Question 1201857: from my 'Vector Equation of a Line' lesson
A line has the equation y = (-5/6)x + 9
a. Give a direction vector for a line that is parallel to this line.
b. Give a direction vector for a line that is perpendicular to this line.
c. Give the coordinates of a point on the given line.
d. In both vector and parametric form, give the equations of the line parallel to the given line and passing through A (7,9)
e. In both vector and parametric form, give the equations of the line perpendicular to the given line and passing through B (-2,1)
Answer by math_tutor2020(3817)   (Show Source):
You can put this solution on YOUR website!
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: < 6,-5 >
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Part (b)
Swap the coordinate positions of the direction vector.
Then flip the sign of exactly one coordinate.
original = < 6,-5 >
swapped = < -5,6 >
change sign of x coord = < 5,6 >
OR
change sign of y coord = < -5,-6 >
You can use the dot product to confirm vectors < 6,-5 > and < 5,6 > are perpendicular (same goes for < 6,-5 > and < -5,-6 > being perpendicular).
If u dot v = 0, then u is perpendicular to v.
Answer: < 5,6 > or < -5,-6 >
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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 "down 5, right 6"
Or along the direction vector < 6,-5 > which means "go right 6, then down 5".
There are infinitely many points on this line.
Answer: (0,9)
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Part (d)
Vector form of the line:
< x,y > = startVector + t*DirectionVector
< x,y > = < 0,9 > + t*< 6,-5 >
The start vector could be any other vector you want, as long as it's on the line.
So you could pick < 6,4 > 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 < 7,9 >
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:
< x,y > = startVector + t*DirectionVector
< x,y > = < 7,9 > + t*< 6,-5 >
Now rewrite things a bit like so
< x,y > = < 7,9 > + t*< 6,-5 >
< x,y > = < 7,9 > + < 6t,-5t >
< x,y > = < 7+6t,9-5t >
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:
| Vector form: | < x,y > = < 7,9 > + t*< 6,-5 > | | Parametric form: | x = 7+6ty = 9-5t |
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Part (e)
Original direction vector = < 6,-5 >
Perpendicular direction vector = < 5,6 >
Refer to part (b)
Vector form of the perpendicular line:
< x,y > = startVector + t*DirectionVector
< x,y > = < -2,1 > + t*< 5,6 >
< x,y > = < -2,1 > + < 5t,6t >
< x,y > = < -2+5t,1+6t >
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:
| Vector form: | < x,y > = < -2,1 > + t*< 5,6 > | | Parametric form: | x = -2+5ty = 1+6t |
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