Question

Equations of diagonals of squares formed by lines \[x=0\], \[y=0\], \[x=1\], and \[y=1\] are
A. \[y=x,\text{ }y+x=1\]
B. \[y=x,\text{ }x+y=2\]
C. \[2y=x,\text{ }y+x=\dfrac{1}{3}\]
D. \[y=2x,\text{ }y+2x=1\]

Answer 1

Hint: In the question, we are to find the equations of diagonals of a square formed by the given lines. Using the given equations of lines, a square is formed and we can able to find out its vertices. Then, the diagonals are calculated by the two points form.

Formula Used: The equation of the line, that is passing through $({{x}_{1}},{{y}_{1}})$ and $({{x}_{2}},{{y}_{2}})$ is
$y-{{y}_{1}}=\dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}(x-{{x}_{1}})$
Where $m=\dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}$ is said to be the slope of the line.

Complete step by step solution: Given that, a square is formed by the lines
\[\begin{align}
  & x=0 \\
 & y=0 \\
 & x=1 \\
 & y=1 \\
\end{align}\]
Plotting these in the graph we get the required vertices as below:

Then, the vertices formed by these lines are:
$\begin{align}
  & A(0,0) \\
 & B(0,1) \\
 & C(1,1) \\
 & D(1,0) \\
\end{align}$
Then, the required diagonals are $\overleftrightarrow{AC}$ and $\overleftrightarrow{BD}$.
By using the equation of line formula (with two points),
$y-{{y}_{1}}=\dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}(x-{{x}_{1}})$
For the diagonal $\overleftrightarrow{AC}$, the vertices are $A(0,0);C(1,1)$
So, the equation of the diagonal is,
$\begin{align}
  & y-0=\dfrac{1-0}{1-0}(x-0) \\
 & \Rightarrow y=x\text{ }...(1) \\
\end{align}$
For the diagonal $\overleftrightarrow{BD}$, the vertices are $B(0,1);D(1,0)$
So, the equation of the diagonal is,
$\begin{align}
  & y-1=\dfrac{0-1}{1-0}(x-0) \\
 & \Rightarrow y-1=-x \\
 & \Rightarrow x+y=1\text{ }...(2) \\
\end{align}$
Therefore, the required diagonal lines are $y=x;x+y=1$.

Option ‘A’ is correct

Note: Here we may go wrong with the vertices. For finding the vertices, a graph is drawn for the given lines. Then, we can able to find the vertices of a diagonal.