midpoint (5sqrt(3), 7sqrt(5)), (sqrt(3),-sqrt(5))

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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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midpoint (5√3, 7√5), (√3, −√5)

Solution

(3√3, 3√5)

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Solution steps

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Midpoint of (x₁, y₁), (x₂, y₂): (x₂+x₁2 , y₂+y₁2 )

(x₁, y₁)=(5√3, 7√5), (x₂, y₂)=(√3, −√5)

=(√3+5√32 , −√5+7√52 )

Refine

=(3√3, 3√5)

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

midpoint (5sqrt(3), 7sqrt(5)), (sqrt(3),-sqrt(5))

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