A square is inscribed in an isosceles right triangle with the length of the leg equal to 2 in. as shown on the figure. Find the
perimeter of the square.
pls help I need it asap!!!
pls help I need it asap!!!
1 answer:
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Answer:
Step-by-step explanation:
We want to find the locus of a point such that the sum of the distance from any point P on the locus to (0, 2) and (0, -2) is 6.
First, we will need the distance formula, given by:
Let the point on the locus be P(x, y).
So, the distance from P to (0, 2) will be:
And, the distance from P to (0, -2) will be:
So sum of the two distances must be 6. Therefore:
Now, by substitution:
Simplify. We can subtract the second term from the left:
Square both sides:
We can cancel the x² terms and continue squaring:
We can cancel the y² and 4 from both sides. We can also subtract 4y from both sides. This leaves us with:
We can divide both sides by -4:
Adding 9 to both sides yields:
And, we will square both sides one final time.
Distribute:
The 36y will cancel. So:
Subtracting 4y² and 36 from both sides yields:
And dividing both sides by 45 produces:
Therefore, the equation for the locus of a point such that the sum of its distance to (0, 2) and (0, -2) is 6 is given by a vertical ellipse with a major axis length of 3 and a minor axis length of √5, centered on the origin.
When we factor out something, we divide the factor by the equation. For example if we were to factor 3 out of 6, It will be 3(2). We got this answer by dividing 3 and 6. Using this concept:
![\frac{-1}{4}[( \frac{-1}{2} / \frac{-1}{4}) - ( \frac{5}{4} y / \frac{-1}{4})]](https://tex.z-dn.net/?f=%20%5Cfrac%7B-1%7D%7B4%7D%5B%28%20%5Cfrac%7B-1%7D%7B2%7D%20%2F%20%5Cfrac%7B-1%7D%7B4%7D%29%20-%20%28%20%5Cfrac%7B5%7D%7B4%7D%20y%20%2F%20%20%5Cfrac%7B-1%7D%7B4%7D%29%5D%20%20)
Simplifying:
Hope this helps!
Simplifying:
Hope this helps!