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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.

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.

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