
One diagonal of a square is along the line \[8x - 15y = 0\] and one of its vertexes is \[(1,2)\] then the equation of the sides of the square passing through this vertex, are
A. \[23x + 7y = 9,7x + 23y = 53\]
B. \[23x - 7y + 9 = 0,7x + 23y + 53 = 0\]
C. \[23x - 7y - 9 = 0,7x + 23y - 53 = 0\]
D. None of these
Answer
232.8k+ views
Hint: In this case, we have a line \[8x - 15y = 0\] and a vertex \[(1,2)\] and we are asked to find the equation of the sides of the square passing through this vertex. For this we have to find the equation of sides \[{\rm{BD}}\] with \[{\rm{DC}}\]and proceed with the calculation to find the desired equation.
Formula Used: To describe the point slope form of a line equation, we use the formula
\[{y_1} = m\left( {x - {x_1}} \right)\]
Complete step by step solution: We have been given the data that,
One diagonal of a square is along the line \[8x - 15y = 0\] and one of its vertexes is \[(1,2)\]
Let us assume the square is \[{\rm{ABCD}}\] and vertex \[{\rm{C}}\] is \[(1,2)\]
We have the slope of \[{\rm{BD}}\] is \[\dfrac{8}{{15}}\]
And the angle made by \[{\rm{BD}}\] with \[{\rm{DC}}\] and \[{\rm{DC}}\] is $45^\circ$
Now let us consider the slope of \[{\rm{DC}}\] be \[{\rm{m}}\]
Now, it becomes
\[\tan {45^\circ } = \pm \dfrac{{{\rm{m}} - \dfrac{8}{{15}}}}{{1 + \dfrac{8}{{15}}\;{\rm{m}}}}\]
Now, we have to solve the fraction to simpler term, we have
\[ \Rightarrow (15 + 8\;{\rm{m}}) = \pm (15\;{\rm{m}} - 8)\]
On solving for the value of \[{\rm{m}}\] we get
\[ \Rightarrow {\rm{m}} = \dfrac{{23}}{7}{\rm{ and }} - \dfrac{7}{{23}}\]
Now, we obtained the equations of \[{\rm{DC}}\] and \[{\rm{BC}}\] as
\[ \Rightarrow {\rm{y}} - 2 = \dfrac{{23}}{7}({\rm{x}} - 1)\]
Now, we have to multiply the denominator with the term on the left side and multiply the numerator with the term on the right side, we get
\[ \Rightarrow 23{\rm{x}} - 7{\rm{y}} - 9 = 0\]
\[{\rm{y}} - 2 = - \dfrac{7}{{23}}({\rm{x}} - 1)\]
\[ \Rightarrow 7{\rm{x}} + 23{\rm{y}} - 53 = 0\]
Therefore, the equation of the sides of the square passing through this vertex, are \[23x - 7y - 9 = 0,7x + 23y - 53 = 0\]
Option ‘C’ is correct
Note: Students are likely to make mistakes in these types of problems because it involves trigonometry expressions and more calculations. On should know the formula to proceed with the answer and also should remember all the formulas to solve these types of problems.
Formula Used: To describe the point slope form of a line equation, we use the formula
\[{y_1} = m\left( {x - {x_1}} \right)\]
Complete step by step solution: We have been given the data that,
One diagonal of a square is along the line \[8x - 15y = 0\] and one of its vertexes is \[(1,2)\]
Let us assume the square is \[{\rm{ABCD}}\] and vertex \[{\rm{C}}\] is \[(1,2)\]
We have the slope of \[{\rm{BD}}\] is \[\dfrac{8}{{15}}\]
And the angle made by \[{\rm{BD}}\] with \[{\rm{DC}}\] and \[{\rm{DC}}\] is $45^\circ$
Now let us consider the slope of \[{\rm{DC}}\] be \[{\rm{m}}\]
Now, it becomes
\[\tan {45^\circ } = \pm \dfrac{{{\rm{m}} - \dfrac{8}{{15}}}}{{1 + \dfrac{8}{{15}}\;{\rm{m}}}}\]
Now, we have to solve the fraction to simpler term, we have
\[ \Rightarrow (15 + 8\;{\rm{m}}) = \pm (15\;{\rm{m}} - 8)\]
On solving for the value of \[{\rm{m}}\] we get
\[ \Rightarrow {\rm{m}} = \dfrac{{23}}{7}{\rm{ and }} - \dfrac{7}{{23}}\]
Now, we obtained the equations of \[{\rm{DC}}\] and \[{\rm{BC}}\] as
\[ \Rightarrow {\rm{y}} - 2 = \dfrac{{23}}{7}({\rm{x}} - 1)\]
Now, we have to multiply the denominator with the term on the left side and multiply the numerator with the term on the right side, we get
\[ \Rightarrow 23{\rm{x}} - 7{\rm{y}} - 9 = 0\]
\[{\rm{y}} - 2 = - \dfrac{7}{{23}}({\rm{x}} - 1)\]
\[ \Rightarrow 7{\rm{x}} + 23{\rm{y}} - 53 = 0\]
Therefore, the equation of the sides of the square passing through this vertex, are \[23x - 7y - 9 = 0,7x + 23y - 53 = 0\]
Option ‘C’ is correct
Note: Students are likely to make mistakes in these types of problems because it involves trigonometry expressions and more calculations. On should know the formula to proceed with the answer and also should remember all the formulas to solve these types of problems.
Recently Updated Pages
JEE Main 2023 April 6 Shift 1 Question Paper with Answer Key

JEE Main 2023 April 6 Shift 2 Question Paper with Answer Key

JEE Main 2023 (January 31 Evening Shift) Question Paper with Solutions [PDF]

JEE Main 2023 January 30 Shift 2 Question Paper with Answer Key

JEE Main 2023 January 25 Shift 1 Question Paper with Answer Key

JEE Main 2023 January 24 Shift 2 Question Paper with Answer Key

Trending doubts
JEE Main 2026: Session 2 Registration Open, City Intimation Slip, Exam Dates, Syllabus & Eligibility

JEE Main 2026 Application Login: Direct Link, Registration, Form Fill, and Steps

Understanding the Angle of Deviation in a Prism

Hybridisation in Chemistry – Concept, Types & Applications

How to Convert a Galvanometer into an Ammeter or Voltmeter

Understanding the Electric Field of a Uniformly Charged Ring

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

NCERT Solutions For Class 11 Maths Chapter 12 Limits and Derivatives (2025-26)

NCERT Solutions For Class 11 Maths Chapter 10 Conic Sections (2025-26)

JEE Advanced Weightage 2025 Chapter-Wise for Physics, Maths and Chemistry

Derivation of Equation of Trajectory Explained for Students

Understanding Electromagnetic Waves and Their Importance

