QuestionFind the slope and equations of the line through points $(6,1)$ and $(10,6)$ .
(A) Slope: A. $m=\square$ or B. Not defined.
(B) Point-slope form.
(C) Slope-intercept form.
(D) Standard form.

Studdy Solution

STEP 1

Assumptions1. The given points are $(6,1)$ and $(10,6)$ . The slope of a line passing through two points $(x1, y1)$ and $(x, y)$ is given by $m = \frac{y - y1}{x - x1}$
3. The point-slope form of the equation of a line is $y - y1 = m(x - x1)$ 4. The slope-intercept form of the equation of a line is $y = mx + b$
5. The standard form of the equation of a line is $Ax + By = C$ , where $A, B,$ and $C$ are integers, and $A$ and $B$ are not both zero.

STEP 2

First, we need to find the slope of the line that passes through the given points. We can do this by using the formula for the slope.
$m = \frac{y2 - y1}{x2 - x1}$

STEP 3

Now, plug in the given values for the points to calculate the slope.
$m = \frac{6 -1}{10 -6}$

STEP 4

Calculate the slope.
$m = \frac{6 -1}{10 -6} = \frac{}{4}$

STEP 5

Now that we have the slope, we can find the point-slope form of the equation of the line. We can do this by using the formula for the point-slope form.
$y - y1 = m(x - x1)$

STEP 6

Plug in the values for the slope and one of the points to calculate the point-slope form of the equation of the line.
$y -1 = \frac{5}{4}(x -6)$

STEP 7

Now that we have the point-slope form of the equation of the line, we can find the slope-intercept form of the equation of the line. We can do this by solving the point-slope form for $y$ .
$y = \frac{5}{4}(x -6) +1$

STEP 8

implify the equation to get the slope-intercept form.
$y = \frac{5}{4}x - \frac{15}{4} +1$ $y = \frac{5}{4}x - \frac{11}{4}$

STEP 9

Now that we have the slope-intercept form of the equation of the line, we can find the standard form of the equation of the line. We can do this by rearranging the slope-intercept form to the standard form.
$4y =5x -11$

STEP 10

Rearrange the equation to get the standard form.
$5x -4y =$ (A) The slope of the line is $\frac{5}{4}$ . (B) The point-slope form of the equation of the line is $y - = \frac{5}{4}(x -6)$ . (C) The slope-intercept form of the equation of the line is $y = \frac{5}{4}x - \frac{}{4}$ . (D) The standard form of the equation of the line is $5x -4y =$ .

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QuestionFind the slope and equations of the line through points $(6,1)$ and $(10,6)$ .
(A) Slope: A. $m=\square$ or B. Not defined.
(B) Point-slope form.
(C) Slope-intercept form.
(D) Standard form.

Studdy Solution

STEP 1

STEP 2

First, we need to find the slope of the line that passes through the given points. We can do this by using the formula for the slope.
$m = \frac{y2 - y1}{x2 - x1}$

STEP 3

Now, plug in the given values for the points to calculate the slope.
$m = \frac{6 -1}{10 -6}$

STEP 4

Calculate the slope.
$m = \frac{6 -1}{10 -6} = \frac{}{4}$

STEP 5

Now that we have the slope, we can find the point-slope form of the equation of the line. We can do this by using the formula for the point-slope form.
$y - y1 = m(x - x1)$

STEP 6

Plug in the values for the slope and one of the points to calculate the point-slope form of the equation of the line.
$y -1 = \frac{5}{4}(x -6)$

STEP 7

Now that we have the point-slope form of the equation of the line, we can find the slope-intercept form of the equation of the line. We can do this by solving the point-slope form for $y$ .
$y = \frac{5}{4}(x -6) +1$

STEP 8

implify the equation to get the slope-intercept form.
$y = \frac{5}{4}x - \frac{15}{4} +1$ $y = \frac{5}{4}x - \frac{11}{4}$

STEP 9

Now that we have the slope-intercept form of the equation of the line, we can find the standard form of the equation of the line. We can do this by rearranging the slope-intercept form to the standard form.
$4y =5x -11$

STEP 10

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QuestionFind the slope and equations of the line through points (6,1)(6,1)(6,1) and (10,6)(10,6)(10,6). (A) Slope: A. m=□m=\squarem=□ or B. Not defined. (B) Point-slope form. (C) Slope-intercept form. (D) Standard form.

Studdy Solution

STEP 1

STEP 2

First, we need to find the slope of the line that passes through the given points. We can do this by using the formula for the slope.m=y2−y1x2−x1m = \frac{y2 - y1}{x2 - x1}m=x2−x1y2−y1​

STEP 3

Now, plug in the given values for the points to calculate the slope.m=6−110−6m = \frac{6 -1}{10 -6}m=10−66−1​

STEP 4

Calculate the slope.m=6−110−6=4m = \frac{6 -1}{10 -6} = \frac{}{4}m=10−66−1​=4​

STEP 5

Now that we have the slope, we can find the point-slope form of the equation of the line. We can do this by using the formula for the point-slope form.y−y1=m(x−x1)y - y1 = m(x - x1)y−y1=m(x−x1)

STEP 6

Plug in the values for the slope and one of the points to calculate the point-slope form of the equation of the line.y−1=54(x−6)y -1 = \frac{5}{4}(x -6)y−1=45​(x−6)

STEP 7

Now that we have the point-slope form of the equation of the line, we can find the slope-intercept form of the equation of the line. We can do this by solving the point-slope form for yyy.y=54(x−6)+1y = \frac{5}{4}(x -6) +1y=45​(x−6)+1

STEP 8

implify the equation to get the slope-intercept form.y=54x−154+1y = \frac{5}{4}x - \frac{15}{4} +1y=45​x−415​+1y=54x−114y = \frac{5}{4}x - \frac{11}{4}y=45​x−411​

STEP 9

Now that we have the slope-intercept form of the equation of the line, we can find the standard form of the equation of the line. We can do this by rearranging the slope-intercept form to the standard form.4y=5x−114y =5x -114y=5x−11

STEP 10

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QuestionFind the slope and equations of the line through points $(6,1)$ and $(10,6)$ .
(A) Slope: A. $m=\square$ or B. Not defined.
(B) Point-slope form.
(C) Slope-intercept form.
(D) Standard form.

First, we need to find the slope of the line that passes through the given points. We can do this by using the formula for the slope.
$m = \frac{y2 - y1}{x2 - x1}$

Now, plug in the given values for the points to calculate the slope.
$m = \frac{6 -1}{10 -6}$

Calculate the slope.
$m = \frac{6 -1}{10 -6} = \frac{}{4}$

Now that we have the slope, we can find the point-slope form of the equation of the line. We can do this by using the formula for the point-slope form.
$y - y1 = m(x - x1)$

Plug in the values for the slope and one of the points to calculate the point-slope form of the equation of the line.
$y -1 = \frac{5}{4}(x -6)$

Now that we have the point-slope form of the equation of the line, we can find the slope-intercept form of the equation of the line. We can do this by solving the point-slope form for $y$ .
$y = \frac{5}{4}(x -6) +1$

implify the equation to get the slope-intercept form.
$y = \frac{5}{4}x - \frac{15}{4} +1$ $y = \frac{5}{4}x - \frac{11}{4}$

Now that we have the slope-intercept form of the equation of the line, we can find the standard form of the equation of the line. We can do this by rearranging the slope-intercept form to the standard form.
$4y =5x -11$