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

Answer 1

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 2

write the equation in standard form

Answer 3

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 4

wutz the slope intercept fowrm

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!