Show that lines 3x − 4y + 5 = 0, 7x − 8y + 5 = 0, and 4x + 5y − 45 = 0 are concurrent. Find their point of concurrence

#### Solution

The number of lines intersecting at a point are called concurrent lines and their point of intersection is called the point of concurrence.

Equations of the given lines are

3x − 4y + 5 = 0 ...(i)

7x − 8y + 5 = 0 ...(ii)

4x + 5y − 45 = 0 ...(iii)

By (i) x 2 − (ii), we get

− x + 5 = 0

∴ x = 5

Substituting x = 5 in (i), we get

3(5) − 4y + 5 = 0

∴ −4y = − 20

∴ y = 5

∴ The point of intersection of lines (i) and (ii) is given by (5, 5).

Substituting x = 5 and y = 5 in L.H.S. of (iii), we get

L.H.S. = 4(5) + 5(5) − 45

= 20 + 25 − 45

= 0

= R.H.S.

∴ Line (iii) also passes through (5, 5).

Hence, the given three lines are concurrent and the point of concurrence is (5, 5).