Question

Image of $\left( {1,2} \right)$ w.r.t $\left( { - 2, - 1} \right)$ is?
A. $\left( {0,5} \right)$
B. $\left( { - 4, - 3} \right)$
C. $\left( { - 4, - 2} \right)$
D. $\left( { - 5, - 4} \right)$

Answer 1

To find the image of given points $\left( {1,2} \right)$ w.r.t $\left( { - 2, - 1} \right)$ we must consider A and B as the variables with respect to the points given and to find the image let C be the point to find the values of x and y in which we can obtain the image of A and B, hence we can find the image with respect to A and B points.

Complete step by step solution:
Let us write the given data:
Let, A = $\left( {1,2} \right)$, B = $\left( { - 2, - 1} \right)$ and C = $\left( {x,y} \right)$

Let image of A is C and B is the midpoint of AC i.e., with respect to the given points:
$\left( { - 2, - 1} \right) = \left( {\dfrac{{1 + x}}{2},\dfrac{{2 + y}}{2}} \right)$
Hence, simplifying the terms
$- 2 = \dfrac{{1 + x}}{2}$
$- 4 - 1 = x$
$x = - 5$
And
$- 1 = \dfrac{{2 + y}}{2}$
$- 2 - 2 = y$
$y = - 4$
Therefore, the Image of $\left( {1,2} \right)$ w.r.t $\left( { - 2, - 1} \right)$ is $\left( { - 5, - 4} \right)$.

Hence, option D is the right answer.

Additional information:
A flat, two-dimensional, which extends infinitely is a plane. It is a 2d analogue of a point, a line and 3-dimensional space. A plane in 3d space is denoted by the equation
$ax + by + cz + d = 0$,
where a, b and c are non-zero.
A point and a vector are perpendicular to the plane, which determines the plane in three-dimensional coordinate space and steps to Find Image of Point in A Plane:

1. Consider the 2 points P and Q. Let $\pi$ be a plane such that there exists a perpendicular line PQ to the plane $\pi$.
2. The midpoint of PQ is on the plane $\pi$. Then, the image of the point is either of the points to one another in the plane $\pi$. The procedure to find the image of a point in a given plane is as follows:
   The equations of the normal to the given plane and the line passing through the point P are written as:
   $\dfrac{{x - {x_1}}}{a} = \dfrac{{y - {y_1}}}{b} = \dfrac{{z - {z_1}}}{c}$

Note:
The key point to find the image of the given points is that we must consider both the given values and then combine to evaluate. As we can see that we have considered $\left( { - 2, - 1} \right) = \left( {\dfrac{{1 + x}}{2},\dfrac{{2 + y}}{2}} \right)$, in which we need to solve – 2 with respect to the first term i.e., $\dfrac{{1 + x}}{2}$and next we must solve taking -1 with respect to the second term i.e., $\dfrac{{2 + y}}{2}$. Hence, in this way we need to solve and find the image.