Answer:
In triangle SHD and triangle STD.
[Side]
Since, a line is said to be perpendicular to another line if the two lines intersect at a right angle.
⇒ 
[leg] [Given]
Reflexive property states that the value is equal to itself.
[Leg] [Reflexive property]
HL(Hypotenuse-leg) theorem states that any two right triangles that have a congruent hypotenuse and a corresponding congruent leg are the congruent triangles.
then, by HL theorem;
Proved!
1x + 4y = 11 ⇒ 5x + 20y = 55
5x - 7y = -10 ⇒ <u>5x - 7y = -10</u>
<u>27y</u> = <u>65</u>
27 27
y = 2¹¹/₂₇
x + 4(2¹¹/₂₇) = 11
x + 8²²/₂₇ = 11
<u> - 8²²/₂₇ - 8²²/₂₇
</u> x = 2⁵/₂₇
(x, y) = (2⁵/₂₇, 2¹¹/₂₇)
This one could be 8/10, or 4/5 simplified
let's multiply both sides by the LCD of all fractions, in this case is 4, just to do away with the denominators for a few seconds
![\bf f(x) = \cfrac{1}{4}x^2+\cfrac{1}{2}x-\cfrac{35}{4}\implies \stackrel{\textit{multiplying both sides by }\stackrel{LCD}{4}}{4[f(x)]=4\left( \cfrac{1}{4}x^2+\cfrac{1}{2}x-\cfrac{35}{4} \right)} \\\\\\ 4f(x) = x^2+2x-35\implies 4f(x) = (x+7)(x-5) \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ ~\hfill f(x) = \cfrac{1}{4}(x+7)(x-5)~\hfill](https://tex.z-dn.net/?f=%5Cbf%20f%28x%29%20%3D%20%5Ccfrac%7B1%7D%7B4%7Dx%5E2%2B%5Ccfrac%7B1%7D%7B2%7Dx-%5Ccfrac%7B35%7D%7B4%7D%5Cimplies%20%5Cstackrel%7B%5Ctextit%7Bmultiplying%20both%20sides%20by%20%7D%5Cstackrel%7BLCD%7D%7B4%7D%7D%7B4%5Bf%28x%29%5D%3D4%5Cleft%28%20%5Ccfrac%7B1%7D%7B4%7Dx%5E2%2B%5Ccfrac%7B1%7D%7B2%7Dx-%5Ccfrac%7B35%7D%7B4%7D%20%5Cright%29%7D%20%5C%5C%5C%5C%5C%5C%204f%28x%29%20%3D%20x%5E2%2B2x-35%5Cimplies%204f%28x%29%20%3D%20%28x%2B7%29%28x-5%29%20%5C%5C%5C%5C%5B-0.35em%5D%20%5Crule%7B34em%7D%7B0.25pt%7D%5C%5C%5C%5C%20~%5Chfill%20f%28x%29%20%3D%20%5Ccfrac%7B1%7D%7B4%7D%28x%2B7%29%28x-5%29~%5Chfill)
