aditya.j wrote:What is the shortest distance between the following 2 lines: x + y = 3 and 2x + 2y = 8?

(A) 0
(B) 1/4
(C) 1/2
(D) √2/2
(E) √2/4

Saw this on the Veritas site, but i didn't quite understand that explanation. Can someone outline a simple way of solving this sum?..

The shortest distance between two parallel lines is always the length of the perpendicular line segment that connects the two lines.

Solving these two equations for y gives us y=-x+3 and y=-x+4, so we have two parallel lines with slope of -1 with y-intercepts of 3 and 4. Draw a perpendicular line segment from the point (0,3) to the line y=-x+4. The length of this line segment is what we're after.

Now, the slope of this line segment must be the negative reciprocal of -1 in order for it to be perpendicular to both lines, so its slope is 1. Also, notice that any line with a slope of 1 forms a 45 degree angle with the y-axis. To see this, think about the line y=x. This line has a slope of 1 and runs diagonally through the first and third quadrants, cutting them right down the middle. Thus, it must also cut the 90 degree angle formed by the x- and y-axis right down the middle, making two 45 degree angles. Moving this line up or down by assigning it different y-intercepts will not change the fact that it forms a 45 degree angle with the y-axis.

So now we have a 45-45-90 triangle formed by the part of the y-axis between 3 and 4, the perpendicular line segment that we drew in, and the line y=-x+4. The hypotenuse is 1, and we're looking for the length of one of the legs. To go from the hypotenuse to a leg, you divide by √2. Thus, the length is 1/√2 or √2/2.

Ans: D

Got it..amazing explanation!

Thanks alot..

There's also a formula for shortest distance between 2 lines

ax + by + c1 = 0 and ax + by +c2 = 0

d = |c1-c2|/√ ( a^2 + b^2 )

= |3-4| / √(1+1) = 1/√2 = √2/2