The midpoint of a line segment with end points as (-10, y1) and (-6, 7) is (-8, 6). What is the value of y1?
-15
2
5
-1
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OpenStudy (anonymous):
Can you tell me what is the sum of y coordinates there??
OpenStudy (anonymous):
uhmm idk
OpenStudy (anonymous):
Sum of y coordinates of end points I am asking..
OpenStudy (anonymous):
how do you find the endpoints?
OpenStudy (anonymous):
See, the sum is :
\[y_1 + 7\]
Divide it by 2, then it will become equal to 6 (y coordinate of midpoint)
\[\frac{y_1 + 7}{2} = 6\]
Multiply 2 both the sides and can you find \(y_1\) from here??
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OpenStudy (anonymous):
ok hold on
OpenStudy (anonymous):
Take your time..
OpenStudy (anonymous):
\[y ^{1}+7\div 2\]
OpenStudy (anonymous):
hold on this is kinda hard
OpenStudy (anonymous):
Hard???
multiply by 2 both the sides and show me what you get??
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OpenStudy (anonymous):
\[2 \times \frac{(y_1 + 7)}{2} = 2 \times 6\]
Can you solve this??
OpenStudy (anonymous):
2y^1 plus 14 over 2 = 12
OpenStudy (anonymous):
Why are you not cancelling 2 with 2???
OpenStudy (anonymous):
ok i will cancel it right now
OpenStudy (anonymous):
2y^1 plus 14 = 12
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OpenStudy (anonymous):
2y^1=-2
OpenStudy (anonymous):
How???
OpenStudy (anonymous):
divide 2 to both sides and i think the answer is y=1
OpenStudy (anonymous):
What are you doing??
OpenStudy (anonymous):
i'm not sure this is confusing
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OpenStudy (anonymous):
Solve for LHS first..
Tell me what do you get:
\[2 \times \frac{(y_1+7)}{2}\]
What you get??
OpenStudy (anonymous):
You have to simply cancel 2 with 2 and answer is the remaining part..
Solve this and tell me what do you get..
OpenStudy (anonymous):
y^1 plus 7
OpenStudy (anonymous):
Yes now you are right..
\[y_1 + 7 = 6 \times 2\]
Solve the RHS now:
What is 6*2 = ???
OpenStudy (anonymous):
12
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OpenStudy (anonymous):
So write the entire now what it becomes after solving.
OpenStudy (anonymous):
y^1 plus 7=12
OpenStudy (anonymous):
Now subtract both the sides by 7
and you will have your answer...
OpenStudy (anonymous):
y^1=5
OpenStudy (anonymous):
Yes this is your answer..
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