<PRE><B>Question 1201857</B>
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Part (a)


The line is of the form y = mx+b
m = slope = -5/6
b = y intercept = 9


The slope tells us how to go from one point to another on the line.
It is the direction vector.


slope = rise/run
rise/run = -5/6
rise = -5
run = 6


The rise tells us how much to move up or down.
In this case we go down 5. This is the change in y.
The run is the change in x. We go 6 units to the right.



Answer: &lt;font color=red size=4&gt; &lt; 6,-5 &gt; &lt;/font&gt; 


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Part (b)


Swap the coordinate positions of the direction vector. 
Then flip the sign of exactly one coordinate. 


original = &lt; 6,-5 &gt;
swapped = &lt; -5,6 &gt;
change sign of x coord = &lt; 5,6 &gt;
OR
change sign of y coord = &lt; -5,-6 &gt;


You can use the dot product to confirm vectors &lt; 6,-5 &gt; and &lt; 5,6 &gt; are perpendicular (same goes for &lt; 6,-5 &gt; and &lt; -5,-6 &gt; being perpendicular).
If u dot v = 0, then u is perpendicular to v. 



Answer: &lt;font color=red size=4&gt; &lt; 5,6 &gt; &lt;/font&gt;  or &lt;font color=red size=4&gt; &lt; -5,-6 &gt; &lt;/font&gt; 


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Part (c)


Plug in x = 0 and find y
y = (-5/6)*x + 9
y = (-5/6)*0 + 9
y = 0 + 9
y = 9
The point (0,9) is on this line


Repeat for x = 6
y = (-5/6)*x + 9
y = (-5/6)*6 + 9
y = -5 + 9
y = 4
The point (6,4) is also on the line


The movement from (0,9) to (6,4) is &quot;down 5, right 6&quot;
Or along the direction vector &lt; 6,-5 &gt; which means &quot;go right 6, then down 5&quot;.
There are infinitely many points on this line.



Answer: &lt;font color=red size=4&gt;(0,9)&lt;/font&gt; 


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Part (d)


Vector form of the line:
&lt; x,y &gt; = startVector + t*DirectionVector
&lt; x,y &gt; = &lt; 0,9 &gt; + t*&lt; 6,-5 &gt;
The start vector could be any other vector you want, as long as it&#39;s on the line. 
So you could pick &lt; 6,4 &gt; for instance.


Now we need to find the vector equation of a line parallel to what was mentioned, but goes through (7,9)
The start vector will be &lt; 7,9 &gt;
The direction vector is the same. 
Parallel vectors are equal or scalar multiples of one another. They point in the same direction (eg: northeast).


Vector form of the parallel line:
&lt; x,y &gt; = startVector + t*DirectionVector
&lt; x,y &gt; = &lt; 7,9 &gt; + t*&lt; 6,-5 &gt;


Now rewrite things a bit like so
&lt; x,y &gt; = &lt; 7,9 &gt; + t*&lt; 6,-5 &gt;
&lt; x,y &gt; = &lt; 7,9 &gt; + &lt; 6t,-5t &gt;
&lt; x,y &gt; = &lt; 7+6t,9-5t &gt;
That breaks down into
x = 7+6t
y = 9-5t
Both of which form a system of equations to define the parametric form of the parallel line.
The t is any real number. It can be thought of as the time value.


What happens at t = 0?
(x,y) = (7+6t,9-5t)
(x,y) = (7+6*0,9-5*0)
(x,y) = (7,9)
Which confirms (7,9) is on the parallel line



Answers: 
&lt;table border = &quot;1&quot; cellpadding = &quot;5&quot;&gt;&lt;tr&gt;&lt;td&gt;Vector form:&lt;/td&gt;&lt;td&gt;&lt;font color=red size=4&gt;&lt; x,y &gt; = &lt; 7,9 &gt; + t*&lt; 6,-5 &gt; &lt;/font&gt; &lt;/td&gt;&lt;/tr&gt;&lt;tr&gt;&lt;td&gt;Parametric form:&lt;/td&gt;&lt;td&gt;&lt;font color=red size=4&gt;x = 7+6t&lt;/br&gt;y = 9-5t&lt;/font&gt;&lt;/td&gt;&lt;/tr&gt;&lt;/table&gt;


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Part (e)


Original direction vector = &lt; 6,-5 &gt;
Perpendicular direction vector = &lt; 5,6 &gt;
Refer to part (b)


Vector form of the perpendicular line:
&lt; x,y &gt; = startVector + t*DirectionVector
&lt; x,y &gt; = &lt; -2,1 &gt; + t*&lt; 5,6 &gt;
&lt; x,y &gt; = &lt; -2,1 &gt; + &lt; 5t,6t &gt;
&lt; x,y &gt; = &lt; -2+5t,1+6t &gt;


Any point on this perpendicular line through (-2,1) is of the general form (x,y) = (-2+5t, 1+6t) where t is any real number. 
Plug t = 0 to find (x,y) = (-2,1)



Answers: 
&lt;table border = &quot;1&quot; cellpadding = &quot;5&quot;&gt;&lt;tr&gt;&lt;td&gt;Vector form:&lt;/td&gt;&lt;td&gt;&lt;font color=red size=4&gt;&lt; x,y &gt; = &lt; -2,1 &gt; + t*&lt; 5,6 &gt; &lt;/font&gt; &lt;/td&gt;&lt;/tr&gt;&lt;tr&gt;&lt;td&gt;Parametric form:&lt;/td&gt;&lt;td&gt;&lt;font color=red size=4&gt;x = -2+5t&lt;/br&gt;y = 1+6t&lt;/font&gt;&lt;/td&gt;&lt;/tr&gt;&lt;/table&gt;
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