Question

If $3a-2b+5c=0$, then family of straight lines $ax+by+c=0$ are always concurrent at a point whose coordinate is $\left(\alpha ,\beta \right)$, then the values of $5\left(\alpha -\beta \right)$.

Answer 1

$\begin{array}{c}3a-2b+5c=0\to \left(i\right)\\ ax+by+c=0\to \left(ii\right)\end{array}$

Make the coefficient of C is 1. so divide by equation (i)

$\frac{3}{5}a-\frac{2}{5}b+c=0\to \left(iii\right)$

So, equation (ii) & (iii) are concurrent,

So, Compare the coefficient, then

$\begin{array}{c}x=\frac{3}{5},y=\frac{-2}{5}.So\left(\alpha ,\beta \right)=\left(\frac{3}{5},\frac{-2}{5}\right)\\ Find5\left(\alpha -\beta \right)\Rightarrow 5\left(\frac{3}{5}+\frac{2}{5}\right)\\ \Rightarrow 5\left(\frac{5}{5}\right)=5\end{array}$