
The image of the point \[\left( {4, - 3} \right)\] with respect to the line \[y = x\] is:
A. \[\left( { - 4, - 3} \right)\]
B. \[\left( {3,4} \right)\]
C. \[\left( { - 4,3} \right)\]
D. \[\left( { - 3,4} \right)\]
Answer
217.8k+ views
Hint: The formula to find the image point with respect to the line is \[\dfrac{{x - {x_1}}}{a} = \dfrac{{y - {y_1}}}{b} = - 2\left[ {\dfrac{{a{x_1} + b{y_1} + c}}{{{a^2} + {b^2}}}} \right]\], where \[\left( {{x_1},{y_1}} \right)\] is the point on the line, \[a\] is coefficient of the \[x\]-variable, is coefficient of the \[y\]-variable and \[c\] be any constant term in the equation of the line.
Apply this formula to find the image point with respect to the line, and then use the given conditions to find the required value.
Complete step-by-step solution
We are given that the equation of the line is \[y = x\] and the point is \[\left( {4, - 3} \right)\].
Rewriting the given equation, we get
\[x - y = 0\]
We know the formula to find the image point with respect to the line is \[\dfrac{{x - {x_1}}}{a} = \dfrac{{y - {y_1}}}{b} = - 2\left[ {\dfrac{{a{x_1} + b{y_1} + c}}{{{a^2} + {b^2}}}} \right]\], where \[\left( {{x_1},{y_1}} \right)\] is the point on the line, \[a\] is coefficient of the \[x\]-variable, \[b\] is coefficient of the \[y\]-variable and \[c\] be any constant term in the equation of the line.
Finding the values of \[{x_1}\], \[{y_1}\], \[a\], \[b\] and \[c\] from the given equation of the line, we get
\[{x_1} = 4\]
\[{y_1} = - 3\]
\[a = 1\]
\[b = - 1\]
\[c = 0\]
Substituting these values in the above formula to find the image point with respect to the line, we get
\[
\Rightarrow \dfrac{{x - 4}}{1} = \dfrac{{y - \left( { - 3} \right)}}{{ - 1}} = - 2\left[ {\dfrac{{1\left( 4 \right) - 1\left( { - 3} \right) + 0}}{{{1^2} + {{\left( { - 1} \right)}^2}}}} \right] \\
\Rightarrow x - 4 = \dfrac{{y + 3}}{{ - 1}} = - 2\left( {\dfrac{{4 + 3}}{{1 + 1}}} \right) \\
\Rightarrow x - 4 = - y - 3 = - 2 \times \dfrac{7}{2} \\
\Rightarrow x - 4 = - y - 3 = - 7 \\
\]
Separating the above equations, we get
\[x - 4 = - 7{\text{ ......}}\left( 1 \right)\]
\[ - y - 3 = - 7{\text{ ......}}\left( 2 \right)\]
Adding the equation \[\left( 1 \right)\] by 4 on each of the sides, we get
\[
\Rightarrow x - 4 + 4 = - 7 + 4 \\
\Rightarrow x = - 3 \\
\]
Adding the equation \[\left( 2 \right)\] by 3 on each of the sides, we get
\[
\Rightarrow - y - 3 + 3 = - 7 + 3 \\
\Rightarrow - y = - 4 \\
\]
Multiplying the above equation by \[ - 1\] on each of the sides, we get
\[ \Rightarrow y = 4\]
Thus, \[x = - 3\] and \[y = 4\].
Therefore, the image point of the given point with respect to the line is \[\left( { - 3,4} \right)\].
Hence, the option D is correct.
Note: In solving these types of questions, you should be familiar with the formula of image point with respect to a line. Then use the given conditions and values given in the question, and substitute the values in this formula, to find the required value. Also, we are supposed to write the values properly to avoid any miscalculation.
Apply this formula to find the image point with respect to the line, and then use the given conditions to find the required value.
Complete step-by-step solution
We are given that the equation of the line is \[y = x\] and the point is \[\left( {4, - 3} \right)\].
Rewriting the given equation, we get
