Question

the point P(5,1)and Q(-2,-2)are reflected along the line x=2. find it's image​

Answer 1

Answer:

Given:

Coordinates of point '\begin{gathered}P\\\end{gathered}P ' = \begin{gathered}(5,1)\\\end{gathered}(5,1)

Coordinates of point '\begin{gathered}Q\\\end{gathered}Q ' = \begin{gathered}(-2,-2)\\\end{gathered}(−2,−2)

The points '\begin{gathered}P\\\end{gathered}P ' and '\begin{gathered}Q\\\end{gathered}Q ' are reflected in the line "\begin{gathered}x=2\\\end{gathered}x=2 ".

After reflection, the two new points formed are denoted by "\begin{gathered}P'\\\end{gathered}P′ " and "\begin{gathered}Q'\\\end{gathered}Q′ ".

To find:

To plot the points \begin{gathered}P, Q, P' and Q'\\\end{gathered}P,Q,P′andQ′ on graph paper.

To name the figure formed by joining the above four points.

To find the area of the figure, so formed.

Note:

An image is attached with the solution showing the graph sheet.

To be recollected:

Reflection of a point about a line means the point on one side of the line forms it's reflection as a point on the other side of the line. Both the point and it's reflection are equidistant from the line.

Step-wise Solution:

Step-1:

This step involves in plotting the points on graph sheet.

The points were plotted on a graph sheet and attached as an image.

Step-2:

This step involves naming the figure formed by joining the '\begin{gathered}P\\\end{gathered}P ', '\begin{gathered}Q\\\end{gathered}Q ' and the reflected points "\begin{gathered}P'\\\end{gathered}P′ " and "\begin{gathered}Q'\\\end{gathered}Q′ ".

The point \begin{gathered}P (5,1)\\\end{gathered}P(5,1) is at a distance of \begin{gathered}3\\\end{gathered}3 units from the line, according to the graph. Hence, it's reflection will also be at the same distance from the line.

The coordinates of so formed reflection are \begin{gathered}P'(-1,1)\\\end{gathered}P′(−1,1) .

Now, the point \begin{gathered}Q(-2,-2)\\\end{gathered}Q(−2,−2) is at a distance of \begin{gathered}4\\\end{gathered}4  units from the line, according to the graph. Hence, it's reflection will also be at the same distance from the line.

The coordinates of so formed reflection are \begin{gathered}Q'(6,-2)\\\end{gathered}Q′(6,−2) .

By joining the four points \begin{gathered}P,Q,P' and Q'\\\end{gathered}P,Q,P′andQ′ , a closed four sided figure is formed with two parallel sides and two non-parallel sides. It is similar to \begin{gathered}trapezium\\\end{gathered}trapezium , which can be seen in the image.

∴ The figure formed by joining the four points is a \begin{gathered}Trapezium\\\end{gathered}Trapezium .

Step-3:

This step involves, finding the area of the figure formed by joining \begin{gathered}P,Q,P' and Q'\\\end{gathered}P,Q,P′andQ′ , i.e., the area of the \begin{gathered}trapezium\\\end{gathered}trapezium .

Area of the trapezium is given by:

\begin{gathered}A=(1/2)h(a+b)\\\end{gathered}A=(1/2)h(a+b) ......(1)

Where, \begin{gathered}A\\\end{gathered}A = Required area of the \begin{gathered}trapezium\\\end{gathered}trapezium

\begin{gathered}h\\\end{gathered}h = Height of the \begin{gathered}trapezium\\\end{gathered}trapezium

\begin{gathered}'a' and 'b'\\\end{gathered}′a′and′b′ are the sides of the \begin{gathered}trapezium\\\end{gathered}trapezium .

The sides of the trapezium can be easily calculated from the graph. It is given by the distance of both the points. Simply, check for the number of units in-between the points in graph.

The distance between \begin{gathered}P\\\end{gathered}P and \begin{gathered}P'\\\end{gathered}P′ is \begin{gathered}6units\\\end{gathered}6units .

The distance between \begin{gathered}Q\\\end{gathered}Q and \begin{gathered}Q'\\\end{gathered}Q′ is \begin{gathered}8units\\\end{gathered}8units .

Height of the \begin{gathered}trapezium = 3units\\\end{gathered}trapezium=3units

\begin{gathered}a=6units\\b=8units\\h=3units\\\end{gathered}a=6unitsb=8unitsh=3units

Using equation (1), the area of the trapezium is calculated in the following way:

\begin{gathered}A=(1/2)*3*(6+8)\\A=(1/2)*3*14\\A=3*7\\A=21square units\\\end{gathered}A=(1/2)∗3∗(6+8)A=(1/2)∗3∗14A=3∗7A=21squareunits

Step-by-step explanation:

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