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# Equation of a straight line

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The points B(1,3) and D(6,8) are two opposite vertices of a square ABCD. Find the equation of the diagonal AC

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1. 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

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2. 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

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