Two supports beams are modeled by the lines y=2x+10 and y=2x+15. What is the distance between these two lines ?
+129839 y=2x+10 and y=2x+15
The lines have the same slope.........only their y intercepts are different.......the first line has a y intercept of 10 and the second has a y intercept of 15....so....the lines are 5 units apart
+129839 I totally blew this one....my answer is true if the lines are horizontal......
Let (0, 10) be a point on the line y = 2x + 10
Now......a line perpendicular to this one will intersect the line y = 2x + 15 at right angles.....and the distance from (0,10) to the intersection of y = 2x + 15 will lie along this perpendicular line...and the perpendicular line will have a slope of -1/2 and will contain (0,10)..and its equation will be y = (-1/)x + 10...so, .let us find the intersection point of these lines
2x + 15 = (-1/2)x + 10 add (1/2)x to both sides and subtract 15 from both sides
2.5x = -5 divide both sides by -2.5
x = -2 this is the x coordinate of the intersection of the lines.....and the y coordinate is y = 2(-2) + 15 = 11
So...we need to find the distance between (0,10) and (-2, 11) .....using the distance formula, we have
√[ (-2)^2 + (11 - 10)^2 ] = √ [4 + 1] = √5 ....and this is the distance between the beams
