coordinate of K is (0,7.5) or (0,[tex]\frac{15}{2}[/tex]}
and
scale factor=[tex]\frac{5}{8}[/tex]
Answer:
solution given:
O(0,0)
B(5,0)
P(8,0)
K(0,x)
S(0,-12)
and
ΔBOK [tex]\sim[/tex] ΔPOS
Now
By using the distance formula
d=[tex]\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}[/tex]
OB=[tex]\sqrt{(5-0)^2-(0-0)^2} =5[/tex]
OP=[tex]\sqrt{(8-0)^2-(0-0)^2} =8[/tex]
OK=[tex]\sqrt{(0-0)^2-(x-o)^2} =x[/tex]
OS=[tex]\sqrt{(0-0)^2-(-12-0)^2} =12[/tex]
since
ΔBOK is similar to ΔPOS
So their side will be proportional.
[tex]\frac{OK}{OS}=\frac{OB}{OP}[/tex]
taking two proportional only
[tex]\frac{x}{12} =\frac{5}{8}[/tex]
[tex]x=\frac{ 12*5}{8}=\frac{15}{2}[/tex]=7.5
Now
coordinate of K is (0,7.5) or (0,[tex]\frac{15}{2}[/tex]}
and
Scale factor = Dimensions of the new shape ÷ Dimensions of the original shape.
scale factor=[tex]\frac{OB}{OP}=\frac{5}{8}[/tex]
