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3 years ago

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Gandalf the Grey started in the Forest of Mirkwood at a point with coordinates (2,1) and arrived in the Iron Hills at the point

with coordinates (3,5). If he began walking in the direction of the vector v=4i+2j and changes direction only once, when he turns at a right angle, what are the coordinates of the point where he makes the turn.

1 answer:

Pavlova-9 [17]3 years ago

4 0

Answer:

(22/5, 11/5)

Step-by-step explanation:

Gandalf starts walking in the vector v = 4i + 2j, the slope of it must be 2/4 = 1/2.

So, calling the turn point (x,y) it must be in that vector, and the start point (2,1) must also be in this line, so:

slope = Δy/Δx

1/2 = (y - 1)/(x - 2)

2y - 2 = x - 2

2y = x

y = x/2

The line that makes a right angle with another is called perpendicular, and it has its slopes equal to the opposite of the inverse of the slope of the other line, so:

slope = - 2

The point (x,y) also pertences to it, such as the point (3,5), so:

-2 = (y - 5)/(x - 3)

y - 5 = -2x +6

y = -2x + 11

So, the turn point is the point where these two lines meet:

x/2 = -2x + 11

x = -4x + 22

5x = 22

x = 22/5

y = (22/5)/2

y = 22/10 = 11/5

So, the point that he makes the turn is (22/5, 11/5).

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2 years ago

A circle has the equation 2x²+12x+2y²−16y−150=0.

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Step-by-step explanation:

The equation of a circle in the center-radius form is:

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Where $(h,k)$ are the coordinates of the center and $r$ is the radius.

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Rearranging the equation:

$x^{2}+6x+y^{2}-8y=75$ (4)

$(x^{2}+6x)+(y^{2}-8y)=75$ (5)

Now we have to complete the square in both parenthesis, in order to have a perfect square trinomial in the form of $(a\pm b)^{2}=a^{2}\pm+2ab+b^{2}$ :

<u>For the first parenthesis:</u>

$x^{2}+6x+b^{2}$

We can rewrite this as:

$x^{2}+2(3)x+b^{2}$

Hence in this case $b=3$ and $b^{2}=9$ :

$x^{2}+2(3)x+3^{2}=x^{2}+6x+9=(x+3)^{2}$

<u>For the second parenthesis:</u>

$y^{2}-8y+b^{2}$

We can rewrite this as:

$y^{2}-2(4)y+b^{2}$

Hence in this case $b=-3$ and $b^{2}=9$ :

$y^{2}-2(4)y+4^{2}=y^{2}-8y+16=(y-4)^{2}$

Then, equation (5) is rewritten as follows:

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<u>Note we are adding 9 and 16 in both sides of the equation in order to keep the equality.</u>

Rearranging:

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At this point we have the circle equation in the center radius form $(x-h)^{2} +(y-k)^{2}=r^{2}$

Hence:

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$k=4$

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