\[x - y = 0\]
We know the formula to find the image point with respect to the line is \[\dfrac{{x - {x_1}}}{a} = \dfrac{{y - {y_1}}}{b} = - 2\left[ {\dfrac{{a{x_1} + b{y_1} + c}}{{{a^2} + {b^2}}}} \right]\], where \[\left( {{x_1},{y_1}} \right)\] is the point on the line, \[a\] is coefficient of the \[x\]-variable, \[b\] is coefficient of the \[y\]-variable and \[c\] be any constant term in the equation of the line.
Finding the values of \[{x_1}\], \[{y_1}\], \[a\], \[b\] and \[c\] from the given equation of the line, we get
\[{x_1} = 4\]
\[{y_1} = - 3\]
\[a = 1\]
\[b = - 1\]
\[c = 0\]
Substituting these values in the above formula to find the image point with respect to the line, we get
\[
\Rightarrow \dfrac{{x - 4}}{1} = \dfrac{{y - \left( { - 3} \right)}}{{ - 1}} = - 2\left[ {\dfrac{{1\left( 4 \right) - 1\left( { - 3} \right) + 0}}{{{1^2} + {{\left( { - 1} \right)}^2}}}} \right] \\
\Rightarrow x - 4 = \dfrac{{y + 3}}{{ - 1}} = - 2\left( {\dfrac{{4 + 3}}{{1 + 1}}} \right) \\
\Rightarrow x - 4 = - y - 3 = - 2 \times \dfrac{7}{2} \\
\Rightarrow x - 4 = - y - 3 = - 7 \\
\]
Separating the above equations, we get
\[x - 4 = - 7{\text{ ......}}\left( 1 \right)\]
\[ - y - 3 = - 7{\text{ ......}}\left( 2 \right)\]
Adding the equation \[\left( 1 \right)\] by 4 on each of the sides, we get
\[
\Rightarrow x - 4 + 4 = - 7 + 4 \\
\Rightarrow x = - 3 \\
\]
Adding the equation \[\left( 2 \right)\] by 3 on each of the sides, we get
\[
\Rightarrow - y - 3 + 3 = - 7 + 3 \\
\Rightarrow - y = - 4 \\
\]
Multiplying the above equation by \[ - 1\] on each of the sides, we get
\[ \Rightarrow y = 4\]
Thus, \[x = - 3\] and \[y = 4\].
Therefore, the image point of the given point with respect to the line is \[\left( { - 3,4} \right)\].
Hence, the option D is correct.
Note: In solving these types of questions, you should be familiar with the formula of image point with respect to a line. Then use the given conditions and values given in the question, and substitute the values in this formula, to find the required value. Also, we are supposed to write the values properly to avoid any miscalculation.
Recently Updated Pages
Area vs Volume: Key Differences Explained for Students

Mutually Exclusive vs Independent Events: Key Differences Explained

JEE Main 2025 Question Papers With Solutions (January and April Sessions)

Adjoint and Inverse of a Matrix Explained for Students

Algebra Formula Guide: Key Equations & Examples for Students

Area Formula for Quadrilateral Explained Simply

Trending doubts
JEE Main 2026: Application Form Open, Exam Dates, Syllabus, Eligibility & Question Papers

JEE Main Syllabus 2026: Download Detailed Subject-wise PDF

JEE Main Previous Year Question Paper with Answer Keys and Solutions

Understanding Newton’s Laws of Motion

JEE Main Cut Off 2026 - Expected Qualifying Marks and Percentile Category Wise

Marks vs Percentile JEE Mains 2026: Calculate Percentile Marks

Other Pages
NCERT Solutions For Class 10 Maths Chapter 12 Surface Area And Volume

NCERT Solutions for Class 10 Maths Chapter Chapter 13 Statistics

NCERT Solutions for Class 10 Maths Chapter 11 Areas Related to Circles 2025-26

Pregnancy Week and Due Date Calculator: Find How Far Along You Are

Complete List of Class 10 Maths Formulas (Chapterwise)

NCERT Solutions for Class 10 Maths Chapter 15 Probability

