Question

Show that A(3, 2), B(6, −2) and C(2, −5) can be the vertices of a square.

1. Find the co-ordinates of its fourth vertex D, if ABCD is a square.
2. Without using the co-ordinates of vertex D, find the equation of side AD of the square and also the equation of diagonal BD.

Answer 1

Using distance formula, we have:

`AB = sqrt((6 - 3)^2 + (-2 - 2)^2`

= `sqrt(9 + 16)`

= 5

`BC = sqrt((2 - 6)^2 + (-5 + 2)^2`

= `sqrt(16 + 9)`

= 5

Thus, AC = BC

Also, Slope of AB = `(-2 - 2)/(6 - 3) = (-4)/3`

Slope of BC = `(-5 + 2)/(2 - 6) = (-3)/(-4) = 3/4`

Slope of AB × Slope of BC = –1

Thus, AB ⊥ BC

Hence, A, B, C can be the vertices of a square…..

i. Slope of AB = `(-2 - 2)/(6 - 3)` Slope of CD

Equation of the line CD is

y – y₁ = m(x – x₁)

`=> y + 5 = (-4)/3 (x - 2)`

`=>` 3y + 15 = −4x + 8

`=>` 4x + 3y = –7   ...(1)

Slope of BC = `((-5 + 2)/(2-6)) = (-3)/(-4) = 3/4` Slope of AD

Equation of the line AD is

y – y₁ = m(x – x₁)

`=> y - 2 = 3/4 (x - 3)`

`=>` 4y – 8 = 3x – 9

`=>` 4y = 3x – 9 + 8

`=>` 4y = 3x – 1   ...(2)

Now, D is the point of intersection of CD and AD

(1) `=>` 16x + 12y = –28

(2) `=>` 9x – 12y = 3

Adding the above two equations we get,

25x = –25

`=>` x = −1

So, 4y = 3x − 1

= −3 − 1

= −4

`=>` y = −1

Thus, the co-ordinates of point D are (−1, −1).

ii. The equation of line AD is found in part (i)

It is 3x – 4y = 1 or 4y = 3x – 1

Slope of BD = `(-1 + 2)/(-1 - 6) = 1/(-7) = (-1)/7`

The equation of diagonal BD is

y − y₁ = m(x − x₁)

`=> y + 1 = (-1)/7 (x + 1)`

`=>` 7y + 7 = −x − 1

`=>` x + 7y + 8 = 0