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3 years ago

Prove that lines 3x-4y=12 and 3y=12-4x are perpendicular.

Mathematics

1 answer:

Svetllana [295]3 years ago

3 0

Answer:

see explanation

Step-by-step explanation:

If 2 lines are perpendicular then the product of their slopes equals - 1

The equation of a line in slope- intercept form is

y = mx + c ( m is the slope and c the y- intercept )

Consider the given equations

3x - 4y = 12 ( subtract 3x from both sides )

- 4y = - 3x + 12 ( divide terms by - 4 )

y = $\frac{3}{4}$ x - 3 ← in slope- intercept form

with slope m = $\frac{3}{4}$

3y = 12 - 4x = - 4x + 12 ( divide terms by 3 )

y = - $\frac{4}{3}$ x + 4 ← in slope- intercept form

with slope m = - $\frac{4}{3}$

Then

$\frac{3}{4}$ × - $\frac{4}{3}$ = - 1

Since the product of their slopes = - 1 then the lines are perpendicular

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4 years ago

Use addition to solve the linear system of equations. Include all of your work in your final answer. $\left\{ \right.$ 3x-y=6

ad-work [718]

Answer:

Solving the given linear system, we get x = 1 and y = -3.

The solution set is: (1,-3)

Step-by-step explanation:

We need to solve the linear system of equations.

$3x-y=6 \\y=x-4$

We can write second equation y=x-4 as: $-x+y=-4$

Let:

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Now, Adding equation 1 and 2

$3x-y=6\\ -x+y=-4\\--------\\2x+0y=2\\2x=2\\x=\frac{2}{2}\\x=1$

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3 years ago

X+4 divided by x^2+16 Why can't it be simplified?

KonstantinChe [14]

Let's try to simplify x^2 + 16. It's a sum of two squares:

x^2 + 16 = 0

x^2 = -16

The problem is, we can't take a square root of a negative. This is where imaginary numbers come in.

Remember that square roots have a plus or minus symbol outside:

±√-16 = ±4i

Our two roots are 4i and -4i. Therefore, the trinomial simplifies to:

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3 years ago