An equation is of the form 7x+17y+3z=54. Find the distance between x intercept and z intercept?

a) 72/7
b) 37/3
c) 122/8
d) 123/7

• for x intercept put y=z=0
  for y intercept put x=z=0
  for z intercept put x=y=0
  so,
  x intercept will be 7x=54 --> x=54/7
  z intercept will be 3z=54 --> z=18
  now distance between them is,
  z-x = 18-54/7
  = (126-54)/7
  = 72/7
• 10 years agoHelpfull: Yes(13) No(2)
• no option matches the answer.
  sqrt of (54/7)^2+(54/3)^2
• 10 years agoHelpfull: Yes(5) No(2)
• option d
  
  sqrt of (54/7)^2+(54/3)^2 =123/7
• 10 years agoHelpfull: Yes(4) No(5)
• for x intercept y,z=0
  so put y,z=0 in 7x+17y+3z=54
  we get x=54/7.
  for z intercept put x,y=0
  we get z=54/7=18
  so distance=18-54/7=72/7
  so ans=72/7
  
• 10 years agoHelpfull: Yes(1) No(2)
• for x intercept y=z=0
  hence x=54/7
  similarly for z intercept x=y=0;
  and z=54/3=18
  distance=18-54/7=72/7
• 10 years agoHelpfull: Yes(0) No(0)