Consider the line passing through the points (2, 1) and (-2, 3). Find the parametric equation for y if x = t + 1.
Find the line that pass through the two points P1(2,1), and P2(-2,3) in rectangular coordinates:
(x2-x1)(y-y1)=(y2-y1)(x-x1)
(-2-2)(y-1) = (3-1)(x-2)
-4(y-1) = 2(x-2)
2y+x=4
Solve for y in terms of x:
y=-x/2 + 2
substitute x=t+1
y=-(t+1)/2+2
=-t/2 + 3/2
So:
x=t+1
y=t/2 + 5/2
To find the parametric equation for y, we need to determine the relationship between x and y in terms of a parameter, which in this case is t.
Let's start by finding the slope of the line passing through the two given points. The slope (m) of a line can be found using the formula:
m = (y2 - y1) / (x2 - x1)
Plugging in the coordinates of the given points, we have:
m = (3 - 1) / (-2 - 2)
m = 2 / (-4)
m = -1/2
So, the slope of the line is -1/2.
Now, let's use the point-slope form of a linear equation to find the equation of the line. The point-slope form is given by:
y - y1 = m(x - x1)
Substituting the coordinates (2, 1) and the slope -1/2, we have:
y - 1 = -1/2(x - 2)
Next, we simplify the equation:
y - 1 = -1/2x + 1
To isolate y, we add 1/2x to both sides:
y = -1/2x + 2
Now, we need to express x in terms of a parameter, t, as given in the question. The equation x = t + 1 already provides the expression of x in terms of t.
So, substituting t + 1 for x in the equation y = -1/2x + 2, we have:
y = -1/2(t + 1) + 2
Finally, we simplify the equation:
y = -1/2t - 1/2 + 2
y = -1/2t + 3/2
Therefore, the parametric equation for y, given x = t + 1, is y = -1/2t + 3/2.