One side of a rectangle lies along the line 4 | vedclass

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(A) Given one side is $4x + 7y + 5 = 0.$ The vertices are $A(-3, 1)$ and $B(1, 1).$ Since $A$ and $B$ do not satisfy the given line equation,the side

Vedclass · Accepted Answer

$7x - 4y + 25 = 0, 4x + 7y = 11$ and $7x - 4y - 3 = 0$

Vedclass · Answer

(A) Given one side is $4x + 7y + 5 = 0.$ The vertices are $A(-3, 1)$ and $B(1, 1).$ Since $A$ and $B$ do not satisfy the given line equation,the side containing $A$ and $B$ is parallel to the given line. Let the side containing $A$ and $B$ be $4x + 7y + k = 0.$ Since it passes through $(-3, 1),$ $4(-3) + 7(1) + k = 0 \implies -12 + 7 + k = 0 \implies k = 5.$ So,the side is $4x + 7y - 5 = 0.$ Wait,the side containing $A$ and $B$ is $4x + 7y - 5 = 0.$ The other side parallel to this is $4x + 7y + k' = 0.$ Passing through $(1, 1),$ $4(1) + 7(1) + k' = 0 \implies k' = -11.$ So,the opposite side is $4x + 7y - 11 = 0.$ The perpendicular sides have slope $7/4.$ Side through $(-3, 1): y - 1 = \frac{7}{4}(x + 3) \implies 4y - 4 = 7x + 21 \implies 7x - 4y + 25 = 0.$ Side through $(1, 1): y - 1 = \frac{7}{4}(x - 1) \implies 4y - 4 = 7x - 7 \implies 7x - 4y - 3 = 0.$ Thus,the equations are $4x + 7y - 11 = 0, 7x - 4y + 25 = 0$ and $7x - 4y - 3 = 0.$ Comparing with options,option $(A)$ matches the form $7x - 4y + 25 = 0, 4x + 7y - 11 = 0$ and $7x - 4y - 3 = 0.$

One side of a rectangle lies along the line $4x + 7y + 5 = 0.$ Two of its vertices are $(-3, 1)$ and $(1, 1).$ Then the equations of the other three sides are

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