Difference between revisions of "004 Sample Final A, Problem 14"
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Kayla Murray (talk | contribs) (Created page with "<span class="exam"> a) Find an equation of the line passing through (-4, 2) and (3, 6).<br> b) Find the slope of any line perpendicular to your answer from a)") |
Kayla Murray (talk | contribs) |
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<span class="exam"> a) Find an equation of the line passing through (-4, 2) and (3, 6).<br> | <span class="exam"> a) Find an equation of the line passing through (-4, 2) and (3, 6).<br> | ||
b) Find the slope of any line perpendicular to your answer from a) | b) Find the slope of any line perpendicular to your answer from a) | ||
+ | {| class="mw-collapsible mw-collapsed" style = "text-align:left;" | ||
+ | ! Foundations | ||
+ | |- | ||
+ | |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>? | ||
+ | |- | ||
+ | |2) What is the equation of a line? | ||
+ | |- | ||
+ | |3) How do you find the slope of a line perpendicular to a line <math> y </math>? | ||
+ | |- | ||
+ | |Answer: | ||
+ | |- | ||
+ | |1) The slope is given by <math>m=\frac{y_2-y_1}{x_2-x_1} </math>. | ||
+ | |- | ||
+ | |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. | ||
+ | |- | ||
+ | |3) The slope is given by <math>-\frac{1}{m}</math> where <math>m</math> is the slope of the line <math>y</math>. | ||
+ | |} | ||
+ | |||
+ | |||
+ | Solution: | ||
+ | |||
+ | {| class = "mw-collapsible mw-collapsed" style = "text-align:left;" | ||
+ | ! Step 1: | ||
+ | |- | ||
+ | |Using the above equation, the slope is equal to <math> m=\frac{6-2}{3-(-4)}=\frac{4}{7}</math>. | ||
+ | |- | ||
+ | | | ||
+ | |} | ||
+ | |||
+ | {|class = "mw-collapsible mw-collapsed" style = "text-align:left;" | ||
+ | ! Step 2: | ||
+ | |- | ||
+ | |The equation of the line is <math> y-6=\frac{4}{7}(x-3)</math>. Solving for <math>y</math>, | ||
+ | |- | ||
+ | |we get <math>y=\frac{4}{7}x+\frac{30}{7}</math>. | ||
+ | |} | ||
+ | |||
+ | {|class = "mw-collapsible mw-collapsed" style = "text-align:left;" | ||
+ | ! Step 3: | ||
+ | |- | ||
+ | |The slope of any line perpendicular to the line in Step 2 is <math>-\frac{1}{(\frac{4}{7})}=-\frac{7}{4}</math>. | ||
+ | |} | ||
+ | |||
+ | {|class = "mw-collapsible mw-collapsed" style = "text-align:left;" | ||
+ | ! Final Answer: | ||
+ | |- | ||
+ | | 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 | ||
+ | |- | ||
+ | |the slope of any line perpendicular to this line is <math>-\frac{7}{4}</math>. | ||
+ | |} | ||
+ | |||
+ | [[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 . |