See attached figure

Equation of Line \(\ell_1\) is y=x. IF we compare that to y=mx+c then we know that slope of line \(\ell_1\) = 1

=> Angle made by line y=x with positive x axis is 45 degree as (tan 45 = 1)

So, we can drop a perpendicular from line \(\ell_2\) to \(\ell_1\) as shown in the attached figure and complete triangle ABC.

Now, Shortest distance between two lines is the perpendicular distance between them.

In Triangle ABC one angle (angle A) is 45 degree and one angle (angle B) is 90 degree. So it becomes a 45-45-90 triangle or isosceles right triangle.

And in Isosceles right triangle sides are in the ratio \(x:x:x\sqrt{2}\), where hypotenuses is \(x\sqrt{2}\)

so, \(x\sqrt{2}\) = 1 => \(x = 1/\sqrt{2}\)

So perpendicular distance = BC = \(x = 1/\sqrt{2}\)

So, Answer is C. (as C is \(\frac{\sqrt{2}}{2}\) which can be simplified as \(1/\sqrt{2}\))

Hope it helps!

Carcass wrote:

Attachment:

The attachment **#greprepclub In the rectangular coordinate system above.jpg** is no longer available

In the rectangular coordinate system above, if the equation of \(\ell_1\) is \(y = x\) and \(\ell_1\) \(\parallel\) \(\ell_2\), what is the shortest distance between \(\ell_1\) and \(\ell_2\) ?

A. \(\sqrt{2}\)

B. \(1\)

C. \(\frac{\sqrt{2}}{2}\)

D. \(\frac{1}{2}\)

E. \(\frac{1}{4}\)

Kudos for the right answer and explanation

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