Question

A (1,3) and C (7,5) are two opposite vertices of a square. Then the equation of a side through A is
\[\begin{align}
  & \text{A}.\text{ x}+\text{2y}-\text{7}=0 \\
 & \text{B}.\text{ x}-\text{2y}+\text{5}=0 \\
 & \text{C}.\text{ 2x}+\text{y}-\text{5}=0 \\
 & \text{D}.\text{ 2x}-\text{y}+\text{1}=0 \\
\end{align}\]

Answer 1

Hint:At first, use the property of square to say that the diagonal bisects the angle at the vertices. Then, use the formula $\tan \theta =\left| \dfrac{{{m}_{1}}-{{m}_{2}}}{1+{{m}_{1}}{{m}_{2}}} \right|$ where $\theta $ is angle between two lines and ${{m}_{1}},{{m}_{2}}$ are slopes of line. Then take value of $\theta $ as ${{45}^{\circ }}$ and ${{m}_{1}}$ as $\dfrac{1}{3}$ and hence find both values of ${{m}_{2}}$. Then, find equation using formula $y-{{y}_{1}}=m\left( x-{{x}_{1}} \right)$ where point is $\left( {{x}_{1}},{{y}_{1}} \right)$ and slope is m.

Complete step by step answer:
In the question, we are said that A (1,3) and C (7,5) are opposite vertices of the square and from this, we have to find the equation of sides of the square with point A in it.
So, at first, we draw the figure,

We know a property of square that its diagonal bisects the angle A, thus we can say $\angle CAB\text{ and }\angle CAD$ is equal to ${{45}^{\circ }}$
So, at first we will find slope of AC by using the formula,
\[m=\dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}\]
If points are $\left( {{x}_{1}},{{y}_{1}} \right)\text{ and }\left( {{x}_{2}},{{y}_{2}} \right)$ so the slope of AC with point A as (1,3) and C as (7,5) is
\[m=\dfrac{3-5}{1-7}=\dfrac{-2}{-6}=\dfrac{1}{3}\]
So, now we can find slope of AB and AD using formula that,
\[\tan \theta =\left| \dfrac{{{m}_{1}}-{{m}_{2}}}{1+{{m}_{1}}{{m}_{2}}} \right|\]
Where $\theta $ is angle between two lines and ${{m}_{1}},{{m}_{2}}$ are slope of the line.
Here, $\theta $ is ${{45}^{\circ }}$, ${{m}_{1}}$ is $\dfrac{1}{3}$ and value of ${{m}_{2}}$ is possible slope for AB and AD.
Thus,
\[\begin{align}
  & \tan {{45}^{\circ }}=\left| \dfrac{\dfrac{1}{3}-{{m}_{2}}}{1+\dfrac{1}{3}{{m}_{2}}} \right| \\
 & \Rightarrow 1=\left| \dfrac{\dfrac{1}{3}-{{m}_{2}}}{1+\dfrac{{{m}_{2}}}{3}} \right| \\
\end{align}\]
As we are removing mod we can say that,
\[\pm 1=\dfrac{1-3{{m}_{2}}}{3+{{m}_{2}}}\]
At first,
\[\dfrac{1-3{{m}_{2}}}{3+{{m}_{2}}}=1\]
On cross multiplication we get:
\[\begin{align}
  & 1-3{{m}_{2}}=3+{{m}_{2}} \\
 & \Rightarrow 1=3+{{m}_{2}}+3{{m}_{2}} \\
 & \Rightarrow 1=3+4{{m}_{2}} \\
 & \Rightarrow 1-3=4{{m}_{2}} \\
 & \Rightarrow 4{{m}_{2}}=-2 \\
 & \Rightarrow {{m}_{2}}=\dfrac{-2}{4}=\dfrac{-1}{4} \\
\end{align}\]
Hence one value of ${{m}_{2}}=\dfrac{-1}{2}$
For,
\[\dfrac{1-3{{m}_{2}}}{3+{{m}_{2}}}=-1\]
On cross multiplication we get:
\[\begin{align}
  & 1-3{{m}_{2}}=-3-{{m}_{2}} \\
 & \Rightarrow 1=-3-{{m}_{2}}+3{{m}_{2}} \\
 & \Rightarrow 1=-3+2{{m}_{2}} \\
 & \Rightarrow 1+3=2{{m}_{2}} \\
 & \Rightarrow 2{{m}_{2}}=4 \\
 & \Rightarrow {{m}_{2}}=\dfrac{4}{2}=2 \\
\end{align}\]
Hence, another value of ${{m}_{2}}=2$
As we know, slopes are 2 and $\dfrac{-1}{2}$ and passes through point (1,3) we can get equations using formula $y-{{y}_{1}}=m\left( x-{{x}_{1}} \right)$ if point is $\left( {{x}_{1}},{{y}_{1}} \right)$ and slope is m.
For slope 2 we get:
\[\begin{align}
  & y-3=2\left( x-1 \right) \\
 & \Rightarrow y-3=2x-2 \\
 & \Rightarrow 2x-2-\left( y-3 \right)=0 \\
 & \Rightarrow 2x-y+1=0 \\
\end{align}\]
For slope $\dfrac{-1}{2}$ we get:
\[y-3=\dfrac{-1}{2}\left( x-1 \right)\]
On cross multiplication we get:
\[\begin{align}
  & 2y-6=-x+1 \\
 & \Rightarrow 2y-6-\left( -x+1 \right)=0 \\
 & \Rightarrow x+2y-7=0 \\
\end{align}\]
Thus, correct options are A and D.
Note:
 Sometimes while using the formula of $\tan \theta =\left| \dfrac{{{m}_{1}}-{{m}_{2}}}{1+{{m}_{1}}{{m}_{2}}} \right|$. Students forget to take both the cases of positive and negative which give them only one of the equation which can make the answer wrong. Thus, they should be careful about that.