please help find answer. very urgent.
triangle PQR is a 30-60-90 triangle with right angle Q and segment PQ as the longest.
Find the possible coordinates of R if (2,6) and Q (2,-6). please help
To find the possible coordinates of point R in a given 30-60-90 triangle with known coordinates of P and Q, we need to understand the ratios of the sides in such a triangle.
In a 30-60-90 triangle, the sides are related by the following ratios:
- The length of the side opposite the 30-degree angle is half the hypotenuse.
- The length of the side opposite the 60-degree angle is √3 times the length of the side opposite the 30-degree angle.
- The hypotenuse is twice the length of the side opposite the 30-degree angle.
In this case, we know that point Q has coordinates (2, -6). Let's assume that point P has coordinates (x, y).
Since Q is at (2, -6), the y-coordinate of point R should be the same as that of Q to maintain the right angle. Therefore, the y-coordinate of point R is also -6.
Now, let's calculate the x-coordinate of point R using the length ratio mentioned earlier.
The length of PQ (the hypotenuse) is given by the distance formula:
PQ = √[(x2 - x1)² + (y2 - y1)²]
= √[(2 - x)² + (-6 - y)²]
The length of QR (opposite the 30-degree angle) is half the length of PQ:
QR = 1/2 * PQ
= 1/2 * √[(2 - x)² + (-6 - y)²]
The length of PR (opposite the 60-degree angle) is √3 times the length of QR:
PR = √3 * QR
= √3 * (1/2 * √[(2 - x)² + (-6 - y)²])
= √[(3/4) * ((2 - x)² + (-6 - y)²)]
Since triangle PQR is a right triangle, the distance from P to Q is equal to the distance from Q to R. Therefore, we can equate the two lengths:
PQ = PR
√[(2 - x)² + (-6 - y)²] = √[(3/4) * ((2 - x)² + (-6 - y)²)]
Now, we can square both sides of the equation to eliminate the square root:
[(2 - x)² + (-6 - y)²] = (3/4) * [(2 - x)² + (-6 - y)²]
Expanding and simplifying the equation:
4[(2 - x)² + (-6 - y)²] = 3[(2 - x)² + (-6 - y)²]
4(4 - 4x + x² + 36 + 12y + y²) = 3(4 - 4x + x² + 36 + 12y + y²)
16 - 16x + 4x² + 144 + 48y + 4y² = 12 - 12x + 3x² + 108 + 36y + 3y²
48 + 32x + x² + 48y + y² = 0
Simplifying the equation further:
x² + y² + 32x + 48y + 48 = 0
This equation represents a circle on the coordinate plane. The possible coordinates of point R lie on the circumference of this circle.
To find the specific coordinates, the equation needs to be rearranged to the standard form of a circle equation, (x - h)² + (y - k)² = r², where (h, k) is the center of the circle and r is the radius.
By completing the square, the equation becomes:
(x² + 32x) + (y² + 48y) = -48
(x² + 32x + 16²) + (y² + 48y + 24²) = -48 + 16² + 24²
(x + 16)² + (y + 24)² = 16² + 24² - 48
The equation is now in the standard form, (x - h)² + (y - k)² = r², where the center of the circle is at (-16, -24) and the radius is √(16² + 24² - 48).
Therefore, the possible coordinates of point R lie on the circumference of the circle centered at (-16, -24) with a radius of √(16² + 24² - 48).