Question

determine the equation of the line that passes through the given points to complete the table below.




4.) (–4,–5) and (2,2)

5.) (4/5,-1/3) and the origin



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Answer 1

Answer:The equation of a line is typically written as y=mx+b where m is the slope and b is the y-intercept.

If you know two points that a line passes through, this page will show you how to find the equation of the line.

Step-by-step explanation:

Answer 2

Answer:

The equation of the line that passes through the given points are given below,

1. (2,5) and (-3,6)

$y=-\dfrac{x}{5}\end{array}+\dfrac{27}{5}$

$\dfrac{x}{5}+y=\dfrac{27}{5}$

2. (3,-2) and (0,4)

$y=-2x+4$

$2x+y=4$

3. (-1,-5) and (2,2)

$y=\dfrac{7x}{3}-\dfrac{8}{3}\end{array}$

$\dfrac{7x}{3} -y=\dfrac{8}{3}$

4. (-4,-5) and (2,2)

$y=\dfrac{7x}{6}-\dfrac{1}{3}\end{array}$

$\dfrac{7x}{6} -y=\dfrac{1}{3}$

5. $(\frac{4}{5} ,\frac{-1}{3})$ and (0,0)

$y=\dfrac{-5x}{12}$

$5x+12y=0$

Step-by-step explanation:

The question is to find the equation of line that passes through the given points. That is,

1. (2,5) and (-3,6)

2. (3,-2) and (0,4)

3. (-1,-5) and (2,2)

4. (-4,-5) and (2,2)

5. $(\frac{4}{5} ,\frac{-1}{3})$ and (0,0)

The formula to find the equation of the line is as follows,

$y-y_{1}=\dfrac{y_{2}-y_{1}}{x_{2}-x_{1}}\left( x-x_{1}\right)$

Here $m=\frac{y_2-y_1}{x_2-x_1}$, this is the slope of the line.

The slope-intercept form is $y=mx+b$. And the standard form is $ax+by=c$. We can find these forms of the equation the line by rearranging the equation.

Solution:

1. (2,5) and (-3,6)

     In this case $x_1=2,\ y_1=5,\ x_2=-3,\ and\ y_{2}=6$. Substitute these values in the equation. Then,

$\begin{array}{l}y-5=\dfrac{6-5}{-3-2}\left(x-2\right)\\\ \ \ \ \ \ \ \ =\dfrac{1}{-5}\left(x-2\right)\\y-5=-\dfrac{x}{5}+\dfrac{2}{5}\end{array}$

∴$\begin{array}{l}y=\dfrac{2}{5}+5-\dfrac{x}{5}\\\ \\\ y=\dfrac{27}{5}-\dfrac{x}{5}\end{array}$ or  $y=-\dfrac{x}{5}\end{array}+\dfrac{27}{5}$

This is the equation for the slope-intercept form of the line.

Now by rearranging the equation $y=-\dfrac{x}{5}\end{array}+\dfrac{27}{5}$  we get,

$\dfrac{x}{5}+y=\dfrac{27}{5}$ → This is the standard form.

2. (3,-2) and (0,4)

      Here,$x_1=3,\ y_1=-2,\ x_2=0,\ and\ y_{2}=4$

By substituting,

⇒   $\ \ \ \ \ \ \ \ y+2=\dfrac{4+2}{0-3}\left( x-3\right) \end{array}$

  $y+2=\dfrac{6}{-3}\left( x-3\right) \end{array}$

  $y+2=-2(x-3)=-2x+6$

⇒  $y=-2x+6-2\\\\y=-2x+4$→ Slope-intercept form

⇒  $2x+y=4$      → Standard form

3. (-1,-5) and (2,2)

       $x_1=-1,\ y_1=-5,\ x_2=2,\ and\ y_{2}=2$

By substituting,

     $y+5=\dfrac{2+5}{2+1}\left(x+1\right)\\\\=\dfrac{7}{3}\left(x+1\right)\\\\y+5=\dfrac{7x}{3}+\dfrac{7 }{3}$

 ⇒    $\begin{array}{l}y=\dfrac{7x}{3}+\dfrac{7}{3}-5\\\\y=\dfrac{7x}{3}-\dfrac{8}{3}\end{array}$ → Slope-intercept form

 ⇒    $-\dfrac{7x}{3} +y=-\dfrac{8}{3} \\\\ \dfrac{7x}{3} -y=\dfrac{8}{3}$ → Standard form

4. (-4,-5) and (2,2)

      $x_1=-4,\ y_1=-5,\ x_2=2,\ and\ y_{2}=2$

By substituting,

      $y+5=\dfrac{2+5}{2+4}\left(x+4\right)\\\\=\dfrac{7}{6}\left(x+4\right)\\\\y+5=\dfrac{7x}{6}+\dfrac{28 }{6}$

 ⇒     $\begin{array}{l}y=\dfrac{7x}{6}+\dfrac{28}{6}-5\\\\y=\dfrac{7x}{6}-\dfrac{1}{3}\end{array}$→ Slope-intercept form

 ⇒        $\dfrac{7x}{6} -y=\dfrac{1}{3}$     → Standard form

5. $(\frac{4}{5} ,\frac{-1}{3})$ and (0,0)

   $x_1=\frac{4}{5} ,\ y_1=\frac{-1}{3} ,\ x_2=0,\ and\ y_{2}=0$

By substituting,

   $y+\dfrac{1}{3}=\dfrac{0+\dfrac{1}{3}}{0-\dfrac{4}{5}}\left( x-\dfrac{4}{5}\right)$

   $\begin{array}{l}y+\dfrac{1}{3}=\dfrac{-5}{12}\left(x-\dfrac{4}{5}\right)\\\\y+\dfrac{1}{3}=\dfrac{-5x}{12}+\dfrac{1}{3}\\\\y=\dfrac{-5x}{12}+\dfrac{1}{3}-\dfrac{1}{3}\end{array}$

  ⇒  $y=\dfrac{-5x}{12}$   → Slope-intercept form

   ⇒   $5x+12y=0$  → Standard form

These are the answers to the questions.