Question

Prove that A(4, 3), B(6, 4), C(5, 6) and D(3, 5) are the angular points of a square.

Answer 1

Now
AB = `sqrt((4 - 6)^2 + (3 - 4)^2)`
= `sqrt(4 + 1) = sqrt(5)"units"`
BC = `sqrt((6 - 5)^2 + (4 - 6)^2) = sqrt(1 + 4)`
BC = `sqrt(5)"units"`.
CD = `sqrt((5 - 3)^2 + (6 - 5)^2) = sqrt(4 + 1)`
CD = `sqrt(5)"units"`
Also DA = `sqrt((4 - 3)^2 + (3 + 5)^2)`
= `sqrt(1 + 4) = sqrt(5)`
DA = `sqrt(5)"units"`
So AB = BC = CD = DA.
Now slope of AB = m₁ - `(4 - 3)/(6 - 4) = (1)/(2)`
Slope of BC = m₂ = `(6 - 4)/(5 - 6) = (2)/(-1)`
Slope of CA = m₃ = `(5 - 6)/(3 - 5) = (1)/(2)`
Slope of DA = m₄ = `(5 - 3)/(3 - 4) = (2)/(-1)`
Since m₁ = m₃ and = m₂ = m₄
So AB || CD
and BC || DA.
Therefore, AB ⊥ BC
∴ ABCD is a square.
Hence proved.