Math / Algebra

Question10. Write the equation of a line parallel to the given line that contains point C . a) $C(-2,1) ; y=2 x-8$ b) $C(2,-5) ; y=-x+4$

Studdy Solution

STEP 1

What is this asking? We need to find the equations of lines that are parallel to the given lines and pass through point C. Watch out! Remember, parallel lines have the same slope!
Don't mix up parallel and perpendicular lines.

STEP 2

1. Find the slope.
2. Use point-slope form.
3. Convert to slope-intercept form.

STEP 3

For part (a), the given line is $y = 2x - 8$ .
The slope here is the coefficient of $x$ , which is 2.
Since parallel lines have the same slope, our new line will also have a slope of 2!

STEP 4

For part (b), the given line is $y = -x + 4$ .
This can be rewritten as $y = -1 \cdot x + 4$ .
The slope of this line is -1.
Our parallel line will also have a slope of -1!

STEP 5

Remember, the point-slope form of a line is $y - y_1 = m(x - x_1)$ , where $m$ is the slope and $(x_1, y_1)$ is a point on the line.
For part (a), our slope is $m = 2$ and our point is $C(-2, 1)$ , so $x_1 = -2$ and $y_1 = 1$ .
Plugging these values into the point-slope form, we get $y - 1 = 2(x - (-2))$ .

STEP 6

For part (b), our slope is $m = -1$ and our point is $C(2, -5)$ , so $x_1 = 2$ and $y_1 = -5$ .
Plugging these values into the point-slope form gives us $y - (-5) = -1(x - 2)$ .

STEP 7

Let's simplify part (a)'s equation: $y - 1 = 2(x + 2)$ .
Distributing the 2, we get $y - 1 = 2x + 4$ .
Adding 1 to both sides gives us $y = 2x + 5$ .
This is our final equation for part (a)!

STEP 8

Now for part (b): $y - (-5) = -1(x - 2)$ simplifies to $y + 5 = -x + 2$ .
Subtracting 5 from both sides, we get $y = -x - 3$ .
This is our final equation for part (b)!

STEP 9

a) $y = 2x + 5$ b) $y = -x - 3$

Was this helpful?

Question10. Write the equation of a line parallel to the given line that contains point C . a) $C(-2,1) ; y=2 x-8$ b) $C(2,-5) ; y=-x+4$

Studdy Solution

STEP 1

What is this asking? We need to find the equations of lines that are parallel to the given lines and pass through point C. Watch out! Remember, parallel lines have the same slope!
Don't mix up parallel and perpendicular lines.

STEP 2

1. Find the slope.
2. Use point-slope form.
3. Convert to slope-intercept form.

STEP 3

For part (a), the given line is $y = 2x - 8$ .
The slope here is the coefficient of $x$ , which is 2.
Since parallel lines have the same slope, our new line will also have a slope of 2!

STEP 4

For part (b), the given line is $y = -x + 4$ .
This can be rewritten as $y = -1 \cdot x + 4$ .
The slope of this line is -1.
Our parallel line will also have a slope of -1!

STEP 5

STEP 6

For part (b), our slope is $m = -1$ and our point is $C(2, -5)$ , so $x_1 = 2$ and $y_1 = -5$ .
Plugging these values into the point-slope form gives us $y - (-5) = -1(x - 2)$ .

STEP 7

Let's simplify part (a)'s equation: $y - 1 = 2(x + 2)$ .
Distributing the 2, we get $y - 1 = 2x + 4$ .
Adding 1 to both sides gives us $y = 2x + 5$ .
This is our final equation for part (a)!

STEP 8

Now for part (b): $y - (-5) = -1(x - 2)$ simplifies to $y + 5 = -x + 2$ .
Subtracting 5 from both sides, we get $y = -x - 3$ .
This is our final equation for part (b)!

STEP 9

a) $y = 2x + 5$ b) $y = -x - 3$

Get Studdy

Studdy solves anything!

