Diagonal of a Rhombus are perpendicular & intersects in their middle point:
Assume the diagonals intersects in H
A(0,-8), B(1,-0), C(8,-4) & D(x, y) are the vertices of the rhombus and we have to calculate D(x, y)
Consider the diagonal AC. Find the coordinate (x₁, y₁) H, the middle of AC
Coordinate (x₁, y₁) of H, middle of A(0,-8), C(8,-4)
x₁ (0+8)/2 & y₁=(-8-4)/2 ==> H(4, -6)
Now let's calculate again the coordinate of H, middle of the diagonal BD
B(1,-0), D(x, y)
x value = (1+x)/2 & y value=(y+0)/2 ==> x= (1+x)/2 & y=y/2
(1+x)/2 & y/2 are the coordinate of the center H, already calculated, then:
H(4, -6) = [(1+x)/2 , y/2]==>(1+x)/2 =4 ==> x=7 & y/2 = -6 ==> y= -12
Hence the coordinates of the 4th vertex D(7, -12)
Answer:
-17
Step-by-step explanation:
First, let's see what the value of w is, by solving -5w - 5 = 5.
-5w - 5 = 5 Add 5 on both sides.
-5w = 10 Divide -5 on both sides.
w = -2
Now, we know that w is equal to -2, let's plug it in to find the value.
9(-2) + 1 Distribute.
-18 + 1 Add.
-17
Assuming that this particular type of question is a system of equations one. More specifically linear equations, you would simply substitute one of the equations into the other one. Then upon doing so, plug in the x and y coordinates for the point into the x and y variables of the equation and then see if it is a solution, by checking if the left hand side of the equation equals that of the right hand side of the equation.
