
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
164.1k+ 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.
Recently Updated Pages
Geometry of Complex Numbers – Topics, Reception, Audience and Related Readings

JEE Main 2021 July 25 Shift 1 Question Paper with Answer Key

JEE Main 2021 July 22 Shift 2 Question Paper with Answer Key

JEE Main 2025 Session 2: Exam Date, Admit Card, Syllabus, & More

JEE Atomic Structure and Chemical Bonding important Concepts and Tips

JEE Amino Acids and Peptides Important Concepts and Tips for Exam Preparation

Trending doubts
Atomic Structure - Electrons, Protons, Neutrons and Atomic Models

Learn About Angle Of Deviation In Prism: JEE Main Physics 2025

Degree of Dissociation and Its Formula With Solved Example for JEE

Instantaneous Velocity - Formula based Examples for JEE

JEE Main Chemistry Question Paper with Answer Keys and Solutions

GFTI Colleges in India - List, Ranking & Admission Details

Other Pages
NCERT Solutions for Class 11 Maths Chapter 6 Permutations and Combinations

JEE Advanced 2025 Notes

Total MBBS Seats in India 2025: Government College Seat Matrix

NEET Total Marks 2025: Important Information and Key Updates

Neet Cut Off 2025 for MBBS in Tamilnadu: AIQ & State Quota Analysis

Karnataka NEET Cut off 2025 - Category Wise Cut Off Marks
