For each of the following, find the parametric equations of the line that passes through the point P with direction vector d. In each case, find two points on the line different from P.
a) P(1,1) d=(4,-4)
b) P(5,0) d=(1,3)
answer at the back of the book is:
a) x=1+t y=1-t ; (2,0) (-2,4)
b) x=5+t y=3t ; ((6,3) (3,-6)
What is 7 1/8×2 5/6= I need help plz 👍👍
To find the parametric equations of a line passing through a point P with a given direction vector d, we can use the following formula:
x = Px + dt
y = Py + dt
where Px and Py are the coordinates of point P, t is a parameter, and dx and dy are the components of the direction vector d.
Let's solve the given examples:
a) For P(1,1) with direction vector d = (4,-4):
x = 1 + 4t
y = 1 - 4t
To find two points on the line different from P, we can substitute different values of t:
When t = 0:
x = 1 + 4(0) = 1
y = 1 - 4(0) = 1
So one point on the line is P(1,1). Let's find another point by substituting a different value of t:
When t = 1:
x = 1 + 4(1) = 5
y = 1 - 4(1) = -3
Therefore, the line passing through P(1,1) with direction vector d = (4,-4) can be represented parametrically as:
x = 1 + 4t
y = 1 - 4t
With two additional points on the line: (2,0) and (-2,4).
b) For P(5,0) with direction vector d = (1,3):
x = 5 + t
y = 0 + 3t
When t = 0:
x = 5 + 0 = 5
y = 0 + 3(0) = 0
So one point on the line is P(5,0). Let's find another point by substituting a different value of t:
When t = 1:
x = 5 + (1) = 6
y = 0 + 3(1) = 3
Therefore, the line passing through P(5,0) with direction vector d = (1,3) can be represented parametrically as:
x = 5 + t
y = 0 + 3t
With two additional points on the line: (6,3) and (3,-6).