Question

The lines defined by Pt = (4+5t, −1+2t) and Qu = (4−2u, −1+5u) intersect perpendicularly. What are the coordinates of the point of intersection?

Answer 1

@agent0smith

Answer 2

You could convert from the parametric equations back to a linear equation, then just find where the two lines intersect.

Answer 3

could i do
y = 2/5x -1 (Pt)
y = -5/2x - 1

Answer 4

y = -5/2x - 1 (Qu)

Answer 5

are the equations correct?

Answer 6

Sorry, gotta go

Answer 7

ok thanks for your help.

Answer 8

@zepdrix can you help me?

Answer 9

I'm a little confused.
Are these vectors?
\(\large\rm \vec P(t) = <4+5t, −1+2t>\)

Or just coordinates using some parameter?
\(\large\rm P(t)=(4+5t,-1+2t)\)

Ahh my brain >.<

Answer 10

\[P_t = (4+5t, -1+2t)\]

I think they are parameters.

Answer 11

Notice that P[0]=Q[0]=(4,-1)
That is the point of intersectio

Answer 12

Oh Wow! Thank you @eliesaab

Answer 13

YW @calculusxy

Answer 14

@eliesaab Would the equation for line Pt y = 2/5x - 1 be incorrect?

Answer 15

No,
The right equation is
\[y=\frac{1}{5} (2 x-13)\]