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?..
Shortest Distance
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The shortest distance between two parallel lines is always the length of the perpendicular line segment that connects the two lines.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?..
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
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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
ax + by + c1 = 0 and ax + by +c2 = 0
d = |c1-c2|/√ ( a^2 + b^2 )
= |3-4| / √(1+1) = 1/√2 = √2/2