For the graph x = -3 find the slope of a line that is parallel to it.

1 answer:

5 0

We have that

the slope of the line that is parallel to x=-3 is classified as undefined.

<span>let's check the answer
</span><span>1) take any two points of the line x = -3
</span>point A (-3,0) point B (-3,5)
2) find the slope of the line x=-3
m=(5-0)/(-3+3)=5/0----------> is undefined

the answer is the option undefined

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Check the picture below.

there are a few angles shaded, I'll call that angle at O ∡O, and that angle at "x" ∡x and the green angle at A, ∡A, and the angle at B, ∡B, just so you know which angle we're referring to.

well, first off let's notice that B and C are points of tangency, meaning we get right-angles as you see in the picture, if we run an angle bisector from ∡x, towards point O, we get two congruent triangles, lemme shorten that up some, 90 + 90 + ∡O +∡x = 360, which means that ∡O + ∡x = 180, meaning that ∡O = 180 - ∡x, if that's so, the angle at point O across the shade, is 360 - ∡O or 360 - (180 - ∡x).

alrighty, by the inscribed angle theorem, ∡A is half of ∡O.

let's now focus on the triangle AOB, we'll be using half of ∡A and half of ∡O and ∡B on that triangle, let's proceed.

$\measuredangle x + \measuredangle O=180\implies \measuredangle O=180-\measuredangle x \\\\\\ \measuredangle A = \cfrac{\measuredangle O}{2}\implies \measuredangle A = \cfrac{180-\measuredangle x}{2} \\\\\\ \stackrel{\textit{the angle across from }\measuredangle O~is}{360-\measuredangle O}\implies 360-(180 - \measuredangle x)\implies 180+\measuredangle x \\\\[-0.35em] ~\dotfill$

$\stackrel{\textit{\large the sum of all three angles at }\triangle AOB}{\stackrel{\textit{half of }\measuredangle A}{\cfrac{1}{2}\cdot \cfrac{180-\measuredangle x}{2}}~~ + ~~\stackrel{\textit{half the angle across from }\measuredangle O}{\cfrac{180+\measuredangle x}{2}}~~ +~~\measuredangle B~~ =~~180} \\\\\\ \cfrac{180-\measuredangle x}{4}+\cfrac{180+\measuredangle x}{2}+\measuredangle B = 180$

now, let's multiply both sides by the LCD of all denominators, hmmmm in this case that'll be 4, so multiplying both sides by the LCD will do away with the denominators, so let's do so.

$4\left( \cfrac{180-\measuredangle x}{4}+\cfrac{180+\measuredangle x}{2}+\measuredangle B \right) = 4(180) \\\\\\ (180-\measuredangle x)+2(180+\measuredangle x)+4\measuredangle B = 720 \\\\\\ (180-\measuredangle x)+(360+2\measuredangle x)+4\measuredangle B = 720\implies 540+\measuredangle x+4\measuredangle B = 720$

$\measuredangle x+4\measuredangle B = 180\implies 4\measuredangle B = 180-\measuredangle x\implies \measuredangle B = \cfrac{180-\measuredangle x}{4} \\\\[-0.35em] ~\dotfill\\\\ ~\hfill \measuredangle B = 45 - \cfrac{\measuredangle x}{4}~\hfill$