GI−→
bisects ∠DGH
so that m∠DGI
is x−3
and m∠IGH
is 2x−13
. Find the value of x.
(2 points)
Since GI bisects ∠DGH, we have:
m∠DGI = m∠IGH.
From the given information, we can set up the following equation:
x - 3 = 2x - 13.
Solving for x, we can subtract x from both sides:
-3 = x - 13.
Then, we add 13 to both sides:
10 = x.
Therefore, the value of x is 10.
The midpoint of UV¯¯¯¯¯
is (5,−11)
. The coordinates of one endpoint are U(3,5)
. Find the coordinates of endpoint V.
We can use the midpoint formula to find the coordinates of endpoint V.
The midpoint formula states that the coordinates of the midpoint (M) between two endpoints (P and Q) is the average of the x-coordinates and the average of the y-coordinates:
M = ((x1 + x2)/2, (y1 + y2)/2).
In this case, we know that:
M = (5, -11).
P = (3, 5).
Let V = (x, y).
Using the midpoint formula, we can set up the following equations:
5 = (3 + x)/2,
-11 = (5 + y)/2.
To solve for x:
Multiply both sides of the first equation by 2:
10 = 3 + x.
Subtract 3 from both sides:
7 = x.
So the x-coordinate of endpoint V is 7.
To solve for y:
Multiply both sides of the second equation by 2:
-22 = 5 + y.
Subtract 5 from both sides:
-27 = y.
So the y-coordinate of endpoint V is -27.
Therefore, the coordinates of endpoint V are V(7, -27).
Find the distance between points M(6,16)
and Z(−1,14)
to the nearest tenth.
To find the distance between two points (x1, y1) and (x2, y2), we can use the distance formula:
d = sqrt((x2 - x1)^2 + (y2 - y1)^2).
In this case, we have:
M(6, 16) -> (x1, y1)
Z(-1, 14) -> (x2, y2)
Plugging these values into the formula, we get:
d = sqrt((-1 - 6)^2 + (14 - 16)^2).
Simplifying further:
d = sqrt((-7)^2 + (-2)^2).
Calculating:
d = sqrt(49 + 4).
d = sqrt(53).
To the nearest tenth, the distance is approximately:
d ≈ 7.3.
Therefore, the distance between points M(6, 16) and Z(-1, 14) is approximately 7.3 units.
To find the value of x, we can use the Angle Bisector Theorem. According to the theorem, if a ray bisects an angle, then it divides the angle into two congruent angles.
In this case, we have the angle DGH, which is bisected by the ray GI. Let's call the angle to the left of the bisector DGI and the angle to the right of the bisector IGH.
According to the problem, the measure of angle DGI is x - 3, and the measure of angle IGH is 2x - 13.
Since the angle is bisected, DGI is congruent to IGH. So we can set up an equation:
x - 3 = 2x - 13
Now, let's solve for x:
x - 2x = -13 + 3
-x = -10
x = 10
Therefore, the value of x is 10.
To find the value of x, we can use the fact that a bisector divides an angle into two congruent angles.
Let's set up an equation using the given information:
m∠DGI = x - 3
m∠IGH = 2x - 13
Since GI bisects ∠DGH, this means that m∠DGI = m∠IGH.
So we can set the two expressions equal to each other:
x - 3 = 2x - 13
Now, we can solve for x.
Subtract x from both sides of the equation:
x - 3 - x = 2x - 13 - x
-3 = x - 13
Add 13 to both sides of the equation:
-3 + 13 = x - 13 + 13
10 = x
Therefore, the value of x is 10.