The point (–4, 5) is the vertex of a square and one of its diagonals is 7x – y + 8 = 0. The equation of the other diagonal is
(A) 7x – y + 23 = 0 (B) 7y + x = 30 (C) 7y + x = 31 (D) x – 7y = 30

Question

Vikas TU · Answer

Equation of perpendicular line to 7x−y+8=0 7x−y+8=0 is x+7y=hx+7y=h which is passes through (−4, 5)(−4, 5) h = 31

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The point (–4, 5) is the vertex of a square and one of its diagonals is 7x – y + 8 = 0. The equation of the other diagonal is (A) 7x – y + 23 = 0 (B) 7y + x = 30 (C) 7y + x = 31 (D) x – 7y = 30

The point (–4, 5) is the vertex of a square and one of its diagonals is 7x – y + 8 = 0. The equation of the other diagonal is
(A) 7x – y + 23 = 0 (B) 7y + x = 30 (C) 7y + x = 31 (D) x – 7y = 30

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The point (–4, 5) is the vertex of a square and one of its diagonals is 7x – y + 8 = 0. The equation of the other diagonal is (A) 7x – y + 23 = 0 (B) 7y + x = 30 (C) 7y + x = 31 (D) x – 7y = 30

The point (–4, 5) is the vertex of a square and one of its diagonals is 7x – y + 8 = 0. The equation of the other diagonal is (A) 7x – y + 23 = 0 (B) 7y + x = 30 (C) 7y + x = 31 (D) x – 7y = 30

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The point (–4, 5) is the vertex of a square and one of its diagonals is 7x – y + 8 = 0. The equation of the other diagonal is
(A) 7x – y + 23 = 0 (B) 7y + x = 30 (C) 7y + x = 31 (D) x – 7y = 30