Find the equation of the line...?

You can eliminate the parameter t, and then find the equation.

x=3+5t

y=2-3t

Multiply the first equation by 3 and the second one by 5, and then add the resulting equations

3x = 9 + 15 t

5y = 10 - 15 t

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3x+5y = 19

5y = -3x +19

y = (-3/5)x + 19/5

The slope of this line is -3/5. So the slope of the line you are looking for is 5/3. And you know that the line passes through (0,4), so its equation is given by

y-4 = (5/3)(x-0)

y-4 = (5/3)x

y = (5/3)x + 4

We can use the vector function to write parametric equations for our line.

x=3+5t

y=2-3t

I now solve 1 of these equations for t and plug it into the other equation

x=3+5t

x-3=5t

(x-3)/5=t

y=2-3(x-3)/5

y=2-3/5x-9/5

y=1/5-3/5x

This line has a slope of -3/5. Any line prepindicular to it must have a slope of 5/3. I use this information and the point given to find the equation of the prepindicular line, (I will use L(x) to represent the prepindicular line)

L(x)-4=5/3(x-0)

L(x)=5/3x+4