Difference between revisions of "004 Sample Final A, Problem 14"

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! Foundations
 
! Foundations
 
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|1) How do you find the slope of a line through points <math>(x_1,y_1)</math> and <math>(x_2,y_2)</math>?
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|2) What is the equation of a line?
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|3) How do you find the slope of a line perpendicular to a line <math> y </math>?
 
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|Answer:
 
|Answer:
 
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|1) The slope is given by <math>m=\frac{y_2-y_1}{x_2-x_1} </math>.
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|2) The equation of a line is <math>y-y_1=m(x-x_1)</math> where <math>(x_1,y_1)</math> is a point on the line.
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|3) The slope is given by <math>-\frac{1}{m}</math> where <math>m</math> is the slope of the line <math>y</math>. 
 
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! Step 1:
 
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|Using the above equation, the slope is equal to <math> m=\frac{6-2}{3-(-4)}=\frac{4}{7}</math>.
 
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! Step 2:
 
! Step 2:
 
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|The equation of the line is <math> y-6=\frac{4}{7}(x-3)</math>. Solving for <math>y</math>,
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|we get <math>y=\frac{4}{7}x+\frac{30}{7}</math>.
 
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! Step 3:
 
! Step 3:
 
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|The slope of any line perpendicular to the line in Step 2 is <math>-\frac{1}{(\frac{4}{7})}=-\frac{7}{4}</math>.
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{|class = "mw-collapsible mw-collapsed" style = "text-align:left;"
 
{|class = "mw-collapsible mw-collapsed" style = "text-align:left;"
! Step 4:
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! Final Answer:
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| The slope is <math> \frac{4}{7}</math>, the equation of the line is <math>y=\frac{4}{7}x+\frac{30}{7}</math>, and
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{|class = "mw-collapsible mw-collapsed" style = "text-align:left;"
 
! Final Answer:
 
 
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|the slope of any line perpendicular to this line is <math>-\frac{7}{4}</math>.
 
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[[004 Sample Final A|<u>'''Return to Sample Exam</u>''']]
 
[[004 Sample Final A|<u>'''Return to Sample Exam</u>''']]

Latest revision as of 21:04, 3 May 2015

a) Find an equation of the line passing through (-4, 2) and (3, 6).
b) Find the slope of any line perpendicular to your answer from a)

Foundations
1) How do you find the slope of a line through points and ?
2) What is the equation of a line?
3) How do you find the slope of a line perpendicular to a line ?
Answer:
1) The slope is given by .
2) The equation of a line is where is a point on the line.
3) The slope is given by where is the slope of the line .


Solution:

Step 1:
Using the above equation, the slope is equal to .
Step 2:
The equation of the line is . Solving for ,
we get .
Step 3:
The slope of any line perpendicular to the line in Step 2 is .
Final Answer:
The slope is , the equation of the line is , and
the slope of any line perpendicular to this line is .

Return to Sample Exam