C(-8,2) and M(0,0) , since M is at the origin. Let x₁ and y₁ be the
coordinates of S →s(x₁ , y₁)
C(-8,2) and S(x₁ , y₁)
The coordinates of M, the midpoint of CS are M(x₂ , y₂)
a) x₂ = (-8 + x₁)/2 , but x₂ = 0, then :
0 = -4+x₁/2 and x₁ = 8
b) y₂ = (2+y₁)/2 , but y₂ = o, then:
0 = 2+ y₁/2 and y₂ = -2
Then the coordinates of S are S(8 , -2)
Answer:
7 square units
Step-by-step explanation:
As with many geometry problems, there are several ways you can work this.
Label the lower left and lower right vertices of the rectangle points W and E, respectively. You can subtract the areas of triangles WSR and EQR from the area of trapezoid WSQE to find the area of triangle QRS.
The applicable formulas are ...
area of a trapezoid: A = (1/2)(b1 +b2)h
area of a triangle: A = (1/2)bh
So, our areas are ...
AQRS = AWSQE - AWSR - AEQR
= (1/2)(WS +EQ)WE -(1/2)(WS)(WR) -(1/2)(EQ)(ER)
Factoring out 1/2, we have ...
= (1/2)((2+5)·4 -2·2 -5·2)
= (1/2)(28 -4 -10) = 7 . . . . square units
The answer is -3
You use the formula y1-y2
———
x1-x2
-1 and 8 are your y’s
1 and -2 are your x’s
Your equation should look like this
-1-8 -9
—— = —— = -3
1-(-2) 3
I plugged the y’s and x’s into the formula.
I hope this helps!!
Hello,
It is true that the sequence is arithmetic. Its common difference, of subtracting 7 each time, is always going to subtract 7 from each number.
Hope this helps!
May
