Find a linear equation whose graph is the straight line with the given property.

Through (2, −7) and (1, 1)

I got -8x-23, I know it doesn't match, maybe its the wrong formula

Find a linear equation whose graph is the straight line with the given property.
Through
(1/4, −1)and parallel to the line
4x − 5y = 5

How do I make it parallel?

Thank you.

slope = (1+7)/(1-2) = -8

I don't know what method or "formula" you have been taught, but using the slope-yintercept form seems to be a popular method these days.

y = mx + b
y = -8x + b
sub in the point (1,1) , (I always use the "easier" point)
1 = -8 + b
b = 9

equation : y = -8x + 9
or
8x + y = 9

The second one is even easier.
Since the new line is parallel to the given line, it must look like
4x - 5y = c , differing only in the constant term
sub in the given point (1/4 , -1)
4(1/4) - 5(-1) = c
c = 1 + 5 = 6

new equation: 4x - 5y = 6

Ah, I see you're dealing with some linear equations! Don't worry, I'll help you out with some humor along the way.

To find a linear equation through the points (2, -7) and (1, 1), we can use the formula for the slope-intercept form of a linear equation:

y - y1 = m(x - x1),

where (x1, y1) is a point on the line and m is the slope of the line. Let's calculate the slope first:

m = (y2 - y1) / (x2 - x1)
= (1 - (-7)) / (1 - 2)
= 8 / (-1)
= -8.

Now we can choose any of the two points to plug into the formula. Let's use (2, -7):

y - (-7) = -8(x - 2)
y + 7 = -8x + 16
y = -8x + 9.

So the linear equation that passes through (2, -7) and (1, 1) is y = -8x + 9, not -8x - 23. But hey, being close is better than being a clown's shoelace!

Now onto making a line parallel to 4x - 5y = 5. To make another line parallel, we need it to have the same slope as the given line. Here's how:

First, let's rearrange the given equation to slope-intercept form (y = mx + b):

4x - 5y = 5
-5y = -4x + 5
y = (4/5)x - 1.

Now we see the slope of the given line is 4/5. Since parallel lines have the same slope, our new line will also have a slope of 4/5.

Using the point-slope form, we can plug in the values of the given point (1/4, -1) and the slope (4/5):

y - (-1) = (4/5)(x - 1/4)
y + 1 = (4/5)x - 4/20
y = (4/5)x - 24/20
y = (4/5)x - 6/5.

And voila! A linear equation parallel to 4x - 5y = 5 through the point (1/4, -1) is y = (4/5)x - 6/5. Parallel lines are like two peas in a pod, strutting their stuff in perfect harmony!

I hope that helps, and remember, if math gets too tough, just add some humor to the equation and it all becomes clearer!

To find a linear equation parallel to a given line, you need to use the same slope as the given line. The slope of a line can be determined by the coefficient of x in the equation of the line.

Given that the line is 4x − 5y = 5, we want to find a line parallel to this. The first step is to rearrange the equation in slope-intercept form (y = mx + b), where m is the slope and b is the y-intercept.

To convert the equation to slope-intercept form, we isolate y:
4x − 5y = 5
-5y = -4x + 5
y = (4/5)x - 1

The slope of the given line is 4/5, so we need to use the same slope for the line parallel to it.

Now, using the point-slope form of a linear equation, we can plug in the values from the given point (1/4, -1) and the slope (4/5). The point-slope form is given by:

y - y₁ = m(x - x₁)

Substituting the values:
y + 1 = (4/5)(x - 1/4)

Expanding and simplifying:
y + 1 = (4/5)x - 1/5
y = (4/5)x - 1 - 1/5
y = (4/5)x - 6/5

Therefore, the linear equation that is parallel to 4x − 5y = 5 and passes through (1/4, -1) is y = (4/5)x - 6/5.

To find a linear equation that represents a line passing through two given points (2, -7) and (1, 1), we can use the point-slope form of a linear equation.

The point-slope form for a linear equation is given by: y - y1 = m(x - x1), where (x1, y1) are the coordinates of a point on the line and m is the slope of the line.

1) Calculate the slope (m) using the formula: m = (y2 - y1) / (x2 - x1)
For the points (2, -7) and (1, 1): m = (1 - (-7)) / (1 - 2) = 8 / (-1) = -8

2) Choose one of the given points, for example, (2, -7), and substitute the values into the point-slope form:
y - (-7) = -8(x - 2)
Simplifying, y + 7 = -8x + 16

3) Rewrite the equation in the standard form (Ax + By = C):
8x + y = 9

Therefore, the equation of the line passing through (2, -7) and (1, 1) is 8x + y = 9.

Now, for the second question, you need to find a linear equation that is parallel to the line 4x - 5y = 5 and passes through the point (1/4, -1).

1) First, rewrite the equation 4x - 5y = 5 in slope-intercept form (y = mx + b):
4x - 5y = 5
-5y = -4x + 5
y = (4/5)x - 1

2) Since parallel lines have the same slope, the slope of the desired line will also be 4/5.

3) The point-slope form can now be used, substituting the known values:
y - (-1) = (4/5)(x - 1/4)
Simplifying, y + 1 = (4/5)x - 4/5
y = (4/5)x - 9/5

So, the linear equation that is parallel to 4x - 5y = 5 and passes through the point (1/4, -1) is y = (4/5)x - 9/5.

I hope this helps explain how to find the linear equations for the given properties! Let me know if you have any further questions.