Question

Find the value of \[\left| {a + b} \right|\] If the mirror image of the point \[P\left( {a,6,9} \right)\] with respect to the line \[\dfrac{{x - 3}}{7} = \dfrac{{y - 2}}{5} = \dfrac{{z - 1}}{{ - 9}}\] is \[\left( {20,b, - a - 9} \right)\] where \[a,b \in R\].
A. \[88\]
B. \[86\]
C. \[84\]
D. \[90\]

Answer 1

Hint: In this question, we need to find the value of \[\left| {a + b} \right|\]. For this, we have to determine the midpoint of point P and the Image of point P with respect to the line first. We can find the values of \[a,b\] and \[c\] as this midpoint lies on the given equation of a line. After that, we will get the desired value.

Formula used: The midpoint of two points such as \[A\left( {{x_1},{y_1},{z_1}} \right)\] and \[B\left( {{x_2},{y_2},{z_2}} \right)\] is
\[M = \left( {\dfrac{{{x_1} + {x_2}}}{2},\dfrac{{{y_1} + {y_2}}}{2},\dfrac{{{z_1} + {z_2}}}{2}} \right)\]
Here, \[M\] is the midpoint of A and B.
Also, the absolute or modulus of a real number x is given by
\[\left| x \right| = \left\{ {\begin{array}{*{20}{c}}{x{\rm{ if }}x \ge 0}\\{ - x{\rm{ if }}x < {\rm{0}}}\end{array}} \right.\]

Complete step-by-step solution:
We have been given the coordinates of the point P as \[P\left( {a,6,9} \right)\].
Also, the equation of a line is \[\dfrac{{x - 3}}{7} = \dfrac{{y - 2}}{5} = \dfrac{{z - 1}}{{ - 9}}\]
Suppose M be the midpoint of point P and the Image of point P with respect to the line(Point Q).
Firstly, we will find the midpoint of points P and Q.
Let us suppose \[\left( {a,6,9} \right) \equiv \left( {{x_1},{y_1},{z_1}} \right)\] and \[\left( {20,b, - a - 9} \right) \equiv \left( {{x_2},{y_2},{z_2}} \right)\]
Thus, we get
\[M = \left( {\dfrac{{a + 20}}{2},\dfrac{{6 + b}}{2},\dfrac{{ - a}}{2}} \right)\]
But the point M lies on the given equation of a line.
So, consider \[\left( {\dfrac{{a + 20}}{2},\dfrac{{6 + b}}{2},\dfrac{{ - a}}{2}} \right) \equiv \left( {x,y,z} \right)\]
As the midpoint lies on the line the coordinates of the midpoint satisfy the equation of line so we will put these values of x, y, and z in the given equation of line.
\[\dfrac{{\left( {\dfrac{{a + 20}}{2}} \right) - 3}}{7} = \dfrac{{\left( {\dfrac{{6 + b}}{2}} \right) - 2}}{5} = \dfrac{{\left( {\dfrac{{ - a}}{2}} \right) - 1}}{{ - 9}}\]
Let us simplify this.
\[\dfrac{{\left( {\dfrac{{a + 20 - 6}}{2}} \right)}}{7} = \dfrac{{\left( {\dfrac{{6 + b - 4}}{2}} \right)}}{5} = \dfrac{{\left( {\dfrac{{ - a - 2}}{2}} \right)}}{{ - 9}}\]
\[\dfrac{{\left( {\dfrac{{a + 14}}{2}} \right)}}{7} = \dfrac{{\left( {\dfrac{{b + 2}}{2}} \right)}}{5} = \dfrac{{\left( {\dfrac{{ - a - 2}}{2}} \right)}}{{ - 9}}\]
\[\left( {\dfrac{{a + 14}}{{2 \times 7}}} \right) = \left( {\dfrac{{b + 2}}{{2 \times 5}}} \right) = \left( {\dfrac{{ - a - 2}}{{2 \times \left( { - 9} \right)}}} \right)\]
By simplifying further, we get
\[\left( {\dfrac{{a + 14}}{{14}}} \right) = \left( {\dfrac{{b + 2}}{{10}}} \right) = \left( {\dfrac{{a + 2}}{{18}}} \right)\]
Now, we will find the values of a and b.
Consider \[\left( {\dfrac{{a + 14}}{{14}}} \right) = \left( {\dfrac{{a + 2}}{{18}}} \right)\]
By simplifying, we get
\[\begin{array}{l}18a + 252 = 14a + 28\\ \Rightarrow 18a - 14a = 28 - 252\\ \Rightarrow 4a = - 224\\ \Rightarrow a = - 56\end{array}\]
Also, consider \[\left( {\dfrac{{b + 2}}{{10}}} \right) = \left( {\dfrac{{a + 2}}{{18}}} \right)\]
Put \[a = - 56\] in the above equation.
Thus, we get
\[\begin{array}{l}18b + 36 = 10\left( { - 56 + 2} \right)\\ \Rightarrow 18b = 10\left( { - 54} \right) - 36\\ \Rightarrow 18b = - 540 - 36\\ \Rightarrow 18b = - 576\\ \Rightarrow b = - 32\end{array}\]
Let us find the value of \[\left| {a + b} \right|\]
So, \[a + b = - 56 - 32\]
\[a + b = - 88\]
Thus, \[\left| {a + b} \right| = \left| { - 88} \right| = 88\]
We know that \[\left| x \right| = \left\{ {\begin{array}{*{20}{c}}{x{\rm{ if }}x \ge 0}\\{ - x{\rm{ if }}x < {\rm{0}}}\end{array}} \right.\]
By applying the rule of modulus of a number, we get
Hence, the value of \[\left| {a + b} \right|\] is \[88\].
Therefore, the correct option is (A).

Note: Many students make mistakes in determining the values of a and b. As the desired answer is completely dependent on the values of a and b, we have to take care of that. Also, students can make mistakes in finding the coordinates of point M.