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Rhombus EFGH is rotated 90 degrees counterclockwise about vertex G. How many pairs of parallel sides does the rotated figure hav
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To find out if a point is inside, on, or outside a circle, we need to substitute the ordered pair into the equation of the circle:
(x-xc)^2+(y-yc)^2=r^2
where (xc,yc) is the centre of the circle, and r=radius of the circle.
If the left-hand side [(x-xc)^2+(y-yc)^2] is less than r^2, then point (x,y) is INSIDE the circle. If the left-hand side is equal to r^2, the point is ON the circle.
Finally, if the left-hand side is greater than r^2, the point is OUTSIDE the circle.
For the given problem, we have xc=3, yc=0, or centre at (3,0), r=sqrt(49)=7
(x-xc)^2+(y-yc)^2=r^2 => (x-3)^2+y^2=7^2
A. (-1,1),
(x-3)^2+y^2=7^2 => (-1-3)^2+1^2=16+1=17 <49 [inside circle]
B. (10,0)
(x-3)^2+y^2=7^2 => (10-3)^2+0^2=49+0=49 [on circle]
C. (4,-8)
(x-3)^2+y^2=7^2 => (4-3)^2+(-8)^2=1+64=65 > 49 [outside circle]
(x-xc)^2+(y-yc)^2=r^2
where (xc,yc) is the centre of the circle, and r=radius of the circle.
If the left-hand side [(x-xc)^2+(y-yc)^2] is less than r^2, then point (x,y) is INSIDE the circle. If the left-hand side is equal to r^2, the point is ON the circle.
Finally, if the left-hand side is greater than r^2, the point is OUTSIDE the circle.
For the given problem, we have xc=3, yc=0, or centre at (3,0), r=sqrt(49)=7
(x-xc)^2+(y-yc)^2=r^2 => (x-3)^2+y^2=7^2
A. (-1,1),
(x-3)^2+y^2=7^2 => (-1-3)^2+1^2=16+1=17 <49 [inside circle]
B. (10,0)
(x-3)^2+y^2=7^2 => (10-3)^2+0^2=49+0=49 [on circle]
C. (4,-8)
(x-3)^2+y^2=7^2 => (4-3)^2+(-8)^2=1+64=65 > 49 [outside circle]
Coordinates of the given triangle ΔPQR P(1,-1), Q(3,-2), R(3,-4).
a) Let us discuss the rule for rotation of 90° clockwise about the origin.
If given coordinate is (h,k) is rotated 90° clockwise about the origin the position of rotated coordinate will be (k, -h).
x and y coordinates would switch their values and after switching x any values, the y value of the coordinates would reverse in sign.
So, the coordinates of rotated ΔPQR P(1,-1), Q(3,-2), R(3,-4) would become
P(1,-1) --> P'(-1,-1)
Q(3,-2) --> Q'( -2,-3)
R(3,-4) --> R'(-4,-3)
b) If given coordinate is (h,k), the coordinates after reflection across the x-axis would besome (h,-k). (Just change in the sign of y-coordinate).
So, the coordinates of rotated ΔPQR P(1,-1), Q(3,-2), R(3,-4) would become
P(1,-1) --> P"(1,1)
Q(3,-2) --> Q"( 3,2)
R(3,-4) --> R"(3,4)
