
The points \[A(1,3)\] and \[C(5,1)\] are the opposite vertices of a rectangle. The equation of the line passing through the other two vertices and of gradient $2$ is
A. \[2x+y-8=0\]
B. \[2x-y-4=0\]
C. \[2x-y+4=0\]
D. \[2x+y+7=0\]
Answer
160.8k+ views
Hint: In this question, we are to find the equation of the line passing through the other two vertices of the given rectangle. Here we have given the two vertices and the gradient(slope) of the line that is formed by these points. So, by using these credentials, we can calculate the required line equation.
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.
So, we can also write the equation of the line in the form of a point-slope as
$y-{{y}_{1}}=m(x-{{x}_{1}})$
Here the slope is also known as the gradient of the line.
The mid-point between two points $({{x}_{1}},{{y}_{1}})$ and $({{x}_{2}},{{y}_{2}})$ is
$M=\left( \dfrac{{{x}_{1}}+{{x}_{2}}}{2},\dfrac{{{y}_{1}}+{{y}_{2}}}{2} \right)$
Complete step by step solution:Given that,
A rectangle has two vertices that are opposite to each other are \[A(1,3)\] and \[C(5,1)\]
Since we know that, the diagonals of a rectangle bisect each other and connect the opposite vertices of the rectangle.
So, the given opposite vertices form a diagonal and the mid-point of this diagonal is also the mid-point of the other diagonal (since they bisect each other).
Thus, the mid-point of the given points is,
$M=\left( \dfrac{{{x}_{1}}+{{x}_{2}}}{2},\dfrac{{{y}_{1}}+{{y}_{2}}}{2} \right)$
$\Rightarrow \left( \dfrac{1+5}{2},\dfrac{3+1}{2} \right)=(3,2)$
Then, the required line passes through the mid-point $(3,2)$ and the gradient i.e., slope \[m=2\] is
$y-{{y}_{1}}=m(x-{{x}_{1}})$
$\begin{align}
& \Rightarrow y-{{y}_{1}}=m(x-{{x}_{1}}) \\
& \text{ }y-2=2(x-3) \\
& \text{ }y-2=2x-6 \\
\end{align}$
$\begin{align}
& \Rightarrow 2x-y-6+2=0 \\
& \Rightarrow 2x-y-4=0 \\
\end{align}$
Option ‘B’ is correct
Note: Here we may go wrong with the mid-point of the diagonal formed by the given opposite vertices. In order to find the equation of the other diagonal line of the rectangle, the mid-point obtained from these points is used. This is because the required line passes through that mid-point.
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.
So, we can also write the equation of the line in the form of a point-slope as
$y-{{y}_{1}}=m(x-{{x}_{1}})$
Here the slope is also known as the gradient of the line.
The mid-point between two points $({{x}_{1}},{{y}_{1}})$ and $({{x}_{2}},{{y}_{2}})$ is
$M=\left( \dfrac{{{x}_{1}}+{{x}_{2}}}{2},\dfrac{{{y}_{1}}+{{y}_{2}}}{2} \right)$
Complete step by step solution:Given that,
A rectangle has two vertices that are opposite to each other are \[A(1,3)\] and \[C(5,1)\]
Since we know that, the diagonals of a rectangle bisect each other and connect the opposite vertices of the rectangle.
So, the given opposite vertices form a diagonal and the mid-point of this diagonal is also the mid-point of the other diagonal (since they bisect each other).
Thus, the mid-point of the given points is,
$M=\left( \dfrac{{{x}_{1}}+{{x}_{2}}}{2},\dfrac{{{y}_{1}}+{{y}_{2}}}{2} \right)$
$\Rightarrow \left( \dfrac{1+5}{2},\dfrac{3+1}{2} \right)=(3,2)$
Then, the required line passes through the mid-point $(3,2)$ and the gradient i.e., slope \[m=2\] is
$y-{{y}_{1}}=m(x-{{x}_{1}})$
$\begin{align}
& \Rightarrow y-{{y}_{1}}=m(x-{{x}_{1}}) \\
& \text{ }y-2=2(x-3) \\
& \text{ }y-2=2x-6 \\
\end{align}$
$\begin{align}
& \Rightarrow 2x-y-6+2=0 \\
& \Rightarrow 2x-y-4=0 \\
\end{align}$
Option ‘B’ is correct
Note: Here we may go wrong with the mid-point of the diagonal formed by the given opposite vertices. In order to find the equation of the other diagonal line of the rectangle, the mid-point obtained from these points is used. This is because the required line passes through that mid-point.
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