**Problem**: If any one pair among the straight lines

ax + by = a + b, bx –(a + b)y = – a, (a + b)x –ay = b

intersect, then show that the three straight lines are concurrent.

**Solution**: Three lines are concurrent if each of them is linear combination of other two & they are not parallel.

Now given one pair intersect that is they are not parallel.

Now

ax + by = a + b … (i)

bx – (a + b)y = – a …(ii)

(a +b)x –ay = b …(iii)

(i) + (ii) – (iii) = 0

So they are concurrent.

*Related*

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