
A vertex of square is $\left( {3,4} \right)$ and diagonal $x + 2y = 1$ then the second diagonal which passes through given vertex will be
1. $2x - y + 2 = 0$
2. $x + 2y = 11$
3. $2x - y = 2$
4. None of these
Answer
233.1k+ views
Hint: In this question, we are given the equation of the first diagonal $x + 2y = 1$ and a point $\left( {3,4} \right)$ from which the second diagonal is passing. We have to find the equation of second line. First step is to find the slope of first diagonal using $y = mx + c$ from the given equation. Then find the slope of second diagonal from the first slope and using slope and point of vertex find the equation of second diagonal by applying the formula $\left( {y - {y_1}} \right) = m\left( {x - {x_1}} \right)$. Solve further.
Formula Used:
General equation of a straight line –
$y = mx + c$
When two lines are perpendicular to each other, and their slopes are ${m_1}$and ${m_2}$then ${m_1}{m_2} = - 1$
If a line is passing through a point $\left( {{x_1},{y_1}} \right)$ and its slope is $m$ then the equation of line is $\left( {y - {y_1}} \right) = m\left( {x - {x_1}} \right)$
Complete step by step Solution:
Given that,
The equation of the first diagonal is $x + 2y = 1$
Equation can also be written as $y = \dfrac{{ - 1}}{2}x + \dfrac{1}{2} - - - - - (1)$
Compare equation (1) with respect to general equation of the straight line $y = mx + c$
Which implies that, slope of the first diagonal $\left( {{m_1}} \right) = \dfrac{{ - 1}}{2}$
Let, the slope of the second diagonal be ${m_2}$
As we know that,
The diagonals of the square are perpendicular to each other.
Therefore, ${m_1}{m_2} = - 1$
${m_2} = \dfrac{{ - 1}}{{{m_1}}}$
${m_2} = \dfrac{{ - 1}}{{\left( {\dfrac{{ - 1}}{2}} \right)}} = 2$
Also, given the second diagonal passes through the point $\left( {3,4} \right)$
$ \Rightarrow {x_1} = 3,{y_1} = 4$
The equation of second diagonal will be
$\left( {y - {y_1}} \right) = {m_2}\left( {x - {x_1}} \right)$
$\left( {y - 4} \right) = 2\left( {x - 3} \right)$
$y - 4 = 2x - 6$
$2x - y = 6 - 4$
$2x - y = 2$
Hence, the correct option is 3.
Note: The key concept involved in solving this problem is a good knowledge of the Equation of a line. Students must know that a line equation is easily understood as a single representation for multiple points on the same line. A line's equation has a general form, which is \[ax + by + c = 0\], and any point on this line satisfies this equation. The slope of the line and any point on the line are two absolutely necessary requirements for forming the equation of a line.
Formula Used:
General equation of a straight line –
$y = mx + c$
When two lines are perpendicular to each other, and their slopes are ${m_1}$and ${m_2}$then ${m_1}{m_2} = - 1$
If a line is passing through a point $\left( {{x_1},{y_1}} \right)$ and its slope is $m$ then the equation of line is $\left( {y - {y_1}} \right) = m\left( {x - {x_1}} \right)$
Complete step by step Solution:
Given that,
The equation of the first diagonal is $x + 2y = 1$
Equation can also be written as $y = \dfrac{{ - 1}}{2}x + \dfrac{1}{2} - - - - - (1)$
Compare equation (1) with respect to general equation of the straight line $y = mx + c$
Which implies that, slope of the first diagonal $\left( {{m_1}} \right) = \dfrac{{ - 1}}{2}$
Let, the slope of the second diagonal be ${m_2}$
As we know that,
The diagonals of the square are perpendicular to each other.
Therefore, ${m_1}{m_2} = - 1$
${m_2} = \dfrac{{ - 1}}{{{m_1}}}$
${m_2} = \dfrac{{ - 1}}{{\left( {\dfrac{{ - 1}}{2}} \right)}} = 2$
Also, given the second diagonal passes through the point $\left( {3,4} \right)$
$ \Rightarrow {x_1} = 3,{y_1} = 4$
The equation of second diagonal will be
$\left( {y - {y_1}} \right) = {m_2}\left( {x - {x_1}} \right)$
$\left( {y - 4} \right) = 2\left( {x - 3} \right)$
$y - 4 = 2x - 6$
$2x - y = 6 - 4$
$2x - y = 2$
Hence, the correct option is 3.
Note: The key concept involved in solving this problem is a good knowledge of the Equation of a line. Students must know that a line equation is easily understood as a single representation for multiple points on the same line. A line's equation has a general form, which is \[ax + by + c = 0\], and any point on this line satisfies this equation. The slope of the line and any point on the line are two absolutely necessary requirements for forming the equation of a line.
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