All categories

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

You might be interested in

If you need more than 60% of the students surveyed to like your product how many out of 500 would you have to survey

Mashutka [201]

Here is your answer.

6 0

3 years ago

Answer pls i need help

nataly862011 [7]

Answer:

What/where is the question you have for me?

Step-by-step explanation:

5 0

3 years ago

What is the base of the exponential function?

jeka94

Answer:

the answer might be 0.86 I don't know though

4 0

3 years ago

If we inscribe a circle such that it is touching all six corners of a regular hexagon of side 10 inches, what is the area of the

Brrunno [24]

Answer:

$\left(100\pi - 150\sqrt{3}\right)$ square inches.

Step-by-step explanation:

<h3>Area of the Inscribed Hexagon</h3>

Refer to the first diagram attached. This inscribed regular hexagon can be split into six equilateral triangles. The length of each side of these triangle will be $10$ inches (same as the length of each side of the regular hexagon.)

Refer to the second attachment for one of these equilateral triangles.

Let segment $\sf CH$ be a height on side $\sf AB$ . Since this triangle is equilateral, the size of each internal angle will be $\sf 60^\circ$ . The length of segment

$\displaystyle 10\, \sin\left(60^\circ\right) = 10 \times \frac{\sqrt{3}}{2} = 5\sqrt{3}$ .

The area (in square inches) of this equilateral triangle will be:

$\begin{aligned}&\frac{1}{2} \times \text{Base} \times\text{Height} \\ &= \frac{1}{2} \times 10 \times 5\sqrt{3}= 25\sqrt{3} \end{aligned}$ .

Note that the inscribed hexagon in this question is made up of six equilateral triangles like this one. Therefore, the area (in square inches) of this hexagon will be:

$\displaystyle 6 \times 25\sqrt{3} = 150\sqrt{3}$ .

<h3>Area of of the circle that is not covered</h3>

Refer to the first diagram. The length of each side of these equilateral triangles is the same as the radius of the circle. Since the length of one such side is $10$ inches, the radius of this circle will also be $10$ inches.

The area (in square inches) of a circle of radius $10$ inches is:

$\pi \times (\text{radius})^2 = \pi \times 10^2 = 100\pi$ .