1 Deepak BoraNewbie Asked: July 10, 20202020-07-10T15:14:57+05:30 2020-07-10T15:14:57+05:30In: ICSE Equation of a straight line 1 The points B(1,3) and D(6,8) are two opposite vertices of a square ABCD. Find the equation of the diagonal AC diagonalequationstraight linevertices Share Facebook 2 Answers Voted Oldest Recent Deepak Bora Newbie 2020-07-10T19:22:10+05:30Added an answer on July 10, 2020 at 7:22 pm equation of line AC is x + y = 9 The point B(1,3) and D(6,8) are two opposite vertices of square ABCD. so, diagonal = √{(6 – 1)² + (8 – 3)²} = 5√2 we know , side length = diagonal/√2 so, side length of ABCD = 5√2/√2 = 5 let point A(a, b) from ∆ABD, slope of AB × slope of AD = -1 [ as both are perpendicular] (b – 3)/(a – 1) × (b – 8)/(a – 6) = -1 ⇒(b² – 11b + 24) = -(a² – 7a + 6) ⇒a² + b² – 11b – 7a + 30 = 0……(1) and (a – 1)² + (b – 3)² = 5² = (a – 6)² + (b – 8)² ⇒-2a – 6b + 10 = -12a – 16b + 36 + 64 ⇒10a + 10b = 36 + 54 = 90 ⇒a + b = 9 ……..(2) so, a² + (9 – a)² – 11(9 – a) – 7a + 30= 0 ⇒a² + 81 + a² – 18a – 99 + 11a – 7a + 30 = 0 ⇒2a² – 14a + 12 = 0 ⇒ a² – 7a + 6 = 0 ⇒a = 1 , 6 and b = 9 – a = 8, 3 so, A(1, 8) and C = (6, 3) now equation of line BC (y – 8) = (8 – 3)/(1 – 6)(x – 1) ⇒y – 8 = -1(x – 1) ⇒x + y – 9 = 0 0 Reply Share Share Share on Facebook Share on Twitter Share on LinkedIn Share on WhatsApp Deepak Bora Newbie 2020-07-14T00:17:10+05:30Added an answer on July 14, 2020 at 12:17 am Because diagonals of the square bisect each other at right angle. ∴ AC ⊥BD & O is the mid point of AC and BD, ∴ coordinates of O(x,y) Using mid point formula x= x1+x2/2 & y= y1+y2/2 x = 6+1/2 & y = 8+3/2 x= 7/2 & y = 11/2 Slope of BD = y2-y1/x2-x1 m1 = 3-8/1-6 m1 = -5/-5 ∴ m1 = 1 BC ⊥ BD (Slope of BD)(Slope of AC) = -1 (1)(m2) = -1 ∴ m2 = -1 Equation of AC,a it passes through the point O(7/2,11/2) (y-y1) = m2(x-x1) (y-11/2) = -1(x-7/2) After Taking LCM and cross multiplying 2y-11 = -1(2x-7) ∴ 2x+2y-18 = 0 0 Reply Share Share Share on Facebook Share on Twitter Share on LinkedIn Share on WhatsApp Leave an answerLeave an answerCancel reply Featured image Select file Browse Add a Video to describe the problem better. Video type Youtube Vimeo Dailymotion Facebook Choose from here the video type. Video ID Put Video ID here: https://www.youtube.com/watch?v=sdUUx5FdySs Ex: "sdUUx5FdySs". Click on image to update the captcha. Save my name, email, and website in this browser for the next time I comment. Related Questions Equation of a straight line Solve the following equation by factorisation. The vertices of a triangle ABC are (3,8) , B(-1,2), C(6-6). Find: 1. The slope of BC , ...

equation of line AC is x + y = 9

The point B(1,3) and D(6,8) are two opposite vertices of square ABCD.

so, diagonal = √{(6 – 1)² + (8 – 3)²} = 5√2

we know , side length = diagonal/√2

so, side length of ABCD = 5√2/√2 = 5

let point A(a, b)

from ∆ABD,

slope of AB × slope of AD = -1 [ as both are perpendicular]

(b – 3)/(a – 1) × (b – 8)/(a – 6) = -1

⇒(b² – 11b + 24) = -(a² – 7a + 6)

⇒a² + b² – 11b – 7a + 30 = 0……(1)

and (a – 1)² + (b – 3)² = 5² = (a – 6)² + (b – 8)²

⇒-2a – 6b + 10 = -12a – 16b + 36 + 64

⇒10a + 10b = 36 + 54 = 90

⇒a + b = 9 ……..(2)

so, a² + (9 – a)² – 11(9 – a) – 7a + 30= 0

⇒a² + 81 + a² – 18a – 99 + 11a – 7a + 30 = 0

⇒2a² – 14a + 12 = 0

⇒ a² – 7a + 6 = 0

⇒a = 1 , 6 and b = 9 – a = 8, 3

so, A(1, 8) and C = (6, 3)

now equation of line BC

(y – 8) = (8 – 3)/(1 – 6)(x – 1)

⇒y – 8 = -1(x – 1)

⇒x + y – 9 = 0

Because diagonals of the square bisect each other at right angle.

∴ AC ⊥BD & O is the mid point of AC and BD,

∴ coordinates of O(x,y)

Using mid point formula

x= x1+x2/2 & y= y1+y2/2

x = 6+1/2 & y = 8+3/2

x= 7/2 & y = 11/2

Slope of BD = y2-y1/x2-x1

m1 = 3-8/1-6

m1 = -5/-5

∴ m1 = 1

BC ⊥ BD

(Slope of BD)(Slope of AC) = -1

(1)(m2) = -1

∴ m2 = -1

Equation of AC,a it passes through the point O(7/2,11/2)

(y-y1) = m2(x-x1)

(y-11/2) = -1(x-7/2)

After Taking LCM and cross multiplying

2y-11 = -1(2x-7)

∴ 2x+2y-18 = 0