You are finding distance, so it would be the absolute value, namely D
Are you using the method
for a point (p,q) to the line Ax + By + C = 0
distance = │Ap + Bq + C│/√(A^2+B^2) ?
Find the distance between P(-4,3) and the line with equation: 2x - 5y = -7
a. (14√29)/29
b. 0
c. (-16√29)/29
d. (16√29)/29
I got C but would it be positive, like D?
2 answers
oh okay, thought so
thanks!
thanks!