Math, asked by prity3659, 1 year ago

Write equation of line passing through the point A(-3,5) ,B(4,-7)

Answers

Answered by krishnasnair

y-5=-12÷7(x+3) is the answer please mark as BRAINLIEST

Answered by SujalBendre

Answer

AnswerWe need to find the equation of the line passing through the points A(−3,4) and B(4,5).

AnswerWe need to find the equation of the line passing through the points A(−3,4) and B(4,5).The equation o

Here (−3,4)≡(x

Here (−3,4)≡(x 1

Here (−3,4)≡(x 1 ,y

Here (−3,4)≡(x 1 ,y 1

Here (−3,4)≡(x 1 ,y 1 ),(4,5)≡(x

Here (−3,4)≡(x 1 ,y 1 ),(4,5)≡(x 2

Here (−3,4)≡(x 1 ,y 1 ),(4,5)≡(x 2 ,y

Here (−3,4)≡(x 1 ,y 1 ),(4,5)≡(x 2 ,y 2

Here (−3,4)≡(x 1 ,y 1 ),(4,5)≡(x 2 ,y 2 )

Here (−3,4)≡(x 1 ,y 1 ),(4,5)≡(x 2 ,y 2 )∴

Here (−3,4)≡(x 1 ,y 1 ),(4,5)≡(x 2 ,y 2 )∴ 4+3

Here (−3,4)≡(x 1 ,y 1 ),(4,5)≡(x 2 ,y 2 )∴ 4+3x+3

Here (−3,4)≡(x 1 ,y 1 ),(4,5)≡(x 2 ,y 2 )∴ 4+3x+3 =

Here (−3,4)≡(x 1 ,y 1 ),(4,5)≡(x 2 ,y 2 )∴ 4+3x+3 = 5−4

Here (−3,4)≡(x 1 ,y 1 ),(4,5)≡(x 2 ,y 2 )∴ 4+3x+3 = 5−4y−4

Here (−3,4)≡(x 1 ,y 1 ),(4,5)≡(x 2 ,y 2 )∴ 4+3x+3 = 5−4y−4 ⇒

Here (−3,4)≡(x 1 ,y 1 ),(4,5)≡(x 2 ,y 2 )∴ 4+3x+3 = 5−4y−4 ⇒ 7

Here (−3,4)≡(x 1 ,y 1 ),(4,5)≡(x 2 ,y 2 )∴ 4+3x+3 = 5−4y−4 ⇒ 7x+3

Here (−3,4)≡(x 1 ,y 1 ),(4,5)≡(x 2 ,y 2 )∴ 4+3x+3 = 5−4y−4 ⇒ 7x+3 =

Here (−3,4)≡(x 1 ,y 1 ),(4,5)≡(x 2 ,y 2 )∴ 4+3x+3 = 5−4y−4 ⇒ 7x+3 = 1

Here (−3,4)≡(x 1 ,y 1 ),(4,5)≡(x 2 ,y 2 )∴ 4+3x+3 = 5−4y−4 ⇒ 7x+3 = 1y−4

Here (−3,4)≡(x 1 ,y 1 ),(4,5)≡(x 2 ,y 2 )∴ 4+3x+3 = 5−4y−4 ⇒ 7x+3 = 1y−4 ⇒x+3=7(y−4)

Here (−3,4)≡(x 1 ,y 1 ),(4,5)≡(x 2 ,y 2 )∴ 4+3x+3 = 5−4y−4 ⇒ 7x+3 = 1y−4 ⇒x+3=7(y−4)⇒x+3=7y−28

Here (−3,4)≡(x 1 ,y 1 ),(4,5)≡(x 2 ,y 2 )∴ 4+3x+3 = 5−4y−4 ⇒ 7x+3 = 1y−4 ⇒x+3=7(y−4)⇒x+3=7y−28⇒x−7y+3+28=0

Here (−3,4)≡(x 1 ,y 1 ),(4,5)≡(x 2 ,y 2 )∴ 4+3x+3 = 5−4y−4 ⇒ 7x+3 = 1y−4 ⇒x+3=7(y−4)⇒x+3=7y−28⇒x−7y+3+28=0⇒x−7y+31=0

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