The definition of a linear equation is an algebraic equation in which each term has an exponent of one and the graphing of the equation results in a straight line

Step-by-step explanation:

u single?

8 0

2 years ago

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Find the equation of the line specified.

OLga [1]

The point slope form of a straight line is y-y1 = m(x-x1)

y -(-3) = -7(x-5)

y + 3 = -7(x-5)

y = -7x + 35 - 3

y = -7x + 32

7 0

3 years ago

Find the slope of the line that passes through (6, 7) and (2, 10). and simplify if needed thx guys.

tekilochka [14]

Answer:

$-\frac{3}{4}$

Step-by-step explanation:

the equation for finding the slope of a line when given two points is $\frac{y_2-y_1}{x_2-x_1}$ , aka the change in y over the change in x.

pick one of your coordinate pairs to be $y_2\\$ and $x_2$ . it doesn't matter which coordinate pair you choose as long as you keep them as $y_2\\$ and $x_2$ . the remaining coordinate pair will be $y_1$ and $x_1$ .

for this example, i'll use (2, 10) for $y_2\\$ and $x_2$ and (6, 7) for $y_1$ and $x_1$ .

**before i begin, i just want to note that you can do these next four steps in any order that you want. i personally prefer to plug in my y-values first and then my x-values, but you can choose to instead plug in the values of each coordinate pair (like starting by plugging in the coordinate pair (2, 10) with 10 for $y_2\\$ and 2 for $x_2$ ). it's up to you. i'm going to explain the steps by plugging in my y-values first and then my x-values because that's the way i normally do it.

first, start by plugging in the y-value from the coordinate pair of your choosing in for $y_2\\$ . since i chose (2, 10) for $y_2\\$ and $x_2$ , i'll plug in 10 for $y_2\\$ .

$\frac{y_2-y_1}{x_2-x_1}$ ⇒ $\frac{10-y_1}{x_2-x_1}$

then plug in the remaining coordinate pair's y-value in for $y_1$ . since the coordinate pair that's left is (6, 7), i will plug in 7 for $y_1$ .

$\frac{y_2-y_1}{x_2-x_1}$ ⇒ $\frac{10-7}{x_2-x_1}$

now i'm going to plug in the x-values. i chose (2, 10) to plug in for $y_2\\$ and $x_2$ , so now i'll plug in 2 for $x_2$ .

$\frac{y_2-y_1}{x_2-x_1}$ ⇒ $\frac{10-7}{2-x_1}$

and all that's left to plug in is the x-value from (6, 7), so i will plug that in for $x_1$ .

$\frac{y_2-y_1}{x_2-x_1}$ ⇒ $\frac{10-7}{2-6}$

after plugging in all the values, you have $\frac{10-7}{2-6}$ .

subtract 10 - 7 as well as 2 - 6.

$\frac{10-7}{2-6}$ ⇒ $\frac{3}{-4}$

$\frac{3}{-4}$ cannot be simplified, therefore the slope of the line is $\frac{3}{-4}$ or $-\frac{3}{4}$ .

i hope this helps! have a lovely day <3

7 0

2 years ago