Question

The points b(1,3) and d(6,8) are two opposite vertices of square abcd. find the equation of diagonal ac

Answer 1

The solution is attached as an image

Answer 2

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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