punto final (-9,4), (-13/2 ,6)

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`x`²	`x`^□	log_□	√☐	^□√☐	≤	≥	□□	·	÷	`x`^◦	π
(☐)^′	`ddx`	∂∂`x`	∫	∫_□^□	lim	∑	∞	`θ`	(`f` ◦ `g`)	`f`(`x`)

☐²

x^☐

√☐

^□√☐

□□

log_□

θ

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∫

ddx

≥	≤	·	÷	`x`^◦	(☐)	\|☐\|	(`f` ◦ `g`)	`f`(`x`)	ln	`e`^☐
(☐)^′	∂∂`x`	∫_□^□	lim	∑	sin	cos	tan	cot	csc	sec

simplify solve for inverse tangent line

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end point (−9, 4), (−132 , 6)

Solution

(−4, 8)

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End point given startpoint (x₁, y₁) and midpoint (x₂, y₂): (2x₂−x₁, 2y₂−y₁)

(x₁, y₁)=(−9, 4), (x₂, y₂)=(−132 , 6)

=(2(−132 )−(−9), 2· 6−4)

Simplify

=(−4, 8)

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−▭▭	<	7	8	9	÷	`AC`
+▭▭	>	4	5	6	×	☐☐☐
×▭▭	(	1	2	3	−	`x`
▭ \|▭	)	.	0	=	+	`y`

punto final (-9,4), (-13/2 ,6)

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