If $4a^2+9b^2-c^2+12ab=0$,the family of straight lines $ax+by+c=0$ is concurrent at which point?
How to solve such problems.Hints please!
2026-04-12 06:41:44.1775976104
If $4a^2+9b^2-c^2+12ab=0$,the family of straight lines ax+by+c=0 is concurrent at which point?
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HINT
What you try to do is complete the square to simplify the expression. Recall that $$(x+z)^2 = x^2 + 2xz + z^2$$ and try to model the first terms in the expression after the pattern on the RHS. We have just fromm those terms: $$ 4a^2+9b^2 + 12ab = (2a)^2 + (3b)^2 + 2 \cdot(2a) \cdot (3b) = (2a+3b)^2. $$
Therefore, $$ 4a^2+9b^2 + 12ab -c^2 = 0 \text{ implies } (2a+3b)^2-c^2 = 0, $$ and we have $$c^2 = (2a+3b)^2.$$