Answer:
The correct option is d.
Step-by-step explanation:
It is given that the figure JKLM is image of ABCD. So the side MJ is the image of DA.
From the figure it is noticed that the A(-8,5) and D(-7,7).
Distance formula is
\(d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\)
\(AD=\sqrt{(-7-(-8))^2+(7-5)^2}\)
\(AD=\sqrt{1+4}\)
\(AD=\sqrt{5}\)
The points of image are J(-6,-8) and M(-3,-2).
\(JM=\sqrt{(-3-(-6))^2+(-2-(-8))^2}\)
\(JM=\sqrt{(3)^2+(6)^2}\)
\(JM=\sqrt{9+36}\)
\(JM=\sqrt{45}\)
\(JM=3\sqrt{5}\)
The scale factor is,
\(k=\frac{JM}{AD}=\frac{3\sqrt{5}}{\sqrt{5}} =3\)
Therefore the corresponding sides of image are 3 times more than the preimage.
From the figure it is noticed that the distance between DC is 1 units.
Since the sides ML is image of DC, therefore the length of ML is 3 units.
The point C is 1 unit right from D, therefore the point L is 3 unit right from M. The y coordinate of M and L are same. To find the x coordinate of L we have to add 3 units in x coordinate of M.
\(L=(-3+3,-2)\)
\(L=(0,-2)\)
Therefore the option d is correct.
