Question

Line ℓ1 has the equation y=x+3 and line ℓ2 has the equation - x+y=0. Find the distance between ℓ1 and ℓ2

Answer 1

`\textbf{The distance between $l_1$ \& $l_2$ is given by $(3)/(√(2))$}\\`

Explanation:
`\textup{Given:}\\$y = x + 3$ \hspace{5mm}  \&  \hspace{5mm} $-x + y = 0$\\\textup{When the lines are of the form:}\\ $ ax + by = c_1 $\\$ ax + by = c_2 $\\\textup{The distance between these two lines is given by:}\\$$ d = (|c_2 - c_1|)/(√(a^2 + b^2)) $$\\\textup{The given lines can be written in the above form as:}\\`

`$$\begin{equation}x - y + 3 = 0 \end{equation}\begin{equation}x - y = 0     \end{equation}\\$$\\\textup{Comparing with the standard form, we get}\\$$ d = (|0 - (-3)|)/(√(1^2 + 1^2)) $$\\$$ \therefore d = (3)/(√(2)) $$`

Answer 2

`(3)/(√(2) )` units

Explanation:

The equation of the line L₁ is y = x + 3 ....... (1) and that of the line L₂ is - x + y = 0 .......... (2)

We have to get the perpendicular distance between the given two lines.

Now, choose any point on the line L₂ from equation (2).

Assume the point be (0,0) {As it satisfies equation (2)}

Now, perpendicular distance from point (x1 ,y1) to a straight line ax + by + c = 0, is given by the formula
`\fracax1 + by1 + c{\sqrt{a^(2)+b^(2)  } }`

Therefore, the distance from (0,0) to the line (1) i.e. x - y + 3 = 0 is

`\frac0-0+3{\sqrt{1^(2)+(-1)^(2)  } } = (3)/(√(2) )` units. (Answer)