A line has parametric equations x=7t-2 and y=-4+3t. What is the slope of the line representing the direct relationship between x and y?
from x=7t-2
t = (x+2)/7
from y=-4+3t
t = (y+4)/3
(x+2)/7 = (y+4)/3
3x + 6 = 7y + 28
3x - 7y = 22 ---> slope = 3/7
To find the slope of the line representing the direct relationship between x and y, we need to find the derivative of y with respect to x.
Given the parametric equations x = 7t - 2 and y = -4 + 3t, we can solve for t in terms of x:
x = 7t - 2
7t = x + 2
t = (x + 2)/7
Now, substitute this value of t into the equation for y:
y = -4 + 3t
y = -4 + 3((x + 2)/7)
y = -4 + (3x + 6)/7
y = (3x - 22)/7
Now we have the equation y = (3x - 22)/7 in terms of x.
To find the slope, we can compare this equation to the slope-intercept form, y = mx + b, where m represents the slope. In this case, the slope m is given by the coefficient of x:
m = 3/7
Therefore, the slope of the line representing the direct relationship between x and y is 3/7.
To determine the slope of the line representing the direct relationship between x and y, we need to find the derivative of y with respect to x.
Given the parametric equations x = 7t - 2 and y = -4 + 3t, we can solve for t in terms of x:
x = 7t - 2
7t = x + 2
t = (x + 2)/7
Substituting this expression for t into the y equation:
y = -4 + 3t
y = -4 + 3((x + 2)/7)
y = (-4/7)x + 2/7
Now we have the equation y = (-4/7)x + 2/7, which is in the slope-intercept form y = mx + b, where m is the slope.
Therefore, the slope of the line representing the direct relationship between x and y is -4/7.