Math / Algebra

QuestionUse the given conditions to write an equation for the line in point-slope form and general form. Passing through $(8,-1)$ and perpendicular to the line whose equation is $x-7 y-9=0$

Studdy Solution

STEP 1

What is this asking? We need to find the equation of a line that's perpendicular to another line and goes through a specific point.
Easy peasy! Watch out! Remember, perpendicular lines have slopes that are negative reciprocals of each other.
Don't forget to flip and change the sign!

STEP 2

1. Find the slope of the given line.
2. Find the slope of the perpendicular line.
3. Write the equation in point-slope form.
4. Write the equation in general form.

STEP 3

Let's rewrite the given equation $x - 7y - 9 = 0$ in slope-intercept form ( $y = mx + b$ ), where $m$ is the slope and $b$ is the y-intercept.
To do this, we want to isolate $y$ .

STEP 4

First, let's add $7y$ to both sides of the equation: $x - 9 = 7y$ .

STEP 5

Now, let's divide both sides by $7$ to isolate $y$ : $\frac{x}{7} - \frac{9}{7} = y$ , or $y = \frac{1}{7}x - \frac{9}{7}$ .

STEP 6

We see that the slope of the given line is $m_1 = \frac{1}{7}$ .
Awesome!

STEP 7

Perpendicular lines have slopes that are negative reciprocals of each other.
So, to find the slope ( $m_2$ ) of the line perpendicular to the given line, we flip the fraction and change the sign.

STEP 8

The negative reciprocal of $\frac{1}{7}$ is $-\frac{7}{1}$ , which simplifies to $-7$ .
So, $m_2 = -7$ .
Perfect!

STEP 9

We know the slope of our perpendicular line ( $m_2 = -7$ ) and a point it passes through $(8, -1)$ .
We can use the point-slope form of a linear equation: $y - y_1 = m(x - x_1)$ , where $(x_1, y_1)$ is the given point and $m$ is the slope.

STEP 10

Substitute the values: $y - (-1) = -7(x - 8)$ .

STEP 11

Simplify: $y + 1 = -7(x - 8)$ .
This is the equation in point-slope form!

STEP 12

The general form of a linear equation is $Ax + By + C = 0$ , where $A$ , $B$ , and $C$ are constants.
Let's convert our point-slope form equation to general form.

STEP 13

Distribute the $-7$ : $y + 1 = -7x + 56$ .

STEP 14

Add $7x$ to both sides: $7x + y + 1 = 56$ .

STEP 15

Subtract $56$ from both sides: $7x + y - 55 = 0$ .
This is the equation in general form!

STEP 16

The equation of the line in point-slope form is $y + 1 = -7(x - 8)$ .
The equation of the line in general form is $7x + y - 55 = 0$ .

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QuestionUse the given conditions to write an equation for the line in point-slope form and general form. Passing through $(8,-1)$ and perpendicular to the line whose equation is $x-7 y-9=0$

Studdy Solution

STEP 1

What is this asking? We need to find the equation of a line that's perpendicular to another line and goes through a specific point.
Easy peasy! Watch out! Remember, perpendicular lines have slopes that are negative reciprocals of each other.
Don't forget to flip and change the sign!

STEP 2

1. Find the slope of the given line.
2. Find the slope of the perpendicular line.
3. Write the equation in point-slope form.
4. Write the equation in general form.

STEP 3

Let's rewrite the given equation $x - 7y - 9 = 0$ in slope-intercept form ( $y = mx + b$ ), where $m$ is the slope and $b$ is the y-intercept.
To do this, we want to isolate $y$ .

STEP 4

First, let's add $7y$ to both sides of the equation: $x - 9 = 7y$ .

STEP 5

Now, let's divide both sides by $7$ to isolate $y$ : $\frac{x}{7} - \frac{9}{7} = y$ , or $y = \frac{1}{7}x - \frac{9}{7}$ .

STEP 6

We see that the slope of the given line is $m_1 = \frac{1}{7}$ .
Awesome!

STEP 7

Perpendicular lines have slopes that are negative reciprocals of each other.
So, to find the slope ( $m_2$ ) of the line perpendicular to the given line, we flip the fraction and change the sign.

STEP 8

The negative reciprocal of $\frac{1}{7}$ is $-\frac{7}{1}$ , which simplifies to $-7$ .
So, $m_2 = -7$ .
Perfect!

STEP 9

We know the slope of our perpendicular line ( $m_2 = -7$ ) and a point it passes through $(8, -1)$ .
We can use the point-slope form of a linear equation: $y - y_1 = m(x - x_1)$ , where $(x_1, y_1)$ is the given point and $m$ is the slope.

STEP 10

Substitute the values: $y - (-1) = -7(x - 8)$ .

STEP 11

Simplify: $y + 1 = -7(x - 8)$ .
This is the equation in point-slope form!

STEP 12

The general form of a linear equation is $Ax + By + C = 0$ , where $A$ , $B$ , and $C$ are constants.
Let's convert our point-slope form equation to general form.

STEP 13

Distribute the $-7$ : $y + 1 = -7x + 56$ .

STEP 14

Add $7x$ to both sides: $7x + y + 1 = 56$ .

STEP 15

Subtract $56$ from both sides: $7x + y - 55 = 0$ .
This is the equation in general form!

