What is the distance between the z-intercept from the x-intercept in the equation ax+by+cz+d=0. (I do not remember the values of a,b,c,d)

• ax+by+cz+d=0
  At the z-intercept, 'x' and 'y' are both zero.
  cz + d = 0 --> z = -d/c --> The z-intercept is the point (0, 0, -d/c).
  At the x-intercept, 'y' and 'z' are zero.
  ax + d = 0 --> x = -d/a --> The x-intercept is the point (-d/a, 0, 0).
  The distance between the points (0, 0, -d/c) and (-d/a, 0, 0) is
  
  { Formula: sqrt((x-x1)^2 + (y-y1)^2+(z-z1)^2) }
  
  => sqrt((0-(-d/a))^2 + (0-0)^2+((-d/c)-0)^2)
  => sqrt((d/a)^2+(d/c))^2
  
  submit the values...
• 9 years agoHelpfull: Yes(10) No(0)
• z intrcpt = -d/c x intrcpt= -d/a distance= sqrt(d2/(c2+a2))
• 12 years agoHelpfull: Yes(6) No(6)
• ans sqrot((da)^2+(db)^2)
  
• 12 years agoHelpfull: Yes(1) No(19)
• ans: sqrt ((d/a) 2+ (d/c) 2)
• 10 years agoHelpfull: Yes(1) No(0)
• z intercept , doing x=y=0 , z= -d/c , point P1(0,0,-d/c)
  x intercept , y=z=0 x= -d/a , point P2(-d/a,0,0,)
  
  Distance P1P2 = I P1-P2I = I (d/a),0, (d/c) I = sqrt ((d/a)^2+ (d/c)^2)
  
  P1P2 = d sqrt ( (1/a)^2 +(1/c)^2)
• 9 years agoHelpfull: Yes(1) No(0)