QuestionFind the equation of line $r$ parallel to $y=\frac{1}{6} x-5$ that passes through the point $(6,2)$ .

Studdy Solution

STEP 1

Assumptions1. The equation of line $q$ is $y=\frac{1}{6} x-5$ . Line $r$ is parallel to line $q$
3. Line $r$ includes the point $(6,)$ 4. The equation of a line is in the form $y = mx + c$ , where $m$ is the slope and $c$ is the y-intercept

STEP 2

Since line $r$ is parallel to line $q$ , they have the same slope. The slope of line $q$ is $\frac{1}{6}$ , so the slope of line $r$ is also $\frac{1}{6}$ .

STEP 3

We can write the equation of line $r$ in the form $y = mx + c$ , where $m$ is the slope and $c$ is the y-intercept. Since we know the slope $m$ is $\frac{1}{6}$ , we can write the equation as $y = \frac{1}{6}x + c$

STEP 4

To find the y-intercept $c$ , we can substitute the point $(6,2)$ into the equation. This gives us $2 = \frac{1}{6} \cdot6 + c$

STEP 5

olve the equation for $c$ $2 =1 + c$

STEP 6

Subtract1 from both sides of the equation to solve for $c$ $c =2 -1$

STEP 7

Calculate the value of $c$ $c =1$

STEP 8

Now that we have the y-intercept $c$ , we can write the equation of line $r$ as $y = \frac{1}{6}x +1$ The equation of line $r$ is $y = \frac{1}{6}x +1$ .

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QuestionFind the equation of line $r$ parallel to $y=\frac{1}{6} x-5$ that passes through the point $(6,2)$ .

Studdy Solution

STEP 1

Assumptions1. The equation of line $q$ is $y=\frac{1}{6} x-5$ . Line $r$ is parallel to line $q$
3. Line $r$ includes the point $(6,)$ 4. The equation of a line is in the form $y = mx + c$ , where $m$ is the slope and $c$ is the y-intercept

STEP 2

Since line $r$ is parallel to line $q$ , they have the same slope. The slope of line $q$ is $\frac{1}{6}$ , so the slope of line $r$ is also $\frac{1}{6}$ .

STEP 3

We can write the equation of line $r$ in the form $y = mx + c$ , where $m$ is the slope and $c$ is the y-intercept. Since we know the slope $m$ is $\frac{1}{6}$ , we can write the equation as $y = \frac{1}{6}x + c$

STEP 4

To find the y-intercept $c$ , we can substitute the point $(6,2)$ into the equation. This gives us $2 = \frac{1}{6} \cdot6 + c$

STEP 5

olve the equation for $c$ $2 =1 + c$

STEP 6

Subtract1 from both sides of the equation to solve for $c$ $c =2 -1$

STEP 7

Calculate the value of $c$ $c =1$

STEP 8

Now that we have the y-intercept $c$ , we can write the equation of line $r$ as $y = \frac{1}{6}x +1$ The equation of line $r$ is $y = \frac{1}{6}x +1$ .

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QuestionFind the equation of line rrr parallel to y=16x−5y=\frac{1}{6} x-5y=61​x−5 that passes through the point (6,2)(6,2)(6,2).

Studdy Solution

STEP 1

Assumptions1. The equation of line qqq is y=16x−5y=\frac{1}{6} x-5y=61​x−5 . Line rrr is parallel to line qqq3. Line rrr includes the point (6,)(6,)(6,)4. The equation of a line is in the form y=mx+cy = mx + cy=mx+c, where mmm is the slope and ccc is the y-intercept

STEP 2

Since line rrr is parallel to line qqq, they have the same slope. The slope of line qqq is 16\frac{1}{6}61​, so the slope of line rrr is also 16\frac{1}{6}61​.

STEP 3

We can write the equation of line rrr in the form y=mx+cy = mx + cy=mx+c, where mmm is the slope and ccc is the y-intercept. Since we know the slope mmm is 16\frac{1}{6}61​, we can write the equation asy=16x+cy = \frac{1}{6}x + cy=61​x+c

STEP 4

To find the y-intercept ccc, we can substitute the point (6,2)(6,2)(6,2) into the equation. This gives us2=16⋅6+c2 = \frac{1}{6} \cdot6 + c2=61​⋅6+c

STEP 5

olve the equation for ccc2=1+c2 =1 + c2=1+c

STEP 6

Subtract1 from both sides of the equation to solve for cccc=2−1c =2 -1c=2−1

STEP 7

Calculate the value of cccc=1c =1c=1

STEP 8

Now that we have the y-intercept ccc, we can write the equation of line rrr asy=16x+1y = \frac{1}{6}x +1y=61​x+1The equation of line rrr is y=16x+1y = \frac{1}{6}x +1y=61​x+1.

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QuestionFind the equation of line $r$ parallel to $y=\frac{1}{6} x-5$ that passes through the point $(6,2)$ .

Assumptions1. The equation of line $q$ is $y=\frac{1}{6} x-5$ . Line $r$ is parallel to line $q$
3. Line $r$ includes the point $(6,)$ 4. The equation of a line is in the form $y = mx + c$ , where $m$ is the slope and $c$ is the y-intercept

Since line $r$ is parallel to line $q$ , they have the same slope. The slope of line $q$ is $\frac{1}{6}$ , so the slope of line $r$ is also $\frac{1}{6}$ .

We can write the equation of line $r$ in the form $y = mx + c$ , where $m$ is the slope and $c$ is the y-intercept. Since we know the slope $m$ is $\frac{1}{6}$ , we can write the equation as $y = \frac{1}{6}x + c$

To find the y-intercept $c$ , we can substitute the point $(6,2)$ into the equation. This gives us $2 = \frac{1}{6} \cdot6 + c$

olve the equation for $c$ $2 =1 + c$

Subtract1 from both sides of the equation to solve for $c$ $c =2 -1$

Calculate the value of $c$ $c =1$

Now that we have the y-intercept $c$ , we can write the equation of line $r$ as $y = \frac{1}{6}x +1$ The equation of line $r$ is $y = \frac{1}{6}x +1$ .