Show that A (3, 2), B (6, -2) and C (2, -5) can be the vertices of a square.
(i) Find the co-ordinates of its fourth vertex D, if ABCD is a square.
(ii) Without using the co-ordinates of vertex D, find the equation of side AD of the square and also the equation of diagonal BD.

Question

Show that A (3, 2), B (6, -2) and C (2, -5) can be the vertices of a square.
(i) Find the co-ordinates of its fourth vertex D, if ABCD is a square.
(ii) Without using the co-ordinates of vertex D, find the equation of side AD of the square and also the equation of diagonal BD.

KnowledgeBoat · Accepted Answer

Given, A (3, 2), B (6, -2) and C (2, -5) By distance formula, Distance between two points = Thus, AC = BC By formula, Slope = Slope of AB = Slope of BC = Slope of AB x Slope of BC = = -1 Hence, AB ⊥ BC Therefore, A, B, C can be the vertices of a square. (i) Slope of CD = Slope of AB = (As they are parallel) So, the equation of CD is ⇒ y – y₁ = m(x – x₁) ⇒ y - (-5) = (x – 2) ⇒ 3(y + 5) = -4(x - 2) ⇒ 3y + 15 = -4x + 8 ⇒ 4x + 3y + 7 = 0 .........(1) Slope of AD = Slope of BC = (As these lines are parallel) So, the equation of the line AD is ⇒ y – y₁ = m(x – x₁) ⇒ y - 2 = (x - 3) ⇒ 4(y – 2) = 3(x – 3) ⇒ 4y - 8 = 3x - 9 ⇒ 3x - 4y = -8 + 9 ⇒ 3x - 4y = 1 ......... (2) Now, D is the point of intersection of CD and AD. Solving (1) and (2), Multiplying equation (1) by 4 and (2) by 3 and adding them we get, ⇒ 16x + 12y + 28 + 9x – 12y = 0 + 3 ⇒ 25x = 3 - 28 ⇒ 25x = -25 ⇒ x = -1 Putting value of x in (1), we get ⇒ 4(-1) + 3y + 7 = 0 ⇒ 3y - 4 + 7 = 0 ⇒ 3y + 3 = 0 ⇒ 3y = -3 ⇒ y = -1 Hence, the co-ordinates of point D are (-1, -1). (ii) From the equation (2) The equation of the line AD is, ⇒ 3x – 4y = 1 ⇒ 4y = 3x - 1. Slope of AC = Since, diagonals of a square are perpendicular to each other. So, product of their slopes = -1. ∴ Slope of AC × Slope of BD = -1 ⇒ Slope of BD = -1 ⇒ Slope of BD = -. The equation of the diagonal BD is ⇒ y – y₁ = m(x – x₁) ⇒ y - (-2) = (x - 6) ⇒ 7(y + 2) = –1(x – 6) ⇒ 7y + 14 = -x + 6 ⇒ x + 7y + 14 - 6 = 0 ⇒ x + 7y + 8 = 0. Hence, equation of line BD is x + 7y + 8 = 0 and AD = 4y = 3x - 1.

Mathematics

Show that A (3, 2), B (6, -2) and C (2, -5) can be the vertices of a square.
(i) Find the co-ordinates of its fourth vertex D, if ABCD is a square.
(ii) Without using the co-ordinates of vertex D, find the equation of side AD of the square and also the equation of diagonal BD.

Straight Line Eq

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Show that A (3, 2), B (6, -2) and C (2, -5) can be the vertices of a square.(i) Find the co-ordinates of its fourth vertex D, if ABCD is a square.(ii) Without using the co-ordinates of vertex D, find the equation of side AD of the square and also the equation of diagonal BD.

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