Geometry Homework Practice Workbook 1st Edition Chapter 1 Points Lines and Planes
Page 5 Problem 1 Answer
Given
![Geometry, Homework Practice Workbook 1st Edition Chapter 1 Points, Lines and Planes 1](https://answerkeyformath.com/wp-content/uploads/2023/03/Geometry-Homework-Practice-Workbook-1st-Edition-Chapter-1-Points-Lines-and-Planes-1.webp)
To find the length of LN.
The absolute value difference between the coordinates of the endpoints will give the length.
Let the coordinate of the endpointL=(l) and the coordinate of N=(n).
The length ofLN=∣l−n∣
=∣3−9∣
=∣−6∣
=6 units.
Hence we can conclude that the length of LN is 6 units.
Page 5 Problem 2 Answer
Here given a number line
![Geometry, Homework Practice Workbook 1st Edition Chapter 1 Points, Lines and Planes 2](https://answerkeyformath.com/wp-content/uploads/2023/03/Geometry-Homework-Practice-Workbook-1st-Edition-Chapter-1-Points-Lines-and-Planes-2-2.png)
By using this number line we have to measure KN.
Now we will consider K as point x1 and N as point x2.
Then we will find the distance between K
And N by using the distance formula that is ∣x2−x1∣.
Given that a number line
![Geometry, Homework Practice Workbook 1st Edition Chapter 1 Points, Lines and Planes 2 1](https://answerkeyformath.com/wp-content/uploads/2023/03/Geometry-Homework-Practice-Workbook-1st-Edition-Chapter-1-Points-Lines-and-Planes-2-1-1.png)
Now we will consider K as point x1 and N as point x2 and to measure the distance we will use the formula of distance that is ∣x2−x1∣.
If we give a closer look at this number line we can say x1 is on −2 and x2 is on 9.
So, KN= ∣x2−x1∣
=∣9−(−2)∣Unit
=∣11∣Unit
=11Unit.
Hence, the length of KN on the number line is 11 units.
Page 5 Problem 3 Answer
Here given a number line
![Geometry, Homework Practice Workbook 1st Edition Chapter 1 Points, Lines and Planes 3](https://answerkeyformath.com/wp-content/uploads/2023/03/Geometry-Homework-Practice-Workbook-1st-Edition-Chapter-1-Points-Lines-and-Planes-3-3.png)
By using this number line we have to measure MN.
Now we will consider M as point x1 and N as point x2.
Then we will find the distance between M and N by using the distance formula that is∣x2−x1.
Given that a number line
![Geometry, Homework Practice Workbook 1st Edition Chapter 1 Points, Lines and Planes 3 1](https://answerkeyformath.com/wp-content/uploads/2023/03/Geometry-Homework-Practice-Workbook-1st-Edition-Chapter-1-Points-Lines-and-Planes-3-1-2.png)
Now we will consider M as point x1 and N as point x2 and to measure the distance we will use the formula of distance that is ∣x2−x1∣.
If we take a closer look at this number line we can say x1 is on 6 x2 is on 9.
So, MN = ∣x2−x1∣
= ∣6−9∣Unit
= ∣−3∣Unit
= 3Unit.
Hence, the length of MN on the number line is 3 units.
Page 5 Problem 4 Answer
Given two points K(2,3) and F(4,4).
Let’s assume (x1,y1) and(x2,y2) be the coordinates of those points respectively.
Finally, we are going to use the formula∣√(x2−x1)2+(y2−y1)2∣to measure the distance between those points.
Given that two points are K(2,3) and F(4,4).
Now to calculate the distance between them from the formula that is∣√(x2−x1)2+(y2−y1)2∣
we will replace the value of x1,y1,x2,y2.
So KF=∣√(4−2)2+(4−3)2Unit
=∣√22+1/2∣Unit
=∣√5∣Unit
=∣2.23∣Unit
=2.23Unit.
Hence, the distance between the two points is 2.23Unit.
Page 5 Problem 5 Answer
Given two points C(−3,−1) and Q(−2,3).
Let’s assume (x1,y1) and (x2,y2) be the coordinates of those points respectively.
Finally, we are going to use the formula∣√(x2−x1)2+(y2−y1)2∣to measure the distance between those points.
Given that two points are C(−3,−1) and Q(−2,3).
Now to calculate the distance between them from the formula that is ∣√(x2−x1)2+(y2−y1)2∣
we will replace the value of x1,y1,x2,y2.
