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Dominik [7]
3 years ago
6

Find the equation of the tangent to the circle x2 + y2 = 109 at point (-10,3)

Mathematics
1 answer:
konstantin123 [22]3 years ago
4 0

Answer:

10x - 3y = -109 .

Step-by-step explanation:

x^2 + y^2 = 109

Implicit Differentiation:

2x + 2y.y' = 0

y' = -2x/2y = -x/y

So at the point (-10,3) the gradient of the tangent is -(-10)/3 = 10/3.

Equation of the tangent:

y - y1 = m(x - x1)

y - 3 = 10/3(x + 10)

y - 3 = 10/3 x + 100/3

3y - 9 = 10x + 100

-109 + 3y = 10x

10x - 3y = -109  

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-2/3 - 5/6 

We need common denominators so, since 3 can go into 6, we only need to multiply the first fraction by 2.

= -2 x 2 / 3x2  -  5/6 
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We only subtract the numbers that are in the numerators, 

= -4-5 / 6 
= -9/6 

Both nine and six are divisible by 3, so to put into lowest terms... 

=-9÷3 / 6÷3
= -3/2  <--- Final Answer 

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Find the midpoint of the segment with the given endpoints. (9.-9) and (2.-10)​
kondor19780726 [428]

Answer:

\boxed{ \bold{ \huge{ \boxed{ \sf{(5.5 \: , \: -  9.5)}}}}}

Step-by-step explanation:

Let the points be A and B

Let A ( 9 , -9 ) be ( x₁ , y₁ ) and B ( 2 , - 10 ) be ( x₂ , y₂ )

<u>Finding</u><u> </u><u>the</u><u> </u><u>midpoi</u><u>nt</u>

\boxed{ \sf{midpoint = ( \frac{x1 + x2}{2} , \frac{y1 + y2}{2} }}

\longrightarrow{ \sf{midpoint = ( \frac{9 + 2}{2} , \:  \frac{ - 9  + ( - 10)}{2}}} )

\longrightarrow{ \sf {midpoint = ( \frac{11}{2} \: , \:  \frac{ - 9 - 10}{2}})}

\longrightarrow{  \sf{midpoint = (5.5 \: , \:  \frac{ - 19}{2}}} )

\longrightarrow{ \sf{midpoint = (5.5 \: , - 9.5}})

Hope I helped!

Best regards! :D

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