If A(3, 6) and C(–1, 2) are two vertices of a rhombus ABCD, then find the equation of straight line that lies along the diagonal BD.

In a rhombus, the diagonals are perpendicular to each other and also bisect each other

Thus, the product of the slope of AC and BD will be –1 and BD will pass through the mid–point of AC

The points are given as A (3,6) and C ( –1, 2)

Slope of AC is

Thus, the slope of BD (m) = –1

The mid–point of AC is

Thus BD passes through the point (1,4)

Now using the slope point form, the equation of the line BD is:

(y –y₁) = m (x – x₁)

Here m = 1 and point is (1, 4)

⇒ y – 4 = (–1) (x –1)

⇒ y – 4 = –x + 1

⇒ x + y –5 = 0