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

$$\boxed{{\displaystyle \frac{1}{3}}}.$$

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:

$$\boxed{{\displaystyle 5x-15y=-59}}$$

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:

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

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