Homework: Equation Of Perpendicular Line (Part I)Write An Equation Of A Line Perpendicular To The Given Line And Passing Through The Given Point. Show Your Complete Solution.1. ${ 3x - 2y = 8\$} ; (6, -4)2. { X - 4y = -20$}$; (12,

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Homework: Equation of Perpendicular Line (Part I)

In this article, we will explore the concept of finding the equation of a line perpendicular to a given line and passing through a given point. This is a fundamental concept in mathematics, particularly in geometry and algebra. We will use the given equations of lines and points to derive the equations of perpendicular lines.

To find the equation of a line perpendicular to a given line and passing through a given point, we need to follow these steps:

  1. Find the slope of the given line.
  2. Find the slope of the perpendicular line using the negative reciprocal of the slope of the given line.
  3. Use the point-slope form of a line to find the equation of the perpendicular line.

Problem 1: Equation of Perpendicular Line

Given Line

The given line is represented by the equation:

3xβˆ’2y=83x - 2y = 8

Given Point

The given point is (6, -4).

Step 1: Find the Slope of the Given Line

To find the slope of the given line, we need to rewrite the equation in the slope-intercept form (y = mx + b), where m is the slope.

3xβˆ’2y=83x - 2y = 8

βˆ’2y=βˆ’3x+8-2y = -3x + 8

y=32xβˆ’4y = \frac{3}{2}x - 4

The slope of the given line is 32\frac{3}{2}.

Step 2: Find the Slope of the Perpendicular Line

The slope of the perpendicular line is the negative reciprocal of the slope of the given line.

mperpendicular=βˆ’1mgivenm_{perpendicular} = -\frac{1}{m_{given}}

mperpendicular=βˆ’132m_{perpendicular} = -\frac{1}{\frac{3}{2}}

mperpendicular=βˆ’23m_{perpendicular} = -\frac{2}{3}

Step 3: Find the Equation of the Perpendicular Line

Using the point-slope form of a line (y - y1 = m(x - x1)), where (x1, y1) is the given point and m is the slope of the perpendicular line, we can find the equation of the perpendicular line.

yβˆ’y1=m(xβˆ’x1)y - y1 = m(x - x1)

yβˆ’(βˆ’4)=βˆ’23(xβˆ’6)y - (-4) = -\frac{2}{3}(x - 6)

y+4=βˆ’23x+4y + 4 = -\frac{2}{3}x + 4

y=βˆ’23xy = -\frac{2}{3}x

The equation of the perpendicular line is y=βˆ’23xy = -\frac{2}{3}x.

Problem 2: Equation of Perpendicular Line

Given Line

The given line is represented by the equation:

xβˆ’4y=βˆ’20x - 4y = -20

Given Point

The given point is (12, -3).

Step 1: Find the Slope of the Given Line

To find the slope of the given line, we need to rewrite the equation in the slope-intercept form (y = mx + b), where m is the slope.

xβˆ’4y=βˆ’20x - 4y = -20

βˆ’4y=βˆ’xβˆ’20-4y = -x - 20

y=14x+5y = \frac{1}{4}x + 5

The slope of the given line is 14\frac{1}{4}.

Step 2: Find the Slope of the Perpendicular Line

The slope of the perpendicular line is the negative reciprocal of the slope of the given line.

mperpendicular=βˆ’1mgivenm_{perpendicular} = -\frac{1}{m_{given}}

mperpendicular=βˆ’114m_{perpendicular} = -\frac{1}{\frac{1}{4}}

mperpendicular=βˆ’4m_{perpendicular} = -4

Step 3: Find the Equation of the Perpendicular Line

Using the point-slope form of a line (y - y1 = m(x - x1)), where (x1, y1) is the given point and m is the slope of the perpendicular line, we can find the equation of the perpendicular line.

yβˆ’y1=m(xβˆ’x1)y - y1 = m(x - x1)

yβˆ’(βˆ’3)=βˆ’4(xβˆ’12)y - (-3) = -4(x - 12)

y+3=βˆ’4x+48y + 3 = -4x + 48

y=βˆ’4x+45y = -4x + 45

The equation of the perpendicular line is y=βˆ’4x+45y = -4x + 45.

In this article, we have explored the concept of finding the equation of a line perpendicular to a given line and passing through a given point. We have used the given equations of lines and points to derive the equations of perpendicular lines. The methodology involves finding the slope of the given line, finding the slope of the perpendicular line using the negative reciprocal of the slope of the given line, and using the point-slope form of a line to find the equation of the perpendicular line. We have applied this methodology to two problems and derived the equations of perpendicular lines.
Homework: Equation of Perpendicular Line (Part II) - Q&A

In the previous article, we explored the concept of finding the equation of a line perpendicular to a given line and passing through a given point. We used the given equations of lines and points to derive the equations of perpendicular lines. In this article, we will provide a Q&A section to help clarify any doubts and provide additional examples.

