Question

Find the equations of the diagonals of a rectangle whose sides are: $$x = -1, x = 4$$, $$y = -1$$ and $$y = 2$$.

[Hint. The equations of sides in order are x = –1, y = –1, x = 4, y = 2. So, its vertices are A(–1, –1), B(4, –1), C(4, 2) and D(–1, 2).]

Answer 1

The vertices of the rectangle formed by the intersection of the lines are $$A(-1, -1)$$, $$B(4, -1)$$, $$C(4, 2)$$ and $$D(-1, 2)$$.

The diagonals are $$AC$$ and $$BD$$.

Equation of diagonal $$AC$$:

Points are $$A(-1, -1)$$ and $$C(4, 2)$$. 

$$m_{AC} = \frac{2 - (-1)}{4 - (-1)}$$

$$= \frac{3}{5}$$ 

Equation of $$AC$$: 

$$y - (-1) = \frac{3}{5}(x - (-1))$$ 

$$5(y + 1) = 3(x + 1)$$ 

$$5y + 5 = 3x + 3$$

$$3x - 5y - 2 = 0$$

Equation of diagonal $$BD$$:

Points are $$B(4, -1)$$ and $$D(-1, 2)$$. 

$$m_{BD} = \frac{2 - (-1)}{-1 - 4}$$

$$= \frac{3}{-5}$$

$$= -\frac{3}{5}$$

Equation of $$BD$$:

$$y - (-1) = -\frac{3}{5}(x - 4)$$ 

$$5(y + 1) = -3(x - 4)$$ 

$$5y + 5 = -3x + 12$$ 

$$3x + 5y - 7 = 0$$

Thus, the equations of the diagonals are $$3x - 5y - 2 = 0$$ and $$3x + 5y - 7 = 0$$.