Question

A vertex of the square is \[\left( 3,4 \right)\] and diagonal \[x+2y=1\], then the second diagonal which passes through the given vertex will be
A. \[2x-y+2=0\]
B. \[x+2y=11\]
C. \[2x-y=2\]
D. None of these

Answer 1

Hint: In this question, we are to find the equation of the second diagonal of the given square. Since we know that the diagonals of a square are perpendicular to each other, then the product of their slopes is $-1$. By using the point-slope form, the required equation of the diagonal is evaluated.

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.
The slope of a line $ax+by+c=0$ is calculated by $m=\dfrac{-a}{b}$
We can also write this in the point-slope form as
$y-{{y}_{1}}=m(x-{{x}_{1}})$
If two lines are perpendicular then the product of their slopes is $-1$ i.e., ${{m}_{1}}\times {{m}_{2}}=-1$

Complete step by step solution: Given that,
The vertex of a square is given as \[\left( 3,4 \right)\] and the equation of one of its diagonals is \[x+2y=1\].
 The slope of the given diagonal is
$\begin{align}
  & {{m}_{1}}=\dfrac{-a}{b} \\
 & \text{ }=\dfrac{-1}{2} \\
\end{align}$
Since we know that, the diagonals of a square are perpendicular to each other, the product of their slopes will be ${{m}_{1}}\times {{m}_{2}}=-1$
So, the required diagonal’s slope is
$\begin{align}
  & {{m}_{2}}=\dfrac{-1}{{{m}_{1}}} \\
 & \text{ }=\dfrac{-1}{\dfrac{-1}{2}} \\
 & \text{ }=2 \\
\end{align}$
Then, the equation of the other diagonal of the given square with the slope ${{m}_{2}}=2$ and passing through the vertex \[\left( 3,4 \right)\] is
$\begin{align}
  & y-{{y}_{1}}=m(x-{{x}_{1}}) \\
 & \Rightarrow y-4=2(x-3) \\
 & \Rightarrow y-4=2x-6 \\
 & \Rightarrow 2x-y=2 \\
\end{align}$
Therefore, the equation of the other diagonal of the given square is \[2x-y=2\].

Option ‘C’ is correct

Note: Here we may go wrong with the slopes of the diagonal. Here we need to remember that the diagonals of a square are perpendicular to each other. So, the product of their slopes are given by ${{m}_{1}}\times {{m}_{2}}=-1$.