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

Question10. Write the equation of a line parallel to the given line that contains point C . a) C(−2,1);y=2x−8C(-2,1) ; y=2 x-8C(−2,1);y=2x−8 b) C(2,−5);y=−x+4C(2,-5) ; y=-x+4C(2,−5);y=−x+4

Studdy Solution

STEP 1

What is this asking? We need to find the equations of lines that are parallel to the given lines and pass through point C. Watch out! Remember, parallel lines have the *same* slope!Don't mix up parallel and perpendicular lines.

STEP 2

1. Find the slope.2. Use point-slope form.3. Convert to slope-intercept form.

STEP 3

For part (a), the given line is y=2x−8y = 2x - 8y=2x−8.The **slope** here is the coefficient of xxx, which is **2**.Since parallel lines have the *same* slope, our new line will *also* have a slope of **2**!

STEP 4

For part (b), the given line is y=−x+4y = -x + 4y=−x+4.This can be rewritten as y=−1⋅x+4y = -1 \cdot x + 4y=−1⋅x+4.The **slope** of this line is **-1**.Our parallel line will *also* have a slope of **-1**!

STEP 5

STEP 6

For part (b), our **slope** is m=−1m = -1m=−1 and our **point** is C(2,−5)C(2, -5)C(2,−5), so x1=2x_1 = 2x1​=2 and y1=−5y_1 = -5y1​=−5.Plugging these values into the point-slope form gives us y−(−5)=−1(x−2)y - (-5) = -1(x - 2)y−(−5)=−1(x−2).

STEP 7

Let's simplify part (a)'s equation: y−1=2(x+2)y - 1 = 2(x + 2)y−1=2(x+2).Distributing the 2, we get y−1=2x+4y - 1 = 2x + 4y−1=2x+4.Adding 1 to both sides gives us y=2x+5y = 2x + 5y=2x+5.This is our **final equation** for part (a)!

STEP 8

Now for part (b): y−(−5)=−1(x−2)y - (-5) = -1(x - 2)y−(−5)=−1(x−2) simplifies to y+5=−x+2y + 5 = -x + 2y+5=−x+2.Subtracting 5 from both sides, we get y=−x−3y = -x - 3y=−x−3.This is our **final equation** for part (b)!

STEP 9

a) y=2x+5y = 2x + 5y=2x+5 b) y=−x−3y = -x - 3y=−x−3

Get Studdy

Studdy solves anything!

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

Question10. Write the equation of a line parallel to the given line that contains point C . a) $C(-2,1) ; y=2 x-8$ b) $C(2,-5) ; y=-x+4$

What is this asking? We need to find the equations of lines that are parallel to the given lines and pass through point C. Watch out! Remember, parallel lines have the same slope!
Don't mix up parallel and perpendicular lines.

1. Find the slope.
2. Use point-slope form.
3. Convert to slope-intercept form.

For part (a), the given line is $y = 2x - 8$ .
The slope here is the coefficient of $x$ , which is 2.
Since parallel lines have the same slope, our new line will also have a slope of 2!

For part (b), the given line is $y = -x + 4$ .
This can be rewritten as $y = -1 \cdot x + 4$ .
The slope of this line is -1.
Our parallel line will also have a slope of -1!

For part (b), our slope is $m = -1$ and our point is $C(2, -5)$ , so $x_1 = 2$ and $y_1 = -5$ .
Plugging these values into the point-slope form gives us $y - (-5) = -1(x - 2)$ .

Let's simplify part (a)'s equation: $y - 1 = 2(x + 2)$ .
Distributing the 2, we get $y - 1 = 2x + 4$ .
Adding 1 to both sides gives us $y = 2x + 5$ .
This is our final equation for part (a)!

Now for part (b): $y - (-5) = -1(x - 2)$ simplifies to $y + 5 = -x + 2$ .
Subtracting 5 from both sides, we get $y = -x - 3$ .
This is our final equation for part (b)!

a) $y = 2x + 5$ b) $y = -x - 3$