Algebra Word problem

Question

Write the equation of a line perpendicular to the line 3x+7y=21 with a y-intercept of (0, 2). Write your answer in standard form.
Follow the steps below.

Step 1. Write the equation of the given line in slope-intercept form and determine its slope.
equation:      
slope:      



Step 2. Use the slope of the given line to determine the slope needed to write the equation of a perpendicular line and write a short paragraph explaining how you determined the slope of the perpendicular line.
slope needed for perpendicular line:      
Reason:      



Step 3. Write the equation of the perpendicular line on slope-intercept form and then write the equation in standard form.
slope-intercept form:      
standard form:      

Answer 1

OK. You're asking for a great deal here. Tutors aren't supposed to be doing your homework for you. LOL

 

But let's start you off.

 

3x + 7y = 21 is in standard form, and if it were slope-intercept form that would help you. I got the y term by itself:

 

7y = -3x + 21 then divide by 7 to have y alone:

 

y = -(3/7)x + 3 slope-intercept form. Now, you want the line that's perpendicular to this, and goes through the point (0,2). The slope of the perpendicular line is the *negative reciprocal* of the slope you have. So that's (7/3). The perpendicular line is something like

 

y = (7/3)x + b and if you substitute 0 for x and 2 for y you get 2 for b.

 

y = (7/3)x + 2 is your perpendicular line. In slope-intercept form. Step 1 done.

 

Now, slope is rise/run, right? By definition (which seems vague, I know), -run/rise is a slope perpendicular to rise/run. Step 2 done. If the teacher needs more detail you will have to consult the textbook.

 

Step 3 calls for standard form for the perpendicular line.

 

y = (7/3)x + 2. Get rid of the fractions. Multiply everything by 3.

 

3y = 7x + 6 Now get x and y on the same side. Subtract 7x from both sides.

 

-7x + 3y = 6

 

Sanity check: sketch the lines, see if they LOOK perpendicular. I did this with Wolfram Alpha.