# Math Help - Scalar Equations

1. ## Scalar Equations

Find the scalar equation of l1 in each case:

a) with point (2,-3) and parallel to l2: (x,y)=(1,3)+t(5,2)
b) with y intercept -2 and perpendicular to l2: (x,y)=(4,-1)+t(6,-5)

can someone check my work? let me know if i have my answers correct or wrong.

a) 5x+2y+c=0
5(2)+2(-3)+c=0
c=-4

therefore s eqtn is 5x+2y-4=0

b) 5x+6y+c=0
5(0)+6(-2)+c=0
c=12

therefore s eqtn is 5x+6y+12=0

2. Hi

You did the wrong way

a) with point (2,-3) and parallel to l2: (x,y)=(1,3)+t(5,2)
(5,2) is a direction vector of l2 and of l1 since l1 is parallel to l2
A Cartesian equation of l1 is therefore 2x-5y+c=0
Substituting (2,-3) gives 2x-5y-19=0

You can check very easily
l1 equation is (x,y)=(2,-3)+t(5,2)
Take t=-1 => (x,y)=(-3,-5) and 2(-3)-5(-5)-19=0

Remember that a line whose equation is ax+by+c=0 is parallel to the vector (-b,a) and perpendicular to the vector (a,b)

b) I let you correct your answer