Question

Whats the slope of a line passing through the points (-4, 13/5), and (-10, 3/5)

Answer 1

write the equation in standard form

Answer 2

wutz the slope intercept fowrm

Answer 3

To find the slope of a line passing through two points \((x_1, y_1)\) and \((x_2, y_2)\), you can use the formula:

\[
m = \frac{y_2 - y_1}{x_2 - x_1}
\]

In this case, the points are \((-4, \frac{13}{5})\) and \((-10, \frac{3}{5})\). So, we have:

\[
(x_1, y_1) = (-4, \frac{13}{5})
\]
\[
(x_2, y_2) = (-10, \frac{3}{5})
\]

Now substituting these values into the slope formula:

\[
m = \frac{\frac{3}{5} - \frac{13}{5}}{-10 - (-4)}
\]

Calculating the numerator:

\[
\frac{3}{5} - \frac{13}{5} = \frac{3 - 13}{5} = \frac{-10}{5} = -2
\]

Now calculating the denominator:

\[
-10 - (-4) = -10 + 4 = -6
\]

Now substituting back into the formula for the slope:

\[
m = \frac{-2}{-6} = \frac{2}{6} = \frac{1}{3}
\]

Thus, the slope of the line passing through the points \((-4, \frac{13}{5})\) and \((-10, \frac{3}{5})\) is

\[
\boxed{\frac{1}{3}}.
\]

Answer 4

To write the equation of the line in standard form \(Ax + By = C\), we can start with the slope-intercept form, which is given by:

\[
y = mx + b
\]

We have already calculated the slope \(m\) to be \(\frac{1}{3}\), and we can use one of the points to find the y-intercept \(b\). Let's use the point \((-4, \frac{13}{5})\):

1. Substitute \(x = -4\) and \(y = \frac{13}{5}\) into the equation \(y = mx + b\):

\[
\frac{13}{5} = \frac{1}{3}(-4) + b
\]

2. Solving for \(b\):

\[
\frac{13}{5} = -\frac{4}{3} + b
\]

To add \(-\frac{4}{3}\) to both sides, first, we convert \(\frac{13}{5}\) to a fraction with a common denominator with \(\frac{4}{3}\). The least common multiple of 5 and 3 is 15.

Converting \(\frac{13}{5}\):

\[
\frac{13}{5} = \frac{13 \cdot 3}{5 \cdot 3} = \frac{39}{15}
\]

Converting \(-\frac{4}{3}\):

\[
-\frac{4}{3} = -\frac{4 \cdot 5}{3 \cdot 5} = -\frac{20}{15}
\]

Now substituting back into the equation gives us:

\[
\frac{39}{15} = -\frac{20}{15} + b
\]

To isolate \(b\), add \(\frac{20}{15}\) to both sides:

\[
b = \frac{39}{15} + \frac{20}{15} = \frac{59}{15}
\]

Now we have the slope \(m = \frac{1}{3}\) and the y-intercept \(b = \frac{59}{15}\). Thus, the equation of the line in slope-intercept form is:

\[
y = \frac{1}{3}x + \frac{59}{15}
\]

To convert this to standard form \(Ax + By = C\), we can rearrange it:

1. Multiply the entire equation by 15 (the least common multiple of the denominators) to eliminate fractions:

\[
15y = 5x + 59
\]

2. Rearranging gives:

\[
-5x + 15y = 59
\]

3. We can multiply through by -1 to keep \(A\) positive (standard convention):

\[
5x - 15y = -59
\]

Thus, the equation of the line in standard form is:

\[
\boxed{5x - 15y = -59}
\]

Answer 5

The slope-intercept form of a linear equation is given by the formula:

\[
y = mx + b
\]

where:
- \(y\) is the dependent variable,
- \(x\) is the independent variable,
- \(m\) is the slope of the line (which indicates how steep the line is),
- \(b\) is the y-intercept (the point where the line crosses the y-axis).

### In Context of the Given Line
From your previous question, we determined the slope \(m\) to be \(\frac{1}{3}\) and the y-intercept \(b\) to be \(\frac{59}{15}\).

Thus, the slope-intercept form of the line passing through the points \((-4, \frac{13}{5})\) and \((-10, \frac{3}{5})\) is:

\[
y = \frac{1}{3}x + \frac{59}{15}
\]

If you need any further assistance, just let me know!