How does the area of triangle RST compare to the area of triangle LMN? The area of △RST is 2 square units less than the area of
△LMN. The area of △RST is equal to the area of △LMN. The area of △RST is 2 square units greater than the area of △LMN. The area of △RST is 4 square units greater than the area of △LMN.
2 answers:
Inscribe triangle RST in the square with dimensions 4×4, as shown in the figure.
from the area of this square, 4*4=16, we remove the triangles with dimensions
3×4, 2×1 and 2×4, whose side lengths are shown in the figure, and we are left with the area of triangle RST.
so
units squared
similarly,
units squared
Thus, the areas are equal.
from the area of this square, 4*4=16, we remove the triangles with dimensions
3×4, 2×1 and 2×4, whose side lengths are shown in the figure, and we are left with the area of triangle RST.
so
similarly,
Thus, the areas are equal.
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So, notice, the focus point is at -7, 5, and the directrix is at y = -11.
keep in mind that the vertex is half-way between those two fellows, and the distance from the vertex to either one of them is "p" units, check the picture below.
with that focus point and that directrix, the half-way over the axis of symmetry will be -7, -3, that's where the vertex is at, and notice the distance "p", is 8 units.
since the parabola is opening upwards, "p" is positive 8.
![\bf \textit{parabola vertex form with focus point distance}\\\\ \begin{array}{llll} 4p(x- h)=(y- k)^2 \\\\ \boxed{4p(y- k)=(x- h)^2} \end{array} \qquad \begin{array}{llll} vertex\ ( h, k)\\\\ p=\textit{distance from vertex to }\\ \qquad \textit{ focus or directrix} \end{array}\\\\ -------------------------------\\\\ \begin{cases} h=-7\\ k=-3\\ p=8 \end{cases}\implies 4(8)[y-(-3)]=[x-(-7)]^2 \\\\\\ 32(y+3)=(x+7)^2\implies y+3=\cfrac{1}{32}(x+7)^2 \\\\\\ y=\cfrac{1}{32}(x+7)^2-3](https://tex.z-dn.net/?f=%5Cbf%20%5Ctextit%7Bparabola%20vertex%20form%20with%20focus%20point%20distance%7D%5C%5C%5C%5C%0A%5Cbegin%7Barray%7D%7Bllll%7D%0A4p%28x-%20h%29%3D%28y-%20k%29%5E2%0A%5C%5C%5C%5C%0A%5Cboxed%7B4p%28y-%20k%29%3D%28x-%20h%29%5E2%7D%0A%5Cend%7Barray%7D%0A%5Cqquad%20%0A%5Cbegin%7Barray%7D%7Bllll%7D%0Avertex%5C%20%28%20h%2C%20k%29%5C%5C%5C%5C%0A%20p%3D%5Ctextit%7Bdistance%20from%20vertex%20to%20%7D%5C%5C%0A%5Cqquad%20%5Ctextit%7B%20focus%20or%20directrix%7D%0A%5Cend%7Barray%7D%5C%5C%5C%5C%0A-------------------------------%5C%5C%5C%5C%0A%5Cbegin%7Bcases%7D%0Ah%3D-7%5C%5C%0Ak%3D-3%5C%5C%0Ap%3D8%0A%5Cend%7Bcases%7D%5Cimplies%204%288%29%5By-%28-3%29%5D%3D%5Bx-%28-7%29%5D%5E2%0A%5C%5C%5C%5C%5C%5C%0A32%28y%2B3%29%3D%28x%2B7%29%5E2%5Cimplies%20y%2B3%3D%5Ccfrac%7B1%7D%7B32%7D%28x%2B7%29%5E2%0A%5C%5C%5C%5C%5C%5C%0Ay%3D%5Ccfrac%7B1%7D%7B32%7D%28x%2B7%29%5E2-3)
keep in mind that the vertex is half-way between those two fellows, and the distance from the vertex to either one of them is "p" units, check the picture below.
with that focus point and that directrix, the half-way over the axis of symmetry will be -7, -3, that's where the vertex is at, and notice the distance "p", is 8 units.
since the parabola is opening upwards, "p" is positive 8.