Question

Two of the vertices of a rectangle ABCD
are B(-3,1) and D(1,1) and the equation
of the line containing one of the side of
the rectangle is 4x+
7y+5=0 . Find the
equations of the lines containing the remaining sides​

Answer 1

Answer:

Since, the side AB is perpendicular to AD

∴ Its equation if of the form 7x−4y+λ=0

Since, it passes through (−3,1)

∴7(−3)−4(1)+λ=0⇒λ=25

∴ Equation of AB is 7x−4y+25=0

Now, BC is parallel to AD

Therefore, its equation is 4x+7y+λ=0

Since, it passes through (1,1).

∴4(1)+7(1)+λ=0⇒λ=−11

∴ equation of BC is 4x+7y−11=0

Now,equation of DC is 7x−4y+λ=0

⇒7(1)−4(1)+λ=0⇒λ=−3

∴7x−4y−3=0

Step-by-step explanation:

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