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
301.2k+ 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
The HCF of two numbers is 96 and their LCM is 1296 class 10 maths JEE_Main

Mutually Exclusive vs Independent Events: Key Differences Explained

Area vs Volume: Key Differences Explained for Students

If the magnetizing field on a ferromagnetic material class 12 physics JEE_Main

Four persons A B C and D initially at the corners of class 11 physics JEE_Main

If a parabola whose length of latus rectum is 4a touches class 11 maths JEE_Main

Trending doubts
JEE Main Marks vs Percentile 2026: Predict Your Score Easily

JEE Main Cutoff 2026: Category-wise Qualifying Percentile

JEE Main 2026: Exam Dates, Session 2 Updates, City Slip, Admit Card & Latest News

JEE Main Marks vs Rank 2026: Expected Rank for 300 to 0 Marks

NIT Cutoff 2026: Tier-Wise Opening and Closing Ranks for B.Tech. Admission

JEE Mains 2027 Subject Wise Percentile Explained

Other Pages
CBSE Class 10 Maths Question Paper 2026 OUT Download PDF with Solutions

Complete List of Class 10 Maths Formulas (Chapterwise)

NCERT Solutions For Class 10 Maths Chapter 11 Areas Related To Circles - 2026-27 Free PDF Download (Login Required)

All Mensuration Formulas with Examples and Quick Revision

NCERT Solutions For Class 10 Maths Chapter 13 Statistics - 2026-27 Free PDF Download (Login Required)

NCERT Solutions For Class 10 Maths Chapter 14 Probability - 2026-27 Free PDF Download (Sign-in Required)

