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Citrus2011 [14]
3 years ago
10

What are the coordinates of the point on the directedline segment from K(-5,-4) to L(5,1) that partitionsthe segment into a rati

o of 3 to 2?

Mathematics
1 answer:
Diano4ka-milaya [45]3 years ago
3 0
You can solve this either just plain algebra or with the use of trigonometry.
In this case, we'll just use algebra.
 
So, if we let M be the the point that partitions the segment into a ratio of 3:2, we have this relation:
KM/ML = 3/2
KM = 1.5 ML

We also have this:
KL = KM + ML
Substituting KM,
KL = (3/2) ML + ML
KL = 2.5 ML

Using the distance formula and the given coordinates of the K and L, we get the length of KL
KL = sqrt ( (5-(-5)^2 + (1-(-4))^2 ) = 5 sqrt(5)

Since,
KL = 2.5 ML

Substituting KL,
ML = (1/2.5) KL = (1/2.5) 5 sqrt(5) = 2 sqrt(5)

Using again the distance formula from M to L and letting (x,y) as the coordinates of the point M
ML = 2 sqrt(5) = sqrt ( (5-x)^2 + (1-y)^2 )   [let this be equation 1]

In order to solve this, we need to find an expression of y in terms of x. We can use the equation of the line KL.
The slope m is:
m = (1-(-4))/(5-(-5) = 0.5

Using the general form of the linear equation:
y = mx +b
We substitue m and the coordinate of K or L. We'll just use K.
-5 = (0.5)(-4) + b
b = -1.5

So equation of the line is
y = 0.5x - 1.5      [let this be equation 2]

Substitute equation 2 to equation 1 and solving for x, we get 2 values of x,
x=1, x=9

Since 9 does not make sense (it does not lie on the line), we choose x=1.
Using the equation of the line, we get y which is -1.

So, we get the coordinates of point M which is (1,-1)
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The maximum value of the function: f(x)= -5 x ^2 +30x-200 is?
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Answer:

\displaystyle    - 155

Step-by-step explanation:

we are given a quadratic function

\displaystyle f(x) =  - 5 {x}^{2}  + 30x - 200

we want to figure out the minimum value of the function

to do so we need to figure out the minimum value of x in the case we can consider the following formula:

\displaystyle x _{ \rm  min} =  \frac{ - b}{2a}

the given function is in the standard form i.e

\displaystyle f(x) = a {x}^{2}  + bx + c

so we acquire:

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thus substitute:

\displaystyle x _{ \rm  min} =  \frac{ - 30}{2. - 5}

simplify multiplication:

\displaystyle x _{ \rm  min} =  \frac{ - 30}{ - 10}

simply division:

\displaystyle x _{ \rm  min} =  3

plug in the value of minimum x to the given function:

\displaystyle f (3)=  - 5 {(3)}^{2}  + 30.3 - 200

simplify square:

\displaystyle f (3)=  - 5 {(9)}^{}  + 30.3 - 200

simplify multiplication:

\displaystyle f (3)=  - 45  + 90- 200

simplify:

\displaystyle f (3)=   - 155

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the minimum value of the function is -155

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