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Citrus2011 [14]
2 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]2 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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Let A(t) = amount of salt (in pounds) in the tank at time t (in minutes). Then A(0) = 11.

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and flows out at a rate

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which I'll solve with the integrating factor method.

\dfrac{dA}{dt} + \dfrac{13}{300-t} A = \dfrac95

-\dfrac1{(300-t)^{13}} \dfrac{dA}{dt} - \dfrac{13}{(300-t)^{14}} A = -\dfrac9{5(300-t)^{13}}

\dfrac d{dt} \left(-\dfrac1{(300-t)^{13}} A\right) = -\dfrac9{5(300-t)^{13}}

Integrate both sides. By the fundamental theorem of calculus,

\displaystyle -\dfrac1{(300-t)^{13}} A = -\dfrac1{(300-t)^{13}} A\bigg|_{t=0} - \frac95 \int_0^t \frac{du}{(300-u)^{13}}

\displaystyle -\dfrac1{(300-t)^{13}} A = -\dfrac{11}{300^{13}} - \frac95 \times \dfrac1{12} \left(\frac1{(300-t)^{12}} - \frac1{300^{12}}\right)

\displaystyle -\dfrac1{(300-t)^{13}} A = \dfrac{34}{300^{13}} - \frac3{20}\frac1{(300-t)^{12}}

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\displaystyle A = 45 \left(1 - \frac t{300}\right) - 34 \left(1 - \frac t{300}\right)^{13}

After 1 hour = 60 minutes, the tank will contain

A(60) = 45 \left(1 - \dfrac {60}{300}\right) - 34 \left(1 - \dfrac {60}{300}\right)^{13} = 45\left(\dfrac45\right) - 34 \left(\dfrac45\right)^{13} \approx 34.131

pounds of salt.

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