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Gemiola [76]
3 years ago
8

AA, BBB, and CCC are collinear, and BBB is between AAA and CCC. The ratio of ABABA, B to BCBCB, C is 1:21:21, colon, 2. If AAA i

s at (7,-1)(7,−1)left parenthesis, 7, comma, minus, 1, right parenthesis and BBB is at (2,1)(2,1)left parenthesis, 2, comma, 1, right parenthesis, what are the coordinates of point CCC?
Mathematics
2 answers:
AURORKA [14]3 years ago
7 0

Answer:

c=(9,4)

Step-by-step explanation:

Point AAA is at (\blue{-7}, \pink{-8})(−7,−8)left parenthesis, start color #6495ed, minus, 7, end color #6495ed, comma, start color #ff00af, minus, 8, end color #ff00af, right parenthesis and point BBB is at (\red{-3}, \gray{-5})(−3,−5)left parenthesis, start color #df0030, minus, 3, end color #df0030, comma, start color gray, minus, 5, end color gray, right parenthesis.

How far do we go in the xxx direction to get from AAA to BBB? We can subtract the xxx coordinate of AAA from the xxx coordinate of BBB to find out:

\red{-3} - (\blue{-7}) = 4−3−(−7)=4start color #df0030, minus, 3, end color #df0030, minus, left parenthesis, start color #6495ed, minus, 7, end color #6495ed, right parenthesis, equals, 4

Hint #22 / 6

We went 444 in the xxx direction to get from AAA to BBB. If 444 is \dfrac{1}{4}  

4

1

​  

start fraction, 1, divided by, 4, end fraction of the horizontal displacement (call it hhh), then we can find the whole displacement this way:

\begin{aligned}4 &= \frac{1}{4} \cdot h\\ \frac{4}{1} \cdot 4 &= h\\ \purple{16} &= \purple{h} \end{aligned}  

4

1

4

​  

⋅4

16

​  

 

=  

4

1

​  

⋅h

=h

=h

​  

 

Hint #33 / 6

How far do we go in the yyy direction to get from AAA to BBB? We can subtract the yyy coordinate of AAA from the yyy-coordinate of BBB to find out:

\gray{-5} - (\pink{-8}) = 3−5−(−8)=3start color gray, minus, 5, end color gray, minus, left parenthesis, start color #ff00af, minus, 8, end color #ff00af, right parenthesis, equals, 3

Hint #44 / 6

We went 333 in the yyy direction to get from AAA to BBB. If 333 is \dfrac{1}{4}  

4

1

​  

start fraction, 1, divided by, 4, end fraction of the vertical displacement (call it vvv), then we can find the whole displacement this way:

\begin{aligned}3&= \frac{1}{4} \cdot v\\ \frac{4}{1} \cdot 3 &= v\\ \green{12} &= \green v\\ \end{aligned}  

3

1

4

​  

⋅3

12

​  

 

=  

4

1

​  

⋅v

=v

=v

​  

 

Hint #55 / 6

To find the coordinates of CCC, we start at AAA (\blue{-7}, \pink{-8})(−7,−8)left parenthesis, start color #6495ed, minus, 7, end color #6495ed, comma, start color #ff00af, minus, 8, end color #ff00af, right parenthesis and add \purple{16}16start color #9d38bd, 16, end color #9d38bd to the xxx coordinate and \green{12}12start color #28ae7b, 12, end color #28ae7b to the yyy coordinate.

xxx coordinate: \blue{-7} + \purple{16}=9−7+16=9start color #6495ed, minus, 7, end color #6495ed, plus, start color #9d38bd, 16, end color #9d38bd, equals, 9

yyy coordinate: \pink{-8} + (\green{12})=4−8+(12)=4start color #ff00af, minus, 8, end color #ff00af, plus, left parenthesis, start color #28ae7b, 12, end color #28ae7b, right parenthesis, equals, 4

C = (9, 4)C=(9,4)C, equals, left parenthesis, 9, comma, 4, right parenthesis

Hint #66 / 6

denis-greek [22]3 years ago
5 0

Answer:

The coordinates of point C are (-8,5).

Step-by-step explanation:

It is given that A, B and C collinear and B is between A and C.

The ratio of AB to BC is 1:2. It means Point divided the line segments AC in 1:2.

Section formula:

(\dfrac{mx_2+nx_1}{m+n},\dfrac{my_2+ny_1}{m+n})

The given points are A(7,-1) and B(2,1).

Let the coordinates of C are (a,b).

Using section formula the coordinates of B are

B=(\dfrac{(1)(a)+(2)(7)}{1+2},\dfrac{(1)(b)+(2)(-1)}{1+2})

B=(\dfrac{a+14}{3},\dfrac{b-2}{3})

We know that point B(2,1).

(2,1)=(\dfrac{a+14}{3},\dfrac{b-2}{3})

On comparing both sides we get

2=\dfrac{a+14}{3}

6=a+14

6-14=a

-8=a

The value of a is -8.

1=\dfrac{b-2}{3}

3=b-2

3+2=b

5=b

The value of b is 5.

Therefore, the coordinates of point C are (-8,5).

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