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prove that the lines 2x+3y+8=0 and 27x-18y+10=0 are perpendicular to each other.

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Answered by cutebrainlystar

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Step-by-step explanation:

2x+3y+7=0⇒3y=−2x−7⇒y=

2x+3y+7=0⇒3y=−2x−7⇒y= 3

2x+3y+7=0⇒3y=−2x−7⇒y= 3−2

2x+3y+7=0⇒3y=−2x−7⇒y= 3−2 x−

2x+3y+7=0⇒3y=−2x−7⇒y= 3−2 x− 3

2x+3y+7=0⇒3y=−2x−7⇒y= 3−2 x− 37

2x+3y+7=0⇒3y=−2x−7⇒y= 3−2 x− 37 27x−18y+25=0⇒18y=27x+25⇒y=

2x+3y+7=0⇒3y=−2x−7⇒y= 3−2 x− 37 27x−18y+25=0⇒18y=27x+25⇒y= 2

2x+3y+7=0⇒3y=−2x−7⇒y= 3−2 x− 37 27x−18y+25=0⇒18y=27x+25⇒y= 23

2x+3y+7=0⇒3y=−2x−7⇒y= 3−2 x− 37 27x−18y+25=0⇒18y=27x+25⇒y= 23 x+

2x+3y+7=0⇒3y=−2x−7⇒y= 3−2 x− 37 27x−18y+25=0⇒18y=27x+25⇒y= 23 x+ 18

2x+3y+7=0⇒3y=−2x−7⇒y= 3−2 x− 37 27x−18y+25=0⇒18y=27x+25⇒y= 23 x+ 1825

2x+3y+7=0⇒3y=−2x−7⇒y= 3−2 x− 37 27x−18y+25=0⇒18y=27x+25⇒y= 23 x+ 1825 ∴m

2x+3y+7=0⇒3y=−2x−7⇒y= 3−2 x− 37 27x−18y+25=0⇒18y=27x+25⇒y= 23 x+ 1825 ∴m 1

2x+3y+7=0⇒3y=−2x−7⇒y= 3−2 x− 37 27x−18y+25=0⇒18y=27x+25⇒y= 23 x+ 1825 ∴m 1 =

2x+3y+7=0⇒3y=−2x−7⇒y= 3−2 x− 37 27x−18y+25=0⇒18y=27x+25⇒y= 23 x+ 1825 ∴m 1 = 3

2x+3y+7=0⇒3y=−2x−7⇒y= 3−2 x− 37 27x−18y+25=0⇒18y=27x+25⇒y= 23 x+ 1825 ∴m 1 = 3−2

2x+3y+7=0⇒3y=−2x−7⇒y= 3−2 x− 37 27x−18y+25=0⇒18y=27x+25⇒y= 23 x+ 1825 ∴m 1 = 3−2 ;m

2x+3y+7=0⇒3y=−2x−7⇒y= 3−2 x− 37 27x−18y+25=0⇒18y=27x+25⇒y= 23 x+ 1825 ∴m 1 = 3−2 ;m 2

2x+3y+7=0⇒3y=−2x−7⇒y= 3−2 x− 37 27x−18y+25=0⇒18y=27x+25⇒y= 23 x+ 1825 ∴m 1 = 3−2 ;m 2 =

2x+3y+7=0⇒3y=−2x−7⇒y= 3−2 x− 37 27x−18y+25=0⇒18y=27x+25⇒y= 23 x+ 1825 ∴m 1 = 3−2 ;m 2 = 2

2x+3y+7=0⇒3y=−2x−7⇒y= 3−2 x− 37 27x−18y+25=0⇒18y=27x+25⇒y= 23 x+ 1825 ∴m 1 = 3−2 ;m 2 = 23

2x+3y+7=0⇒3y=−2x−7⇒y= 3−2 x− 37 27x−18y+25=0⇒18y=27x+25⇒y= 23 x+ 1825 ∴m 1 = 3−2 ;m 2 = 23 ⇒m

2x+3y+7=0⇒3y=−2x−7⇒y= 3−2 x− 37 27x−18y+25=0⇒18y=27x+25⇒y= 23 x+ 1825 ∴m 1 = 3−2 ;m 2 = 23 ⇒m 1

2x+3y+7=0⇒3y=−2x−7⇒y= 3−2 x− 37 27x−18y+25=0⇒18y=27x+25⇒y= 23 x+ 1825 ∴m 1 = 3−2 ;m 2 = 23 ⇒m 1 m

2x+3y+7=0⇒3y=−2x−7⇒y= 3−2 x− 37 27x−18y+25=0⇒18y=27x+25⇒y= 23 x+ 1825 ∴m 1 = 3−2 ;m 2 = 23 ⇒m 1 m 2

2x+3y+7=0⇒3y=−2x−7⇒y= 3−2 x− 37 27x−18y+25=0⇒18y=27x+25⇒y= 23 x+ 1825 ∴m 1 = 3−2 ;m 2 = 23 ⇒m 1 m 2 =

2x+3y+7=0⇒3y=−2x−7⇒y= 3−2 x− 37 27x−18y+25=0⇒18y=27x+25⇒y= 23 x+ 1825 ∴m 1 = 3−2 ;m 2 = 23 ⇒m 1 m 2 = 3

2x+3y+7=0⇒3y=−2x−7⇒y= 3−2 x− 37 27x−18y+25=0⇒18y=27x+25⇒y= 23 x+ 1825 ∴m 1 = 3−2 ;m 2 = 23 ⇒m 1 m 2 = 3−2

2x+3y+7=0⇒3y=−2x−7⇒y= 3−2 x− 37 27x−18y+25=0⇒18y=27x+25⇒y= 23 x+ 1825 ∴m 1 = 3−2 ;m 2 = 23 ⇒m 1 m 2 = 3−2 ×

2x+3y+7=0⇒3y=−2x−7⇒y= 3−2 x− 37 27x−18y+25=0⇒18y=27x+25⇒y= 23 x+ 1825 ∴m 1 = 3−2 ;m 2 = 23 ⇒m 1 m 2 = 3−2 × 2

2x+3y+7=0⇒3y=−2x−7⇒y= 3−2 x− 37 27x−18y+25=0⇒18y=27x+25⇒y= 23 x+ 1825 ∴m 1 = 3−2 ;m 2 = 23 ⇒m 1 m 2 = 3−2 × 23

2x+3y+7=0⇒3y=−2x−7⇒y= 3−2 x− 37 27x−18y+25=0⇒18y=27x+25⇒y= 23 x+ 1825 ∴m 1 = 3−2 ;m 2 = 23 ⇒m 1 m 2 = 3−2 × 23 =−1

2x+3y+7=0⇒3y=−2x−7⇒y= 3−2 x− 37 27x−18y+25=0⇒18y=27x+25⇒y= 23 x+ 1825 ∴m 1 = 3−2 ;m 2 = 23 ⇒m 1 m 2 = 3−2 × 23 =−1the given lines are perpendicular to each other

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prove that the lines 2x+3y+8=0 and 27x-18y+10=0 are perpendicular to each other.​

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prove that the lines 2x+3y+8=0 and 27x-18y+10=0 are perpendicular to each other.