Question

Use graph paper for this question. (i) The point P (-2, 4) is reflected as P' along the line AB parallel to X-axis at a distance of 1 unit on the negative side of the Y- axis. Plot P and P' and write the co-ordinates of P' (ii) The point P' is mapped to P'' on reflection in Y- axis. Find the co-ordinates of P''.(iii) P''' is the image of P'' when reflected in the origin. Find the co-ordinates of P''' (iv) Name the geometric figure P'P''P'''and find its area.

Answer 1

Given the coordinates of the point P:

$P=(-2,4)$

(i)

A reflection across the line y = -1 (that is, parallel to the x-axis at a distance of 1 unit on the negative side of the y-axis) is expressed by the transformation:

$(x,y)\rightarrow(x,-y-2)$

Then, using the coordinates of P, we can find the coordinates of P':

$\begin{gathered} (-2,4)\rightarrow(-2,-4-2) \\ \\ \Rightarrow P^{\prime}=(-2,-6) \end{gathered}$

Plotting both points and the line y = -1:

(ii)

A reflection across the y-axis is given by the following transformation:

$(x,y)\rightarrow(-x,y)$

Using the coordinates of P', we can find the coordinates of P'':

$\begin{gathered} (-2,-6)\rightarrow(2,-6) \\ \\ \Rightarrow P^{\prime}^{\prime}=(2,-6) \end{gathered}$

(iii)

A reflection across the origin can be expressed by the transformation rule:

$(x,y)\rightarrow(-x,-y)$

Then, using this rule, we can go from the coordinates of P'' to the coordinates of P''':

$\begin{gathered} (2,-6)\rightarrow(-2,6) \\ \\ \Rightarrow P^{\prime}^{\prime}^{\prime}=(-2,6) \end{gathered}$

(iv)

We plot the figure generated by P', P'', and P''':

As we can see, it is a right triangle with a height of 12 units and a base of 4 units. The area of this triangle is:

$\begin{gathered} A=\frac{12\cdot4}{2} \\ \\ \therefore A=24 \end{gathered}$