Problem: If (a,b) and (c,d) are two points on the line whose equation is y=mx+k, then the distance between (a,b) and (c,d), in terms of a,c and m, is
Answer Choices:
A. ∣a−c∣1+m2
B. ∣a+c∣1+m2
C. 1+m2∣a−c∣
D. ∣a−c∣(1+m2)
E. ∣a−c∣∣m∣
Solution:
Since ( a,b ) and ( c,d) are on the same line, y=mx+k, they satisfy the same equation. Therefore.
b=ma+k
d=mc+k
Now the distance between (a,b) and (c,d) is (a−c)2+(b−d)2. We obtain from the above two equations (b−d)=m(a−c),so that
(a−c)2+(b−d)2=(a−c)2+m2(a−c)2=∣a−c∣1+m2
Note we are using the fact that x2=∣x∣ for all real x.