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.
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.
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