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This is one of our many quizzes on Transformations. Click on this to access more such competitive quizzes on 3D Coordinate Geometry.

Transformations Quiz 2

mathematics

The below figure consists of an image. It is a combination of rotation and reflection. At what angle it is rotated? <img src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/67e98576-90a5-4e62-8107-b53bf9bd6d8c3512151988036180845.png"/>

 Observe the figure, and think about rotation and reflection about clockwise and anti-clockwise direction.

 The image is first rotated and reflected. The rotation is taken about 90°. Then the image is reflected about BC. Therefore, the image is formed.

What will be the coordinates of $C'$in the reflected image about y –axis of the triangle given below: <img src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/98555f8c-a3c7-47b5-8086-88da07ce2af83751526142189548370.png"/>

 Check the reflected point by comparing it with the initial point in the respected quadrant.

If the image is reflected by about y- axis then the image will be formed in the second quadrant. In the second quadrant y-axis is positive and x-axis is negative. Initially, the point C is (4, 6) after reflection the point will become (-4, 6).

What will be the coordinates A’B’C’ when ABC is translated to $\left( \begin{matrix} &nbsp;&nbsp;&nbsp;2 \\ &nbsp;&nbsp;&nbsp;4 \\ \end{matrix} \right)$ and reflected about the line f? <img src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/d8564f41-dbf4-4c15-8d0e-73a37ffb8c6d2204062615275702283.png"/>

 Translate the figure according to given information then reflect it and find the coordinates of the final image.

When ABC is translated to $\left( \begin{matrix} &nbsp;&nbsp;&nbsp;2 \\ &nbsp;&nbsp;&nbsp;4 \\ \end{matrix} \right)$ we get, <img src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/abd92cb4-bb97-4931-a61e-ca8162d62d3e6476174975934851229.png"/> Now reflecting it along the line we get, <img src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/e03e1cd8-54b4-4999-ae43-ca508eaf41d62600835998888905157.png"/> The coordinates will be $\text{A }\!\!'\!\!\text{ }\left( 15,10 \right),\text{B }\!\!'\!\!\text{ }\left( 10,5 \right),\text{C }\!\!'\!\!\text{ }\left( 10,10 \right)$.

What is the equation of line of reflection in the given figure? <img src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/25aff35b-1cae-4247-b516-de220ac1e5683261644758940635147.png"/>

By using the equation, $(y-{{y}_{1}})=\dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}(x-{{x}_{1}})$ From the figure the values are (2,8) (6,2). By substituting the value in the equation, $\begin{align} &nbsp;&nbsp;& (y-8)=\dfrac{2-8}{6-2}(x-2) \\ &nbsp;& \Rightarrow 2y-16=-3x+6 \\ &nbsp;& \Rightarrow y=\dfrac{-3}{2}x+11 \\ \end{align}$

What is the product of distances between the point of intersection of ABC and the image when it is translated to $\left( \begin{matrix} &nbsp;&nbsp;&nbsp;0 \\ &nbsp;&nbsp;&nbsp;-3 \\ \end{matrix} \right)$ and when it is translated to $\left( \begin{matrix} &nbsp;&nbsp;&nbsp;-3 \\ &nbsp;&nbsp;&nbsp;0 \\ \end{matrix} \right)$. <img src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/d7873efc-9efb-4394-9e81-9fe8c700abe01392092195880443304.png"/>

 Translate the given triangle to $\left( \begin{matrix}   0  \\   -3  \\\end{matrix} \right)$ and $\left( \begin{matrix}   -3  \\   0  \\\end{matrix} \right)$ to find the point of intersections by translating the coordinates of a given triangle. Then determine the distances using the distance formula and find the product.

When ABC is translated to $\left( \begin{matrix} &nbsp;&nbsp;&nbsp;0 \\ &nbsp;&nbsp;&nbsp;-3 \\ \end{matrix} \right)$ we get A’B’C’ as, <img src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/52a531f2-e13d-4bc7-9dea-354ccf953b353172401766349380302.png"/> The points of intersection are $\left( -5,-2 \right)\text{ and }\left( -7,-3 \right)$. The distance between the points is $\sqrt{{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}+{{\left( {{y}_{2}}-{{y}_{1}} \right)}^{2}}}$. &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;$\begin{align} &nbsp;&nbsp;& \Rightarrow \sqrt{{{\left( -7+5 \right)}^{2}}+{{\left( -3+2 \right)}^{2}}} \\ &nbsp;& \Rightarrow \sqrt{{{\left( -2 \right)}^{2}}+{{\left( -1 \right)}^{2}}} \\ &nbsp;& \Rightarrow \sqrt{5} \\ \end{align}$ When ABC is translated to $\left( \begin{matrix} &nbsp;&nbsp;&nbsp;-3 \\ &nbsp;&nbsp;&nbsp;0 \\ \end{matrix} \right)$ we get A’B’C’ as, <img src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/83d49d4b-05f1-4fec-8164-77f4dab0dacc823206481618885387.png"/> The points of intersection are $\left( -7,0 \right)\text{ and }\left( -8,-2 \right)$. The distance between the points is $\sqrt{{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}+{{\left( {{y}_{2}}-{{y}_{1}} \right)}^{2}}}$. &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;$\begin{align} &nbsp;&nbsp;& \Rightarrow \sqrt{{{\left( -8+7 \right)}^{2}}+{{\left( -2+0 \right)}^{2}}} \\ &nbsp;& \Rightarrow \sqrt{{{\left( -1 \right)}^{2}}+{{\left( -2 \right)}^{2}}} \\ &nbsp;& \Rightarrow \sqrt{5} \\ \end{align}$ $\Rightarrow $The product will be$\sqrt{5}\times \sqrt{5}$$=5$