The lines 3x + 4y = 12 and 3x + 4y = 72 are parallel. Find the distance that separates these lines.
1. Line has the y-intercept .
Line has the y-intercept .
2. The (perpendicular) distance of a point to a straight line with the equation is calculated by
3. Calculate the distance of P(0, 3) to :
4. So the distance you are looking for has the value 12. The negative sign of d indicates that the point P and the origin are on the same side of the straight line.
Here's how I would do that- because I am terrible at memorizing formulas! The slope of the two lines is -3/4 so a line perpendicular to both will have slope 4/3. The equation of a line with slope 4/3 and passing through (0, 3) is . That will cross the second line where . Then and . And, of course, .
That is, the line perpendicular to both given lines intersects them at and . I have left the y value in that form because we need to subtract 3 from it to find the distance between those points.
The distance is .
(Added after thinking it over) However, note Prove It's post. The first line passes through the point (0, 3) and the second through (0, 18). The difference between the two y intercepts is 18- 3= 15. Now that is NOT the distance between the lines (which is always measured perpendicular to both lines) but if we were to drop a perpendicular from, say, (0, 3) to the second line, it is the hypotenuse of a right triangle that has the common perpendicular to both lines as a leg. Further, a little geometry shows that those two legs in the ratio of 3 to 4 since the slope of both lines is -3/4. That is, if the perpendicular distance is "x" then the other leg has length (3/4)x so we must have or ..