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Question
Show that A(3, 2), B(6, −2) and C(2, −5) can be the vertices of a square.
- Find the co-ordinates of its fourth vertex D, if ABCD is a square.
- Without using the co-ordinates of vertex D, find the equation of side AD of the square and also the equation of diagonal BD.
Solution
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
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