4. Perpendicular to x - 3y = 2 and passes thru (-3,5)

Algebra and Trigonometry (6th Edition)
6th Edition
ISBN:9780134463216
Author:Robert F. Blitzer
Publisher:Robert F. Blitzer
ChapterP: Prerequisites: Fundamental Concepts Of Algebra
Section: Chapter Questions
Problem 1MCCP: In Exercises 1-25, simplify the given expression or perform the indicated operation (and simplify,...
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**Transcription for Educational Website:**

4. Perpendicular to \( x - 3y = 2 \) and passes through \((-3, 5)\)

5. Parallel to the \( x \)-axis and through \((-7, 3)\)

6. Parallel to the \( y \)-axis and through \((-7, 3)\)

**Explanation:**

- **Problem 4** involves finding a line perpendicular to a given line, \( x - 3y = 2 \), that passes through the point \((-3, 5)\).

- **Problem 5** requires determining a line parallel to the \( x \)-axis, passing through the point \((-7, 3)\). Lines parallel to the \( x \)-axis have a slope of 0.

- **Problem 6** demands finding a line parallel to the \( y \)-axis, passing through the point \((-7, 3)\). Lines parallel to the \( y \)-axis are vertical lines with an undefined slope.
Transcribed Image Text:**Transcription for Educational Website:** 4. Perpendicular to \( x - 3y = 2 \) and passes through \((-3, 5)\) 5. Parallel to the \( x \)-axis and through \((-7, 3)\) 6. Parallel to the \( y \)-axis and through \((-7, 3)\) **Explanation:** - **Problem 4** involves finding a line perpendicular to a given line, \( x - 3y = 2 \), that passes through the point \((-3, 5)\). - **Problem 5** requires determining a line parallel to the \( x \)-axis, passing through the point \((-7, 3)\). Lines parallel to the \( x \)-axis have a slope of 0. - **Problem 6** demands finding a line parallel to the \( y \)-axis, passing through the point \((-7, 3)\). Lines parallel to the \( y \)-axis are vertical lines with an undefined slope.
Expert Solution
Step 1

given that a line is perpendicular to the line x-3y=2 and passes through the point (-3,5).

we need to find the equation of that line.

If two lines with slope m1 and m2 are perpendicular then products of their slope is equal to -1.

that is  

m1.m2=-1.

so let us find the slope for the given line.

the given line is 

x-3y=23y=x-2y=x-23y=x3-23    ------(1)

the equation is of the form y=mx+c (standard equation for slope intercept form)

where m is the slope of the line.

So from the equation 1 m1=1/3.

 

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