STEP 16

The equation of the line in point-slope form is $y + 1 = -7(x - 8)$ .
The equation of the line in general form is $7x + y - 55 = 0$ .

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QuestionUse the given conditions to write an equation for the line in point-slope form and general form. Passing through (8,−1)(8,-1)(8,−1) and perpendicular to the line whose equation is x−7y−9=0x-7 y-9=0x−7y−9=0

Studdy Solution

STEP 1

What is this asking? We need to find the equation of a line that's perpendicular to another line and goes through a specific point.Easy peasy! Watch out! Remember, perpendicular lines have slopes that are *negative reciprocals* of each other.Don't forget to flip and change the sign!

STEP 2

1. Find the slope of the given line.2. Find the slope of the perpendicular line.3. Write the equation in point-slope form.4. Write the equation in general form.

STEP 3

Let's **rewrite** the given equation x−7y−9=0x - 7y - 9 = 0x−7y−9=0 in slope-intercept form (y=mx+by = mx + by=mx+b), where mmm is the **slope** and bbb is the **y-intercept**.To do this, we want to isolate yyy.

STEP 4

First, let's **add** 7y7y7y to both sides of the equation: x−9=7yx - 9 = 7yx−9=7y.

STEP 5

Now, let's **divide** both sides by 777 to isolate yyy: x7−97=y\frac{x}{7} - \frac{9}{7} = y7x​−79​=y, or y=17x−97y = \frac{1}{7}x - \frac{9}{7}y=71​x−79​.

STEP 6

We see that the **slope** of the given line is m1=17m_1 = \frac{1}{7}m1​=71​.Awesome!

STEP 7

Perpendicular lines have slopes that are *negative reciprocals* of each other.So, to find the slope (m2m_2m2​) of the line perpendicular to the given line, we **flip** the fraction and **change** the sign.

STEP 8

The **negative reciprocal** of 17\frac{1}{7}71​ is −71-\frac{7}{1}−17​, which simplifies to −7-7−7.So, m2=−7m_2 = -7m2​=−7.Perfect!

STEP 9

STEP 10

**Substitute** the values: y−(−1)=−7(x−8)y - (-1) = -7(x - 8)y−(−1)=−7(x−8).

STEP 11

**Simplify**: y+1=−7(x−8)y + 1 = -7(x - 8)y+1=−7(x−8).This is the equation in **point-slope form**!

STEP 12

The **general form** of a linear equation is Ax+By+C=0Ax + By + C = 0Ax+By+C=0, where AAA, BBB, and CCC are constants.Let's convert our point-slope form equation to general form.

STEP 13

**Distribute** the −7-7−7: y+1=−7x+56y + 1 = -7x + 56y+1=−7x+56.

STEP 14

**Add** 7x7x7x to both sides: 7x+y+1=567x + y + 1 = 567x+y+1=56.

STEP 15

**Subtract** 565656 from both sides: 7x+y−55=07x + y - 55 = 07x+y−55=0.This is the equation in **general form**!

STEP 16

The equation of the line in **point-slope form** is y+1=−7(x−8)y + 1 = -7(x - 8)y+1=−7(x−8).The equation of the line in **general form** is 7x+y−55=07x + y - 55 = 07x+y−55=0.

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QuestionUse the given conditions to write an equation for the line in point-slope form and general form. Passing through $(8,-1)$ and perpendicular to the line whose equation is $x-7 y-9=0$

What is this asking? We need to find the equation of a line that's perpendicular to another line and goes through a specific point.
Easy peasy! Watch out! Remember, perpendicular lines have slopes that are negative reciprocals of each other.
Don't forget to flip and change the sign!

1. Find the slope of the given line.
2. Find the slope of the perpendicular line.
3. Write the equation in point-slope form.
4. Write the equation in general form.

Let's rewrite the given equation $x - 7y - 9 = 0$ in slope-intercept form ( $y = mx + b$ ), where $m$ is the slope and $b$ is the y-intercept.
To do this, we want to isolate $y$ .

First, let's add $7y$ to both sides of the equation: $x - 9 = 7y$ .

Now, let's divide both sides by $7$ to isolate $y$ : $\frac{x}{7} - \frac{9}{7} = y$ , or $y = \frac{1}{7}x - \frac{9}{7}$ .

We see that the slope of the given line is $m_1 = \frac{1}{7}$ .
Awesome!

Perpendicular lines have slopes that are negative reciprocals of each other.
So, to find the slope ( $m_2$ ) of the line perpendicular to the given line, we flip the fraction and change the sign.

The negative reciprocal of $\frac{1}{7}$ is $-\frac{7}{1}$ , which simplifies to $-7$ .
So, $m_2 = -7$ .
Perfect!

Substitute the values: $y - (-1) = -7(x - 8)$ .

Simplify: $y + 1 = -7(x - 8)$ .
This is the equation in point-slope form!

The general form of a linear equation is $Ax + By + C = 0$ , where $A$ , $B$ , and $C$ are constants.
Let's convert our point-slope form equation to general form.

Distribute the $-7$ : $y + 1 = -7x + 56$ .

Add $7x$ to both sides: $7x + y + 1 = 56$ .

Subtract $56$ from both sides: $7x + y - 55 = 0$ .
This is the equation in general form!

The equation of the line in point-slope form is $y + 1 = -7(x - 8)$ .
The equation of the line in general form is $7x + y - 55 = 0$ .