1. Find the slope of the line that passes through the

points (-1, 2), (0, 5).

2. Suppose y varies directly with x, and y = 15 and x = 5.
Write a direct variation equation that relates x and y.
What is the value of y when x = 9?

3. Write an equation in slope-intercept form of the line
that passed through (-3, 4) and (1. 4).

4. Use point-slope form to write the equation of a line
that has a slope of 2/3
and passes through (-3, -1).
Write your final equation in slope-intercept form.

5. Write the equation in standard form using integers
(no fractions or decimals): = −2/3 − 1

I'm NOT asking for answers. I'm simply asking for someone to explain the concepts of these because I don't understand at all

#1 the slope of a line is the change in y, divided by the change in x. You can pick any two points on the line. Then just apply the formula you have:

slope = (y2-y1)/(x2-x1)

which divides y-change by x-change

#2 y varies directly with x, means that
y = kx
for some value of k. Or, you can see that this means that

y/x = k

is a constant for any two values of x and y involved. So, you want y such that

y/9 = 15/5

You don't even need to know what k is, though clearly k=3 in this problem.

#3 y = mx+b
So, just plug in your two points to get two equations to solve for m and b. However, see #4 if you want to see another way. (it uses #1 to find the slope)

#4 Since the slope of a line is constant, it is the same between any two points on the line. In particular, if one of the points is (-3,-1) and the other is any (x,y) we have (from #1)

[y-(-1)]/[x-(-3)] = 2/3
Or, as it is more usually written,
y+1 = 2/3 (x+3)

By now you should see how all these problems use the idea of slopes and points on lines.

#5 I can't tell; something's missing. But whatever form you come up with, just rearrange the terms to standard form by clearing fractions. For example,

y = -2/3 x - 1 can be rearranged via
3y = -2x-3
2x + 3y = -3

this is U6L9 not U6L6

OHHHHHHHH ok that makes so much more since, THANK YOU SO MUCH Steve

If u guys are from Connexus ALEGRBA 1A UNIT 6 LESSON 6 Parallel and Perpendicular Lines practice is use this.

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GOOD LUCK!
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1. The slope of a line can be found using the formula: rise over run. It represents how steep or flat the line is. To find the slope between two points, you subtract the y-coordinates of the two points and divide it by the difference in x-coordinates.

Using the points (-1, 2) and (0, 5), we can calculate the slope:
slope = (5 - 2) / (0 - (-1))
slope = 3 / 1
slope = 3

2. In direct variation, two variables are related by a constant ratio. The equation that describes direct variation is y = kx, where k is the constant of variation.

Given that y = 15 and x = 5, we can substitute these values into the equation to find k:
15 = k * 5
k = 15 / 5
k = 3

So, the direct variation equation is y = 3x. To find the value of y when x = 9, we substitute x = 9 into the equation:
y = 3 * 9
y = 27

3. The slope-intercept form of a line's equation is y = mx + b, where m represents the slope and b represents the y-intercept.

To find the equation using the points (-3, 4) and (1, 4), we can observe that the y-coordinate remains the same for both points. This means that the line is horizontal and its slope is 0. The equation would be:
y = 0x + b
y = b

Since the y-intercept is 4, the equation is y = 4.

4. The point-slope form of a line's equation is y - y1 = m(x - x1), where m represents the slope and (x1, y1) represents a point on the line.

Given that the slope is 2/3 and it passes through the point (-3, -1), we can substitute these values into the point-slope form:
y - (-1) = (2/3)(x - (-3))
y + 1 = (2/3)(x + 3)

To convert it into slope-intercept form, we simplify the equation:
y + 1 = (2/3)x + 2
y = (2/3)x + 1

5. To write an equation in standard form, it should have the format Ax + By = C, where A, B, and C are integers.

The given equation is -2/3 - 1. We need to convert it to standard form:
Multiply every term by 3 to eliminate the fraction:
-2 - 3 = -3C
-5 = -3C

To have integers, we multiply both sides by -1:
5 = 3C

So, the equation in standard form is 3C - 5 = 0.

1. To find the slope of a line passing through two points, you can use the formula:

s = (y2 - y1) / (x2 - x1)

Where (x1, y1) and (x2, y2) are the coordinates of the two points.

In this case, the coordinates of the two points are (-1, 2) and (0, 5). Substitute these values into the formula:

s = (5 - 2) / (0 - (-1))
s = 3 / 1
s = 3

Therefore, the slope of the line passing through these two points is 3.

2. A direct variation equation expresses a relationship where one variable is directly proportional to another. It can be written in the form y = kx, where k is the constant of variation.

In this scenario, we are given that y = 15 when x = 5. To find k, substitute these values into the equation:

15 = k * 5
k = 15 / 5
k = 3

Now that we have the value of k, we can write the direct variation equation as y = 3x.

To find the value of y when x = 9, substitute this value into the equation:

y = 3 * 9
y = 27

Therefore, when x = 9, y = 27.

3. The slope-intercept form of a linear equation is y = mx + b, where m represents the slope of the line, and b represents the y-intercept (the point where the line crosses the y-axis).

To write an equation in slope-intercept form, we need to find the values of m and b. We are given two points that the line passes through: (-3, 4) and (1, 4).

The slope of the line can be calculated using the formula mentioned earlier:

m = (y2 - y1) / (x2 - x1)

Substituting the values of the given points:

m = (4 - 4) / (1 - (-3))
m = 0 / 4
m = 0

Since the slope is 0, the line is horizontal. In this case, the y-intercept is the same as all the points the line passes through, which is y = 4.

Therefore, the equation of the line in slope-intercept form is y = 0x + 4, which simplifies to y = 4.

4. The point-slope form of a linear equation is y - y1 = m(x - x1), where (x1, y1) is a point on the line, and m is the slope.

We are given that the line has a slope of 2/3 and passes through the point (-3, -1). To write the equation of the line, substitute these values into the point-slope form:

y - (-1) = (2/3)(x - (-3))
y + 1 = (2/3)(x + 3)

To write the equation in slope-intercept form, we can simplify the equation further:

y + 1 = (2/3)x + 2
y = (2/3)x + 1

Therefore, the equation of the line in slope-intercept form is y = (2/3)x + 1.

5. The equation is not properly provided in your question. Please revise the equation, and I will be happy to explain how to write it in standard form.

Find the slope of a line that passes through the point (3,5) and (–6,–10)