Let alpha, beta, gamma be three non-zero real | vedclass

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(D) Given the condition $\alpha a + \beta b + \gamma c = 0$. Since $\gamma 
eq 0$,we can write $c = -\frac{\alpha}{\gamma} a - \frac{\beta}{\gamma} b

Vedclass · Accepted Answer

$\left(\frac{\alpha}{\gamma}, \frac{\beta}{\gamma}\right)$

Vedclass · Answer

(D) Given the condition $\alpha a + \beta b + \gamma c = 0$. Since $\gamma 
eq 0$,we can write $c = -\frac{\alpha}{\gamma} a - \frac{\beta}{\gamma} b$. Substitute this into the equation of the family of lines $ax + by + c = 0$: $ax + by + (-\frac{\alpha}{\gamma} a - \frac{\beta}{\gamma} b) = 0$. Rearranging the terms,we get: $a(x - \frac{\alpha}{\gamma}) + b(y - \frac{\beta}{\gamma}) = 0$. For this equation to hold for all arbitrary $a$ and $b$,the coefficients of $a$ and $b$ must be zero independently. Therefore,$x - \frac{\alpha}{\gamma} = 0 \Rightarrow x = \frac{\alpha}{\gamma}$ and $y - \frac{\beta}{\gamma} = 0 \Rightarrow y = \frac{\beta}{\gamma}$. Thus,the point of concurrence is $\left(\frac{\alpha}{\gamma}, \frac{\beta}{\gamma}ight)$.

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 $ax + by + c = 0$ is

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