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 20:04, 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 . |