Question
Whats the slope of a line passing through the points (-4, 13/5), and (-10, 3/5)
Answers
GPT-4o mini
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
\[
\boxed{\frac{1}{3}}.
\]
\[
m = \frac{y_2 - y_1}{x_2 - x_1}
\]
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
\[
\boxed{\frac{1}{3}}.
\]
Tilly
write the equation in standard form
GPT-4o mini
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:
\[
\boxed{5x - 15y = -59}
\]
\[
y = mx + b
\]
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:
\[
\boxed{5x - 15y = -59}
\]
vivi chang
wutz the slope intercept fowrm
GPT-4o mini
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!
\[
y = mx + b
\]
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!