State the coordinates of two points on the following line
x=3-8t
y=4t
For one point I got (3,0). In the back of the book it says (11, -4) for another point. How would I get that?
b) r = (4,0) + t(0, 5)
I got (4,0) for one point. In the back of the book it says (4,10) for another point. How would I get that?
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State a direction vector with integer components.
x = (1/3) + 2t, y = 3-(2/3)t
State the integer components and name a point on the line with integer coordinates.
r = (1/3, 1/2) + t(1/3, 1/4)
x=3-8t
y=4t
let t=-1
b) for r = (4,0) + t(0, 5)
let t=2
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x = (1/3) + 2t, y = 3-(2/3)t
so r= (1/3,3) + t(2,2/3)
a direction vector is
(2,2/3) or (4,2) or (8,4), or ....
do you follow?
for the last one, what do you think should be value of t to have integer solutions?
For the first equation:
To find another point on the line x = 3 - 8t and y = 4t, you can substitute the value of t from the known point (3,0) into the equations.
For x = 3 - 8t:
3 = 3 - 8t
0 = -8t
t = 0
Thus, another point on the line is when t = 0. Plug this back into the equations x = 3 - 8t and y = 4t:
x = 3 - 8(0) = 3
y = 4(0) = 0
Therefore, the other point on the line is (3,0).
For the second equation:
To find another point on the line r = (4,0) + t(0, 5), you can once again substitute different values of t into the equation.
When t = 0:
r = (4,0) + 0(0, 5) = (4,0)
However, this is the same point you already found. It is possible that there is an error in the book, as there seems to be no other point on the line according to the given equation.
For the third question:
A direction vector with integer components can be obtained from the x and y coefficients of any equation. In this case, the given line is x = (1/3) + 2t and y = 3 - (2/3)t.
The direction vector can be written as <2, -2/3> or as fractions: (6/3, -2/3).
To find a point on the line with integer coordinates, substitute t = 0 into the equations:
x = (1/3) + 2(0) = 1/3
y = 3 - (2/3)(0) = 3
Therefore, a point on the line with integer coordinates is (1,3).
To find the coordinates of another point on the given line x = 3 - 8t and y = 4t, you can substitute the value of t into the equations and solve for x and y.
For the point (11, -4), substitute the x-coordinate (11) into the equation x = 3 - 8t:
11 = 3 - 8t
Rearrange the equation to isolate t:
8t = 3 - 11
8t = -8
t = -1
Now substitute the value of t into the equation y = 4t:
y = 4(-1)
y = -4
Therefore, the point (11, -4) lies on the given line.
Similarly, for the point (4,10), substitute the x-coordinate (4) into the equation x = 4 + t(0):
4 = 4
Since t can be any value, the value of y does not matter. Thus, any y-value can be chosen, and (4, 10) is one of the infinite points on the line given by r = (4, 0) + t(0, 5).
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A direction vector for a line with integer components can be found by subtracting the coordinates of one point on the line from the coordinates of another point on the line.
For example, given the line x = (1/3) + 2t and y = 3 - (2/3)t, let's find a direction vector with integer components:
Choose two points on the line, such as (1/3, 1/2) and (1, 1/6).
Subtracting the coordinates, we have:
(1 - 1/3, 1/6 - 1/2) = (2/3, -1/3)
Therefore, the direction vector with integer components for the given line is (2/3, -1/3).
To find a point on the line with integer coordinates, you can substitute an appropriate value of t into the equation. For example, choosing t = 1:
x = (1/3) + 2(1) = 7/3
y = 3 - (2/3)(1) = 7/3
Thus, a point on the line with integer coordinates is (7/3, 7/3).