bearing in mind that perpendicular lines have negative reciprocal slopes, so
![\bf \begin{array}{|c|ll} \cline{1-1} slope-intercept~form\\ \cline{1-1} \\ y=\underset{y-intercept}{\stackrel{slope\qquad }{\stackrel{\downarrow }{m}x+\underset{\uparrow }{b}}} \\\\ \cline{1-1} \end{array}~\hspace{10em}\stackrel{slope}{y=\stackrel{\downarrow }{-\cfrac{1}{3}}x-1} \\\\[-0.35em] ~\dotfill](https://tex.z-dn.net/?f=%5Cbf%20%5Cbegin%7Barray%7D%7B%7Cc%7Cll%7D%20%5Ccline%7B1-1%7D%20slope-intercept~form%5C%5C%20%5Ccline%7B1-1%7D%20%5C%5C%20y%3D%5Cunderset%7By-intercept%7D%7B%5Cstackrel%7Bslope%5Cqquad%20%7D%7B%5Cstackrel%7B%5Cdownarrow%20%7D%7Bm%7Dx%2B%5Cunderset%7B%5Cuparrow%20%7D%7Bb%7D%7D%7D%20%5C%5C%5C%5C%20%5Ccline%7B1-1%7D%20%5Cend%7Barray%7D~%5Chspace%7B10em%7D%5Cstackrel%7Bslope%7D%7By%3D%5Cstackrel%7B%5Cdownarrow%20%7D%7B-%5Ccfrac%7B1%7D%7B3%7D%7Dx-1%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill)
so we're really looking for a line whose slope is 3 and runs through (1,5)
Yeah the answer tells you right there lol, 113
Answer:
∠MLP = 72° , ∠LJK = 22° , ∠JKL = 72° , ∠KLJ = 86°
Step-by-step explanation:
Here, given In ΔJLK and ΔMLP
Here, JK II ML, LM = MP
∠JLM = 22° and ∠LMP = 36°
Now, As angles opposite to equal sides are equal.
⇒ ∠MLP = ∠MPL = x°
Now, in ΔMLP
By <u>ANGLE SUM PROPERTY</u>: ∠MLP + ∠MPL + ∠LMP = 180°
⇒ x° + x° + 36° = 180°
⇒ 2 x = 180 - 36 = 144
or, x = 72°
⇒ ∠MLP = ∠MPL = 72°
Now,as JK II ML
⇒ ∠LJK = ∠JLM = 22° ( Alternate pair of angles)
Now, by the measure of straight angle:
∠MLP + ∠JLM + ∠JLK = 180° ( Straight angle)
⇒ 72° + 22° + ∠JLK = 180°
or, ∠JLK = 86°
In , in ΔJLK
By <u>ANGLE SUM PROPERTY</u>: ∠JKL + ∠JLK + ∠LJK = 180°
⇒ ∠JKL + 86° + 22° = 180°
⇒ ∠JKL = 180 - 108 = 72 , or ∠JKL = 72°
Hence, from above proof , ∠MLP = 72° , ∠LJK = 22° , ∠JKL = 72° ,
∠KLJ = 86°
Step-by-step explanation:
Opposite rays is the answer....