So, CQ= ∣√{(−2)−(−3)2+{3−(−1)}2∣Unit
= ∣√12+4/2∣Unit
= ∣√17∣Unit
= ∣4.12∣Unit
= 4.12Unit.
Hence, the distance between the two points is 4.12Unit.
Page 5 Problem 6 Answer
Given two pointsY(2,0) and P(2,6).
Let assume (x1,y1) and (x2,y2) be the coordinates of those points respectively.
Finally, we are going to use the formula∣√(x2−x1)2+(y2−y1)2∣to measure the distance between those points.
Given that two points are Y(2,0) and P(2,6).
Now to calculate the distance between them from the formula that is∣√(x2−x1)2+(y2−y1)2∣
we will replace the value of x1,y1,x2,y2.
So, YP=∣√(2−2)2+(6−0)2∣Unit
=∣√6/2∣Unit
=∣6∣Unit
=6Unit.
Hence, the distance between the two points is 6 units.
Page 5 Problem 7 Answer
Given two points W(−2,2) and R(5,2).
Let’s assume (x1,y1) and (x2,y2) be the coordinates of those points respectively.
Finally, we are going to use the formula∣√(x2−x1)2+(y2−y1)2| to measure the distance between those points.
Given that two points are W(−2,2) and R(5,2).
Now to calculate the distance between them from the formula that is∣√(x2−x1)2+(y2−y1)2∣
we will replace the value of x1,y1,x2,y2.
So,WR=∣√{5−(−2)}2+(2−2)2∣Unit
= ∣√72+0/2∣Unit
= ∣√49∣Unit
= ∣7∣Unit
= 7Unit
Hence, the distance between the two points is 7 units.
Page 5 Problem 8 Answer
Given two points A(−7,−3) and B(5,2). Let’s assume (x1,y1) and(x2,y2) be the coordinates of those points respectively.
Finally, we are going to use the formula∣(x2−x1)2+(y2−y1)2∣to measure the distance between those points.
Given that two points are A(−7,−3) and B(5,2).
Now to calculate the distance between them from the formula that is∣√(x2−x1)2+(y2−y1)2∣
we will replace the value of x1,y1,x2,y2.
So, AB=∣√{5−(−7)}2+{2−(−3)}2∣Unit
= ∣√122+5/2∣Unit
= ∣√144+25∣Unit
= ∣√169∣Unit
= ∣13∣Unit
= 13 Unit.
Hence, the distance between the two points is 13 Units.
Page 5 Problem 9 Answer
Given two points C(−3,1) and Q(2,6).
Let’s assume(x1,y1) and (x2,y2) be the coordinates of those points respectively.
Finally, we are going to use the formula∣√(x2−x1)2+(y2−y1)2∣to measure the distance between those points.
Given that two points are C(−3,1) and Q(2,6).
Now to calculate the distance between them from the formula that is∣√(x2−x1)2+(y2−y1)2∣
we will replace the value of x1,y1,x2,y2.
So CQ=∣√{2−(−3)}2+(6−1)2∣Unit
= ∣√52+5/2∣Unit
= ∣√25+25∣Unit
= ∣√50∣Unit
= ∣7.07∣Unit
= 7.07Unit.
Hence, the distance between the two points is 7.07 Units.
Page 5 Problem 10 Answer
Given
![Geometry, Homework Practice Workbook 1st Edition Chapter 1 Points, Lines and Planes 10](https://answerkeyformath.com/wp-content/uploads/2023/03/Geometry-Homework-Practice-Workbook-1st-Edition-Chapter-1-Points-Lines-and-Planes-10.webp)
To find the coordinate of the midpoint of DE.
The midpoint of a line with end coordinates x1 and x2 is given by the formula \(\frac{x_1+y_1}{2}\)
The coordinate of the endpoint D is x1=7.
The coordinate of the endpoint E is y1=11.
Hence the coordinate of the midpoint of the line DE is \(\frac{x_1+y_1}{2}\)=\(\frac{7+11}{2}\)
=\(\frac{8}{2}\)
=9.
Hence we can conclude that the coordinate of the midpoint of DE is 9.
Page 5 Problem 11 Answer
Given
![Geometry, Homework Practice Workbook 1st Edition Chapter 1 Points, Lines and Planes 11](https://answerkeyformath.com/wp-content/uploads/2023/03/Geometry-Homework-Practice-Workbook-1st-Edition-Chapter-1-Points-Lines-and-Planes-11-1.webp)
To find the coordinate of the midpoint of BC.
