What is the equation of the line which passes through the point (6,4) and is parallel to the line with the parametric equations x=5t+4 and y=t-7?
I think the answer is y = 1/5x + 14/5.
For the parametric equation:
dy/dt = 1
dx/dt = 5
So, dy/dx = (1/5)
This is the slope of the tangent to the line with the given parametric equation.
So, m = (1/5)
Point-slope form of a line:
(y - y') = m(x - x')
(y - 4) = (1/5)(x - 6)
y = (1/5)x + 14/5
You are correct.
To find the equation of the line passing through the point (6,4) and parallel to the line with the parametric equations x=5t+4 and y=t-7, we can follow these steps:
Step 1: Determine the slope of the parallel line.
A line parallel to the given line will have the same slope. In this case, the slope can be found by taking the coefficient of t in the y equation of the given line. So, the slope is 1.
Step 2: Use the point-slope form of a linear equation.
The point-slope form of a linear equation is given by y - y1 = m(x - x1), where (x1, y1) is a point on the line and m is the slope. In this case, we know the point (6,4) and the slope (1), so we can substitute these values into the equation to find the line's equation.
Substituting the values, we have:
y - 4 = 1(x - 6)
Simplifying the equation, we get:
y - 4 = 1x - 6
Step 3: Convert the equation to slope-intercept form.
To get the equation in the form y = mx + c, where m is the slope, and c is the y-intercept, we can rearrange the equation.
y - 4 = x - 6
y = x - 6 + 4
y = x - 2
Therefore, the equation of the line passing through the point (6,4) and parallel to the line with the parametric equations x=5t+4 and y=t-7 is y = x - 2.
It seems that the expression y = 1/5x + 14/5 you mentioned is incorrect. The correct equation is y = x - 2.