Point P(4,-2) undergoes a translation given by (x,y) - (x+3, x-a) followed by another translation (x,y) - (x-b, x+7) to produce the image of P” (-5,8). find the values of a and b and point P’.

• In this problem, we are moving from point P, to point P', and then to point P''.
,• To move from one point to another, we make translations of the points.
,• The ticks of the points are simply a notation. One tick denotes the point where you are after the first translation. Two ticks denote the point where you are after the second translation.
• Mathematically, a translation from a point P to a new point P' consists in summing numbers to the coordinates of the point P to get the coordinates of the new point P'.
,• For example, if we have the point P with coordinates (x,y), and we move that point 3 units to the right, and 4 units up, we get the point P' with coordinates:
[tex]P^{\prime}=(x+3,y+4)\text{.}[/tex]This problem consists of the following:
1) We start with a point P(4,-2) with coordinates x = 4 and y = -2.
2) We make the first translation, which consists in going from point P to point P'.
The coordinates of point P' are given by:
[tex]P^{\prime}=(x+3,y-a).[/tex]The coordinates x and y in that formula are the values of x and y of the original point. So we must replace x = 4 and y = -2 in the formula above. Doing that we have the following coordinates for the point P':
[tex]P^{\prime}=(4+3,-2-a)=(7,-2-a)\text{.}[/tex]3) We made the first translation. Now we will do another translation, from point P' with coordinates x = 7 and y = -2 - a, to the point P'' with coordinates:
[tex]P^{\prime\prime}=(x-b,y+7).[/tex]Replacing the values x = 7 and y = -2 - a we get the following coordinates for P'':
[tex]\begin{gathered} P^{\prime\prime}=(7-b,-2-a+7), \\ P^{\prime\prime}=(7-b,5-a), \end{gathered}[/tex]4) Finally, doing the translations we get the following coordinates for point P'':
[tex]P^{\prime\prime}=(7-b,5-a)\text{.}[/tex]But from the statement of the problem, we know that the coordinates of point P'' are:
[tex]P^{\prime\prime}=(-5,8)\text{.}[/tex]Comparing each coordinate we have the following equations:
[tex]7-b=-5,\text{ and }^{}5-a=8.[/tex]Solving the equation of b, we get:
[tex]\begin{gathered} 7-b=-5, \\ 7=-5+b, \\ b=7+5, \\ b=12. \end{gathered}[/tex]Solving the equation of a, we get:
[tex]\begin{gathered} 5-a=8, \\ 5=8+a, \\ a=5-8, \\ a=-3. \end{gathered}[/tex]So the values of a and b are:
[tex]\begin{gathered} a=-3, \\ b=12. \end{gathered}[/tex]5) Using the value a = -3, the coordinates of point P' are:
[tex]P^{\prime}=(7,-2-(-3))=(7,-2+3)=(7,1)\text{.}[/tex]Answers
[tex]\begin{gathered} a=-3, \\ b=12, \\ P^{\prime}(7,1)\text{.} \end{gathered}[/tex]