The midpoint of a line with end coordinates x1 and y1 is given by the formula \(\frac{x_1+y_1}{2}\)
The coordinate of the endpoint B is −1.
The coordinate of the endpoint C is 3.
Hence the coordinate of the midpoint of line BC is
⇒ \(\frac{x_1+y_1}{2}\)
= \(\frac{-1+3}{2}\)
= \(\frac{2}{2}\)
=1.
Hence we can conclude that the coordinate of the midpoint of BC is 1.
Page 5 Problem 12 Answer
Given
![Geometry, Homework Practice Workbook 1st Edition Chapter 1 Points, Lines and Planes 12](https://answerkeyformath.com/wp-content/uploads/2023/03/Geometry-Homework-Practice-Workbook-1st-Edition-Chapter-1-Points-Lines-and-Planes-12-2.webp)
To find the coordinate of the midpoint of BD.
The midpoint of a line with end coordinates x1 and y1 is given by the formula \(\frac{x_1+y_1}{2}\)
The coordinate of the endpoint B is −1.
The coordinate of the endpoint D is 7.
Hence the coordinate of the midpoint of the line BD is \(\frac{x_1+y_1}{2}\)
= \(\frac{-1+7}{2}\)
= \(\frac{6}{2}\)
= 3.
Hence we can conclude that the coordinate of the midpoint of BD is 3.
Page 5 Problem 13 Answer
Given
![Geometry, Homework Practice Workbook 1st Edition Chapter 1 Points, Lines and Planes 13](https://answerkeyformath.com/wp-content/uploads/2023/03/Geometry-Homework-Practice-Workbook-1st-Edition-Chapter-1-Points-Lines-and-Planes-13-2.webp)
To find the coordinate of the midpoint of AD.
The midpoint of a line with end coordinates x1 and y1 is given by the formula \(\frac{x_1+y_1}{2}\)
The coordinate of the endpoint A is −4.
The coordinate of the endpoint D is 7.
Hence the coordinate of the midpoint of the line AD is \(\frac{x_1+y_1}{2}\)
= \(\frac{-4+7}{2}\)
= \(\frac{3}{2}\)
=1.5
Hence we can conclude that the coordinate of the midpoint of AD is 1.5.
Page 5 Problem 14 Answer
Given endpoints T(3,1) and U(5,3).
To find the coordinates of the midpoint of a segment with the given endpoints.
The midpoint of a line with end coordinates(x1,y1)and(x2,y2) is given by the formula \(\frac{x_1+x_2}{2}\), \(\frac{y_1+y_2}{2}\)
The coordinates of the endpoint T is (3,1).
The coordinates of the endpoint U is (5,3).
Hence the coordinates of the midpoint of the line TU are \(\frac{x_1+x_2}{2}\), \(\frac{y_1+y_2}{2}\)
= \(\frac{3+5}{2}\), \(\frac{1+3}{2}\)
= \(\frac{8}{2}\), \(\frac{4}{2}\)
=(4,2).
Hence we can conclude that the coordinates of the midpoint of TU is (4,2).
Page 5 Problem 15 Answer
Given endpoints J(−4,2) and F(5,−2).
To find the coordinates of the midpoint of a segment with the given endpoints.
The midpoint of a line with end coordinates(x1,y1) and (x2,y2) is given by the formula \(\frac{x_1+x_2}{2}\), \(\frac{y_1+y_2}{2}\)
The coordinates of the endpoint J are (−4,2).
The coordinates of the endpoint F is (5,−2).
Hence the coordinates of the midpoint of the line JF are \(\frac{x_1+x_2}{2}\), \(\frac{y_1+y_2}{2}\)
= \(\frac{-4+5}{2}\), \(\frac{2-2}{2}\)
= \(\frac{1}{2},0\)
Hence we can conclude that the coordinates of the midpoint of JF are \(\frac{1}{2},0\)
Page 5 Problem 16 Answer
Given endpoint N (2,0) and the midpoint of the line NQ is P(5,2).
To find the coordinates of the missing endpoint.
The midpoint of a line with end coordinates(x1,y1) and (x2,y2) is given by the formula \(\frac{x_1+x_2}{2}\), \(\frac{y_1+y_2}{2}\)
Let the coordinates of Q be (x,y).
The coordinates of the endpoint N are (2,0).
The coordinates of the midpoint P is (5,2).