Q1: What is the slope of a line perpendicular to a given line?

A1: The slope of a line perpendicular to a given line is the negative reciprocal of the slope of the given line.

Q2: How do I find the equation of a line perpendicular to a given line and passing through a given point?

A2: To find the equation of a line perpendicular to a given line and passing through a given point, you need to follow these steps:

  1. Find the slope of the given line.
  2. Find the slope of the perpendicular line using the negative reciprocal of the slope of the given line.
  3. Use the point-slope form of a line to find the equation of the perpendicular line.

Q3: What is the point-slope form of a line?

A3: The point-slope form of a line is given by the equation:

yβˆ’y1=m(xβˆ’x1)y - y1 = m(x - x1)

where (x1, y1) is the given point and m is the slope of the line.

Q4: How do I find the equation of a line perpendicular to a given line and passing through a given point if the given line is in the form ax + by = c?

A4: To find the equation of a line perpendicular to a given line and passing through a given point if the given line is in the form ax + by = c, you need to follow these steps:

  1. Rewrite the equation in the slope-intercept form (y = mx + b), where m is the slope.
  2. Find the slope of the perpendicular line using the negative reciprocal of the slope of the given line.
  3. Use the point-slope form of a line to find the equation of the perpendicular line.

Q5: Can I find the equation of a line perpendicular to a given line and passing through a given point if the given line is in the form y = mx + b?

A5: Yes, you can find the equation of a line perpendicular to a given line and passing through a given point if the given line is in the form y = mx + b. You can use the point-slope form of a line to find the equation of the perpendicular line.

Example 1: Equation of Perpendicular Line

Given Line

The given line is represented by the equation:

2x+3y=62x + 3y = 6

Given Point

The given point is (3, 1).

Step 1: Find the Slope of the Given Line

To find the slope of the given line, we need to rewrite the equation in the slope-intercept form (y = mx + b), where m is the slope.

2x+3y=62x + 3y = 6

3y=βˆ’2x+63y = -2x + 6

y=βˆ’23x+2y = -\frac{2}{3}x + 2

The slope of the given line is βˆ’23-\frac{2}{3}.

Step 2: Find the Slope of the Perpendicular Line

The slope of the perpendicular line is the negative reciprocal of the slope of the given line.

mperpendicular=βˆ’1mgivenm_{perpendicular} = -\frac{1}{m_{given}}

mperpendicular=βˆ’1βˆ’23m_{perpendicular} = -\frac{1}{-\frac{2}{3}}

mperpendicular=32m_{perpendicular} = \frac{3}{2}

Step 3: Find the Equation of the Perpendicular Line

Using the point-slope form of a line (y - y1 = m(x - x1)), where (x1, y1) is the given point and m is the slope of the perpendicular line, we can find the equation of the perpendicular line.

yβˆ’y1=m(xβˆ’x1)y - y1 = m(x - x1)

yβˆ’1=32(xβˆ’3)y - 1 = \frac{3}{2}(x - 3)

yβˆ’1=32xβˆ’92y - 1 = \frac{3}{2}x - \frac{9}{2}

y=32xβˆ’72y = \frac{3}{2}x - \frac{7}{2}

The equation of the perpendicular line is y=32xβˆ’72y = \frac{3}{2}x - \frac{7}{2}.

Example 2: Equation of Perpendicular Line

Given Line

The given line is represented by the equation:

y=2xβˆ’1y = 2x - 1

Given Point

The given point is (2, 3).

Step 1: Find the Slope of the Given Line

The slope of the given line is 2.

Step 2: Find the Slope of the Perpendicular Line

The slope of the perpendicular line is the negative reciprocal of the slope of the given line.

mperpendicular=βˆ’1mgivenm_{perpendicular} = -\frac{1}{m_{given}}

mperpendicular=βˆ’12m_{perpendicular} = -\frac{1}{2}

Step 3: Find the Equation of the Perpendicular Line

Using the point-slope form of a line (y - y1 = m(x - x1)), where (x1, y1) is the given point and m is the slope of the perpendicular line, we can find the equation of the perpendicular line.

yβˆ’y1=m(xβˆ’x1)y - y1 = m(x - x1)

yβˆ’3=βˆ’12(xβˆ’2)y - 3 = -\frac{1}{2}(x - 2)

yβˆ’3=βˆ’12x+1y - 3 = -\frac{1}{2}x + 1

y=βˆ’12x+4y = -\frac{1}{2}x + 4

The equation of the perpendicular line is y=βˆ’12x+4y = -\frac{1}{2}x + 4.

In this article, we have provided a Q&A section to help clarify any doubts and provide additional examples on finding the equation of a line perpendicular to a given line and passing through a given point. We have used the given equations of lines and points to derive the equations of perpendicular lines. The methodology involves finding the slope of the given line, finding the slope of the perpendicular line using the negative reciprocal of the slope of the given line, and using the point-slope form of a line to find the equation of the perpendicular line.