1.

Show that the lines 3 – 4y + 5 = 0, lx – 8y + 5 = 0 and 4JC + 5y – 45 = 0 are concurrent. Find their point of concurrence.

Answer»

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).


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