Hence the coordinates of the midpoint of the line NQ
are \(\frac{x_1+x_2}{2}\), \(\frac{y_1+y_2}{2}\) =(5,2)
⇒ \(\frac{2+x}{2}\), \(\frac{0+y}{2}\) =(5,2)
⇒ 2+x=10, y=4
⇒ x=8, y=4.
Thus, the coordinates of Q are (8,4).
Hence we can conclude that the coordinates of the missing endpoint are(8,4).
Page 5 Problem 17 Answer
Given endpoint N(5,4) and the midpoint of the line NQ is P(6,3).
To find the coordinates of the missing endpoint.The midpoint of a line with end coordinates(x1,y1)
And (x2,y2) is given by the formula \(\frac{x_1+x_2}{2}\), \(\frac{y_1+y_2}{2}\)
Let the coordinates of Q be (x,y).
The coordinates of the endpoint N are (5,4).
The coordinates of the midpoint P are (6,3).
Hence the coordinates of the midpoint of the line NQ
are \(\frac{x_1+x_2}{2}\), \(\frac{y_1+y_2}{2}\) =(6,3)
⇒ \(\frac{5+x}{2}\), \(\frac{4+y}{2}\) =(6,3)
⇒ 5+x=12,4+y=6
⇒ x=7,y=2.
Thus, the coordinates of Q are (7,2).
Hence we can conclude that the coordinates of the missing endpoint are(7,2).
Page 5 Problem 18 Answer
Given endpoint Q(3,9) and the midpoint of the line NQ is P(−1,5).
To find the coordinates of the missing endpoint.
The midpoint of a line with end coordinates (x1,y1) and (x2,y2) is given by the formula \(\frac{x_1+x_2}{2}\), \(\frac{y_1+y_2}{2}\)
Let the coordinates of N be(x,y).
The coordinates of the endpoint Q are (3,9).
The coordinates of the midpoint P are (−1,5).
Hence the coordinates of the midpoint of the line NQ are \(\frac{x_1+x_2}{2}\), \(\frac{y_1+y_2}{2}\)=(−1,5)
⇒ \(\frac{x+3}{2}\), \(\frac{y+9}{2}\) =(−1,5)
⇒ x+3=−2,y+9=10
⇒ x=−5,y=1.
Thus, the coordinates of N are (−5,1).
Hence we can conclude that the coordinates of the missing endpoint are (−5,1).
Page 6 Problem 19 Answer
Given
![Geometry, Homework Practice Workbook 1st Edition Chapter 1 Points, Lines and Planes 19](https://answerkeyformath.com/wp-content/uploads/2023/03/Geometry-Homework-Practice-Workbook-1st-Edition-Chapter-1-Points-Lines-and-Planes-19-2.webp)
To find the length of VW.
The absolute value difference between the coordinates of the endpoints will give the length.
Let the coordinate of the endpoint V=(v) and the coordinate of W=(w).
The length of VW=∣v−w∣
= ∣1−5∣
= ∣−4∣
= 4 units.
Hence we can conclude that the length of VW is 4 units.
Page 6 Problem 20 Answer
Here given a number line By using this number line we have to measure TV.
Now we will consider T as point x1 and V as point x2.
Then we will find the distance between T and V by using the distance formula that is ∣x2−x1∣.
Given that a number line
![Geometry, Homework Practice Workbook 1st Edition Chapter 1 Points, Lines and Planes 20](https://answerkeyformath.com/wp-content/uploads/2023/03/Geometry-Homework-Practice-Workbook-1st-Edition-Chapter-1-Points-Lines-and-Planes-20-2.png)
Now we will consider T as point x1 and V as point x2 and to measure the distance we will use the formula of distance that is ∣x2−x1∣.
If we give a closer look at this number line we can say T is on −4 and V is on 1.
So, TV= ∣x2−x1∣
= ∣−4−1∣Unit
= ∣−5∣Unit
= 5Unit.
Hence the length of TV on the number line is 5 units.
![Geometry, Homework Practice Workbook 1st Edition Chapter 1 Points, Lines and Planes 20 1](https://answerkeyformath.com/wp-content/uploads/2023/03/Geometry-Homework-Practice-Workbook-1st-Edition-Chapter-1-Points-Lines-and-Planes-20-1.png)
Page 6 Problem 21 Answer
Here given a number line By using this number line we have to measure ST
Now we will consider S as point x1 and T as point x2.
Then we will find the distance between S And T by using the distance formula that is ∣x2−x1∣.
