Question

If the vertices of a rectangle are `A(1, 3), B (3, -4),` and `C( 1,-4)`. What is its area?

Answer 1

Area of rectangle is `14`

Explanation:

As we are given three vertices, `A(1,3)`, `B(3,-4)` and `C(1,-4)`, observe that as `A` and `C` have common abscissa, the line `AC` is parallel to `y`-axis and its length is `3-(-4)=7`.

Further`B(3,-4)` and `C(1,-4)` have common ordinate so line `BC` is parallel to `x`-axis and its length is `3-1=2`.

Hence area of rectangle is `7xx2=14`

Note that fourth point will be `D(3,3)` and our vertices appear as shown below:

graph{((x-1)^2+(y-3)^2-0.02)((x-3)^2+(y-3)^2-0.02)((x-3)^2+(y+4)^2-0.02)((x-1)^2+(y+4)^2-0.02) =0 [-10, 10, -5, 5]}