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

thanks in advance