Given that a number line
![Geometry, Homework Practice Workbook 1st Edition Chapter 1 Points, Lines and Planes 21](https://answerkeyformath.com/wp-content/uploads/2023/03/Geometry-Homework-Practice-Workbook-1st-Edition-Chapter-1-Points-Lines-and-Planes-21.png)
Now we will consider S as point x1 and T as point x2 and to measure the distance we will use the formula of distance that is ∣x2−x1∣.
If we give a closer look at this number line we can say x1 is on−7 and x2 is on−4.
So, ST= ∣x2−x1∣
= ∣−7−(−4)∣Unit
= ∣−3∣Unit
= 3Unit.
Hence the length of ST on the number line is 3 Units.
![Geometry, Homework Practice Workbook 1st Edition Chapter 1 Points, Lines and Planes 21 1](https://answerkeyformath.com/wp-content/uploads/2023/03/Geometry-Homework-Practice-Workbook-1st-Edition-Chapter-1-Points-Lines-and-Planes-21-1.png)
Page 6 Problem 22 Answer
Here given a number line By using this number line we have to measure SV.
Now we will consider S as point x1 and V as point x2.
Then we will find the distance between S and V
by using the distance formula that is ∣x2−x1∣.
Given that a number line
![Geometry, Homework Practice Workbook 1st Edition Chapter 1 Points, Lines and Planes 22](https://answerkeyformath.com/wp-content/uploads/2023/03/Geometry-Homework-Practice-Workbook-1st-Edition-Chapter-1-Points-Lines-and-Planes-22-.png)
Now we will consider S as point x1 and V as point x2 and to measure the distance we will use the formula of distance that is ∣x2−x1∣.
If we give a closer look at this number line we can say S is on −7 and V is on 1.
![Geometry, Homework Practice Workbook 1st Edition Chapter 1 Points, Lines and Planes 22 1](https://answerkeyformath.com/wp-content/uploads/2023/03/Geometry-Homework-Practice-Workbook-1st-Edition-Chapter-1-Points-Lines-and-Planes-22-1.png)
So SV=∣x2−x1∣
= ∣−7−1∣Unit
= ∣−8∣Unit
= 8 Unit.
Hence the length of SV on the number line is 8Unit.
Page 6 Problem 23 Answer
Given that
![Geometry, Homework Practice Workbook 1st Edition Chapter 1 Points, Lines and Planes 23](https://answerkeyformath.com/wp-content/uploads/2023/03/Geometry-Homework-Practice-Workbook-1st-Edition-Chapter-1-Points-Lines-and-Planes-23.png)
By using this graph we have to measure the distance between two given points.
Now find the coordinates of those points and find out the distance between them using∣√(x2−x1)2+(y2−y1)2∣
where(x1,y1) and (x2,y2) are the coordinates of S and E respectively.
Given that
![Geometry, Homework Practice Workbook 1st Edition Chapter 1 Points, Lines and Planes 23 1](https://answerkeyformath.com/wp-content/uploads/2023/03/Geometry-Homework-Practice-Workbook-1st-Edition-Chapter-1-Points-Lines-and-Planes-23-1.png)
By using this graph we have to measure the distance between two given points.
From the graph the coordinates of S and E respectively (4,4) and (−3,−4).
Now we are going to use the formula ∣√(x2−x1)2+(y2−y1)2∣ to calculate the distance between each pair of points.
Putting the value of x1,x2,y1,y2
The distance of SE
GF= ∣√(−3−4)2+(−4−4)2∣Unit
= ∣√(−7)2+(−8)2∣Unit
= ∣√49+64∣Unit
= ∣√113∣Unit
= 10.630Unit
≈10.6Unit.
Hence, the distance between each pair of points is 10.6 Units.
Page 6 Problem 24 Answer
The given points are.
We need to find the distance between the two points.
The distance can be calculated as d=2√(x2−x1)2+(y2−y1)2.
Using the formula d=2√(x2−x1)2+(y2−y1)2
we can say that,
d= 2 √(5+7)2+(9−0)2
= 2√(12)2+(9)2
= √144+81
= √225
= 15
The total distance is found to be 15 units.
Page 6 Problem 25 Answer
The given points are U(1,3),B(4,6). We need to find the distance between the two points.
The distance can be calculated as d = 2/√(x2−x1)2+(y2−y1)2.
Using the formula d = 2/√(x2−x1)2+(y2−y1)2
we can say that,
d=2/√(4−1)2+(6−3)2
=2√32+32
=√9+9
=√18
=3/√2
The total distance is found to be 3/√2 units.
