Let, coordinate of point A' is (x,y).
Since, A' is the symmetric point A(3, 2) with respect to the line 2x + y - 12 = 0.
So, slope of line containing A and A' will be perpendicular to the line 2x + y - 12 = 0 and also their center lies in the line too.
Now, their center is given by :
Also, product of slope will be -1 .( Since, they are parallel )
x = 2y - 1
So,
Also, C satisfy given line :
Also,
Therefore, the symmetric points is .
Answer:
1) d) Square
2) Proofs that PWRS is a rhombus are
Length of QS ≠ PR and
Slope of segment QR and PS is -1/2 and Slope of segment RS and QP is -2.
Step-by-step explanation:
The given points (x, y) of the parallelogram are;
E(-2, -4), F(0, -1), G(-3, 1), H(-5, -2)
The slope, m, of the segments are found as follows;
By computation, the slope of segment EF = 1.5
The slope of segment FG = -0.67
The slope of segment GH = 1.5
The slope of segment HE = -0.67
Therefore, EF is parallel to GH and FG is parallel to HE
The length of the sides are;
By computation, the length of segment EF = 3.61
The length of segment FG = 3.61
The length of segment GH = 3.61
The length of segment HE = 3.61
The diagonals are;
EG and FH
The length of segment EG = 5.099
The length of segment FH = 5.099
Therefore, the diagonals are equal and the parallelogram is a square
2) The given dimensions are;
P(-1, 3), Q(-2, 5), R(0, 4), S(1, 2)
A rhombus has all sides equal
The length of segment PQ = 2.24
The length of segment QR = 2.24
The length of segment RS = 2.24
The length of segment PS = 2.24
The diagonals are;
QS and PR
The length of segment QS = 4.24
The length of segment PR = 1.41
The slope of segment QR = -0.5
The slope of segment PS = -0.5
The slope of segment RS = -2
The slope of segment QP = -2
Therefore, QS≠QR the parallelogram is a rhombus
The correct option ;
Length of QS ≠ PR and
Slope of segment QR and PS is -1/2 and Slope of segment RS and QP is -2.
Where there are acute angles in parallelogram PQRS, then the correct option is d) Length of QR and PS is 2.2 and Length of RS and QP is 2.2