Question

The points A(1, 3) and C(6, 8) are two opposite vertices of a square ABCD. Find the equation of the diagonal BD.

[Hint. $$BD$$ is the right-bisector of $$AC$$.]

Answer 1

In a square, the diagonals are perpendicular bisectors of each other. Therefore, diagonal $$\mathrm{BD}$$ is the perpendicular bisector of diagonal $$\mathrm{AC}$$.

Find the mid-point of $$\mathrm{AC}$$, which lies on $$\mathrm{BD}$$:

$$\mathrm{M} = \left(\frac{1 + 6}{2}, \frac{3 + 8}{2}\right)$$

$$= \left(\frac{7}{2}, \frac{11}{2}\right)$$

Slope of $$\mathrm{AC}$$: 

$$m_{\mathrm{AC}} = \frac{8 - 3}{6 - 1}$$

$$= \frac{5}{5}$$

$$= 1$$

Since $$\mathrm{BD} \perp \mathrm{AC}$$, the slope of $$\mathrm{BD}$$ is: 

$$m = -\frac{1}{m_{\mathrm{AC}}}$$

$$= -\frac{1}{1}$$

$$= -1$$

Equation of $$\mathrm{BD}$$ passing through $$\left(\frac{7}{2}, \frac{11}{2}\right)$$:

$$y - \frac{11}{2} = -1\left(x - \frac{7}{2}\right)$$

$$y - \frac{11}{2} = -x + \frac{7}{2}$$

$$x + y = \frac{11}{2} + \frac{7}{2}$$

$$x + y = \frac{18}{2} = 9$$

$$x + y - 9 = 0$$