Page 6 Problem 26 Answer
The given points are V(−2,5),M(0,−4). We need to find the distance between the two points.
The distance can be calculated as d = 2/√(x2−x1)2+(y2−y1)2.
Using the formula we can say that,
d = 2/√(x2−x1)2+(y2−y1)2
=2/√(0+2)2+(−4−5)2
=2/√(2)2+(−9)2
=2/√4+81
=√85
The total distance is found to be √85 units.
Page 6 Problem 27 Answer
The given points are C(−2,−1),K(8,3). We need to find the distance between the two points.
The distance can be calculated as d=2√(x2−x1)2+(y2−y1)2.
Using the formula we can say that,
d=2√(x2−x1)2+(y2−y1)2 or,d=2√(8+2)2+(3+1)2=2
√(10)2+(4)2=√100+16
=√116
=2/√29
The total distance is found to be 2/√29 units.
Page 6 Problem 28 Answer
The given line is RT.
We must find the endpoints of the line.
The midpoint can be found using the formula m= \(\frac{x_1+x_2}{2}\)
We can see that the line has endpoints as−2 and 4.
Thereby using the formula m=\(\frac{x_1+x_2}{2}\),
m= \(\frac{-2+4}{2}\)
m=\(\frac{2}{2}\)
m=1.
The midpoint is found to be (1,0).
Page 6 Problem 29 Answer
The given line is QR.
We must find the endpoints of the line.
The midpoint can be found using the formula m= \(\frac{x_1+x_2}{2}\)
We can see that the line has endpoints as−6and−2,
Thereby using the formula m=\(\frac{x_1+x_2}{2}\)
m= \(\frac{-6-2}{2}\)
= \(\frac{-8}{2}\)
=−4
The midpoint is found to be (−4,0).
Page 6 Problem 30 Answer
The given line is ST.
We must find the endpoints of the line.
The midpoint can be found using the formula m= x1+x2/2.
We can see that the line has endpoints 1 and 4.
Thereby using the formula m= \(\frac{x_1+x_2}{2}\)
m= \(\frac{1+4}{2}\)
= \(\frac{5}{2}\)
The midpoint is found to be \(\frac{5}{2},0\)
Page 6 Problem 31 Answer
The given line is PR.
We must find the endpoints of the line.
The midpoint can be found using the formula m = \(\frac{x_1+x_2}{2}\)
We can see that the line has endpoints as −9 and −2.
Thereby using the formula m = \(\frac{x_1+x_2}{2}\),m = \(\frac{-9-2}{2}\)
= \(\frac{-11}{2}\)
The midpoint is found to be \(\frac{-11}{2},0\)
Page 6 Problem 32 Answer
The given points are K(−9,3) and H(5,7).
We must find the endpoints of the line.
The midpoint can be found using the formula m= \(\frac{x_1+x_2}{2}\), \(\frac{y_1+y_2}{2}\)
Using the formula,
m = \(\frac{x_1+x_2}{2}\), \(\frac{y_1+y_2}{2}\) or,m=(−9+5)
(2,3+\(\frac{7}{2}\)) or,m= \(\frac{-4}{2}\), \(\frac{10}{2}\)or,m=(−2,5)
The midpoint is found to be (−2,5).
Page 6 Problem 33 Answer
The given points are W(−12,−7);T(−8,−4).We must find the endpoints of the line.
The midpoint can be found using the formula m=\(\frac{x_1+x_2}{2}\), \(\frac{y_1+y_2}{2}\)
Using the formula,
m= \(\frac{x_1+x_2}{2}\), \(\frac{y_1+y_2}{2}\) or,m=(−12−8/2,−7−4/2)
or,m= \(\frac{-20}{2}\)\(\frac{-11}{2}\)or,m=(−10,\(\frac{-11}{2}\))
The midpoint is found to be (−10,−11/2).
Page 6 Problem 34 Answer
The given points are R(−1,3),S(3,3),T(5,−1),U(−2,−1) and.
We must find the length of each side.
The perimeter can be found by summing up all the lengths of the sides.
The perimeter The length found is given below,
RS= √(3+1)2+(3−3)2
=√42
=4
ST=√(5−3)2+(−1−3)2
=√22+(−4)2
=√20
TU=√(5+2)2+(1−1)2
=√72
=7
RU=√(−2+1)2+(−1−3)2
=√1+42
=√17
Perimeter of quadrilateral RSTU=4+7+√20+√17
=19.59
The perimeter is found to be 19.59 sq. units.