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Let $\alpha, \beta, \gamma$ be three non-zero real constants and $a, b, c$ be three arbitrary real numbers which satisfy $\alpha a+\beta b+\gamma c$ $=0$. Then the point of concurrence of the family of lines $a x+b y+c=0$ is
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Verified Answer
The correct answer is:
$\left(\frac{\alpha}{\gamma}, \frac{\beta}{\gamma}\right)$
Given $\alpha \mathrm{a}+\beta \mathrm{b}+\gamma \mathrm{c}=0$
$$
\Rightarrow \mathrm{c}=\left(\frac{-\alpha}{\gamma}\right) \mathrm{a}-\left(\frac{\beta}{\gamma}\right) \mathrm{b}
$$
how, family of lines is given by an equation
$$
c=-a x-b y
$$
$$
\Rightarrow \mathrm{x}=\frac{\alpha}{\gamma}, \mathrm{y}=\frac{\beta}{\gamma}
$$
Thus $\left(\frac{\alpha}{\gamma}, \frac{\beta}{\gamma}\right)$ is the point of concurrency.
$$
\Rightarrow \mathrm{c}=\left(\frac{-\alpha}{\gamma}\right) \mathrm{a}-\left(\frac{\beta}{\gamma}\right) \mathrm{b}
$$
how, family of lines is given by an equation
$$
c=-a x-b y
$$
$$
\Rightarrow \mathrm{x}=\frac{\alpha}{\gamma}, \mathrm{y}=\frac{\beta}{\gamma}
$$
Thus $\left(\frac{\alpha}{\gamma}, \frac{\beta}{\gamma}\right)$ is the point of concurrency.
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