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trapecia [35]
3 years ago
11

If PQ=8 and Q lies at -13 where could P be located?

Mathematics
1 answer:
choli [55]3 years ago
7 0

Answer:

In the given figure the point on segment PQ is twice as from P as from Q is. What is the point? Ans is (2,1).

Step-by-step explanation:

There is really no need to use any quadratics or roots.

( Consider the same problem on the plain number line first.  )

How do you find the number between 2 and 5 which is twice as far from 2 as from 5?

You take their difference, which is 3. Now splitting this distance by ratio 2:1 means the first distance is two thirds, the second is one third, so we get

4=2+23(5−2)

It works completely the same with geometric points (using vector operations), just linear interpolation: Call the result R, then

R=P+23(Q−P)

so in your case we get

R=(0,−1)+23(3,3)=(2,1)

Why does this work for 2D-distances as well, even if there seem to be roots involved? Because vector length behaves linearly after all! (meaning |t⋅a⃗ |=t|a⃗ | for any positive scalar t)

Edit: We'll try to divide a distance s into parts a and b such that a is twice as long as b. So it's a=2b and we get

s=a+b=2b+b=3b

⇔b=13s⇒a=23s

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Answer:

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Step-by-step explanation:

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The vertex of this function is

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The function is reflected in the x-axis.

The vertex is therefore the maximum point on the graph of the function.

The range is therefore;

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y=-\frac{1}{2} |0-1|-2

y=-\frac{1}{2} (1)-2

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The y-intercept is b=-2.5

The graph is shown in the attachment.

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