Answer :
Answer:
[tex]\frac{3}{\sqrt{2} }[/tex] units
Step-by-step 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 [tex]\frac{|ax1 + by1 + c|}{\sqrt{a^{2}+b^{2} } }[/tex]
Therefore, the distance from (0,0) to the line (1) i.e. x - y + 3 = 0 is
[tex]\frac{|0-0+3|}{\sqrt{1^{2}+(-1)^{2} } } = \frac{3}{\sqrt{2} }[/tex] units. (Answer)
Answer:
[tex]\textbf{The distance between $l_1$ \& $l_2$ is given by $\frac{3}{\sqrt{2}}$}\\[/tex]
Step-by-step explanation:[tex]\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 = \frac{|c_2 - c_1|}{\sqrt{a^2 + b^2}} $\\\textup{The given lines can be written in the above form as:}\\[/tex]
[tex]$\begin{equation}x - y + 3 = 0 \end{equation}\begin{equation}x - y = 0 \end{equation}\\$\\\textup{Comparing with the standard form, we get}\\$ d = \frac{|0 - (-3)|}{\sqrt{1^2 + 1^2}} $\\$ \therefore d = \frac{3}{\sqrt{2}} $[/tex]