{4}{3}x+5$

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(4, -2) (see attached)

Step-by-step explanation:

Vector addition on a graph is accomplished by placing the tail of one vector on the nose of the one it is being added to. The negative of a vector is in the direction opposite to the original.

<h3>vector components</h3>

The components of the vectors are ...

u = (1, -2)

v = (-6, -6)

Then the components of the vector sum are ...

2u -1/3v = 2(1, -2) -1/3(-6, -6) = (2 +6/3, -4 +6/3)

2u -1/3v = (4, -2)

<h3>graphically</h3>

The sum is shown graphically in the attachment. Vector u is added to itself by putting a copy at the end of the original. Then the nose of the second vector is at 2u.

One-third of vector v is subtracted by adding a vector to 2u that is 1/3 the length of v, and in the opposite direction. The nose of this added vector is the resultant: 2u-1/3v.

The resultant is in red in the attachment.