3, 7×3=21+5=26, there is your answer
A tranversal is a line that cuts and divides other lines proportionately
<h3>
Answer: point Q is located at (-1, 1)</h3>
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Explanation:
Check out the diagram below.
Plot R(-3,7) and T(3,-11) on the same xy grid.
Draw a vertical line through R and a horizontal line through T. A right triangle forms. At the intersection point is point S(-3,-11)
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Now measure the distance from point S to point T. You can count out the spaces or subtract the x coordinates and use absolute value.
|x1-x2| = |-3-3| = |-6| = 6
From point S to point T is 6 units.
We want to subdivide this horizontal length in the ratio 1:2
What this means is that we want to plot a point U somewhere such that SU:UT = 1:2
In other words,
SU = x
UT = 2x
SU+UT = ST
x+2x = 6
3x = 6
x = 6/3
x = 2
So we must move 2 spaces to the right from point S to get to U(-1,-11)
Going from point U(-1,-11) to T(3,-11) is 4 spaces
We have SU:UT = 2:4 = 1:2 to help confirm we have the correct location for point U
From point U, we then move straight up to the line segment RT
We'll land on Q(-1,1)
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Another way to find the y coordinate of point Q is to subdivide the segment RS into the ratio 1:2 similar to how we divided ST up
Segment RS is 18 units long since we go from y = 7 to y = -11 when going from R to S.
If V was on segment RS such that
RV:VS = 1:2
and RV = y
then
RV+VS = RS
y+2y = 18
3y = 18
y = 6
RV = y = 6
VS = 2y = 2*6 = 12
So you'll move 6 units down from y = 7 to land on y = 1 (when going from the y coordinate of R to the y coordinate of Q)
It is convenient to do part b first, then use that result to do part a.
b. For some pair of points (x1, y1) and (x2, y2), you want to find a point (a, b) such that (a, b) - (x1, y1) = (x2, y2) - (a, b). That is, the differences of coordinates from one end to the center are the same as the differences from the center to the other end.
Adding (a, b) to the above equation gives
2(a, b) - (x1, y1) = (x2, y2)
Adding (x1, y1) then gives
2(a, b) = (x1, y1) + (x2, y2)
Finally, dividing by 2 gives a formula for (a, b):
(a, b) = ((x1, y1) + (x2, y2))/2
The midpoint is the average of the end points.
a. Using the result from part b, the midpoint is
midpoint = ((2, 3) + (6, 7))/2 = (8, 10)/2
midpoint = (4, 5)
Answer:
x = 1
y = 5
z = 2
Step-by-step explanation:
System of equations:
6x - 2y + z = -2
2x + 3y - 3z = 11
x + 6y = 31
Isolate one variable in any of the equations:
x + 6y = 31
x = 31 - 6y
Plug in this value for x in another equation:
6(31 - 6y) - 2y + z = -2
186 - 36y - 2y + z = -2
186 - 38y + z = -2
-38y + z = -188
z = -188 + 38y
Plug in these values in the remaining equation:
2(31 - 6y) + 3y - 3(-188 + 38y) = 11
62 - 12y + 3y + 564 - 114y = 11
626 - 12y + 3y - 114y = 11
626 - 9y - 114y = 11
626 - 123y = 11
-123y = -615
y = 5
Plug in value of y into our other answers to solve for x and z:
x = 31 - 6(5)
x = 31 - 30
x = 1
z = -188 + 38(5)
z = -188 + 190
z = 2
Check your work:
6x - 2y + z = -2
6(1) - 2(5) + 2 = -2
6 - 10 + 2 = -2
-4 + 2 = -2
-2 = -2
Correct!
*Note there are several ways to solve for these types of problems. I used substitution*