A(7,2)and C(1,4) are two vertices of a square ABCD,find the equation of the diagonal BD

Question

THIS QUESTION IS FROM THE ADVANCED LEVEL MATHEMATICS PURE 1 OF HUGH NEILL AND DOUGLAS QUADLING.ITS FROM THE CHAPTER 1 MISCELLANEOUS EXERCISE 1 OF QUESTION 4

Answer 1

Since AC is the other diagonal of square ABCD, AC and BD perpendicularly bisect each other at the midpoint of AC, O(4,3). So, OA=OB=OC=OD, and ∠AOB=90°. 1). Thru O, draw a horizontal y=3, and a vertical x=4. From A and C, draw 2 verticals toward the horizontal y=3. Label each foot on the line, E(7,3) and F(1,3), respectively. 2). From B and D, 2 horizontals toward the vertical x=4. Label each foot on the line, G and H, respectively. These ΔAOE, ΔBOG, ΔCOF and ΔDOH are right triangles. 3). In ΔAOE and ΔCOF, AE//CF. So, ∠OAE=OCF (alternate angles), ∠AOE=∠COF (vertical angles) and OA=OC. Thus, ΔAOE≡ΔCOF (A.S.A). In the same way, ΔBOG≡ΔDOH (A.S.A). 4). In ΔAOE and ΔBOG, AC⊥BD and EF⊥GH. So, ∠AOE=90°－∠EOB=∠BOG. OA=OB, OE//BG and ∠OAE=90°－∠AOE=∠EOB=∠OBG (alternate angles). Thus,ΔAOE≡ΔBOG (A.S.A). Therfore, AE=BG=CF=DH=1, OE=OG=OF=OH=3, and the coordinates of B and D are B(5,6) and D(3,0). 5). A linear equation that passes thru 2 points, P(x1,y1) and Q(x2,y2), is expressed as follows: (y－y1) / (y2－y1)=(x－x1) / (x2－x1). To find a equation that passes thru B and D, plug x1=5, y1=6, x2=3 and y2=0 into the equation and simplify it. (y－6) / (0－6)=(x－5) / (3－5), y=3x－9. CK this at point O(4,3). Plug x=4 into the equation. y=3x－9=3x4－9=3. Points B,O and D are on a straight line y=3x－9. CKD. The equation of diagonal BD is y=3x－9 (3≦x≦5). 　

Answer 2

First, find the coordinates of points B(x1,y1) and D(x2,y2). 1). Label the midpoint of diagonal AC: O(4,3). Thru O, draw a horizontal y=3, and mark a point P on the line just above vertex A. Assume pt.O and OP be the origin and the initial side, respectively, for the measurement of angles that rotate around O. 2). Define m∠AOP=θ and the length of OA=r. Each coordinate of A(7,2) is expressed, respectively, with θ and r, and simplified as follows: 7=rcos(-θ)+4=rcosθ+4, and 2=rsin(-θ)+3=-rsinθ+3. So, rcosθ=3 and rsinθ=1. 4). AC⊥BD, OA=OB=OC=OD=r, and m∠AOB=90°. Each coordinate of B and D is also expressed respectively as follows: x1=rcos(90°－θ)+4=rsinθ+4, y1=rsin(90°－θ)+3=rcosθ+3, and x2=rcos(270°－θ)+4=-rsinθ+4, y2=rsin(270°－θ)+3=-rcosθ+3. Plug rcosθ=3 and rsinθ=1 into the equations. x1=5, y1=6 and x2=3, y2=0. So, the coordinates of 2 vertices are B(5,6) and D(3,0). 5). An equation that goes thru 2 points B and D is expressed as follows: (y－y1)/(y2－y1)=(x－x1)/(x2－x1). Plug x1=5, y1=6 and x2=3, y2=0 and simplify it. y=3x-9. CK this at pt.O plugging x=4 into the equation. y=3×4－9=3. y=3x-9 goes thru points B,O and D. CKD. The equation of the diagonal BD is y=3x－9 (3≦x≦5).

Answer 3

Mark the midpoint of diagonal AC as pt.O(4,3). Draw a circle centered at O thru 2 points A(7,2) and C(1,4). The slope of AC, m, can be written as follows: m=(rise / run)=tanθ, where the angle θ is measured around O, assuming O as the origin and a rightwards horizontal ray from O as the initial side. So, n, the slope of a line that crosses AC perpendicularly at O, can be written as follows: n=tan(θ＋90°)=－cotθ=－1/tanθ=－1/m. This states n is the negative reciprocal of m: n=－1/m=3. Thus, the equation of a line, that bisects AC perpendicularly at O(4,3), can be written and simplified as follows: y－3=3(x－4), y=3x－9. This perpendicular crosses circle O at E above O, and at F below. Connect 4 points A,E,C and F. EF⊥AC and OA=OE=OC=OF= AC/2 Thus, AC and EF perpendicularly bisect each other at O. This states quarilateral AECF is a square that has AC as its diagonal. So, square AECF is identical to square ABCD, where 2 vertices E and F correspond to B and D respectively. Two points B and D lie on the y=3x－9. The equation of diagonal BD is y=3x－9 (3≦x≦5).