TS EAMCET · Maths · Straight Lines
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
- A \(\left(\frac{\alpha}{\beta}, \frac{\beta}{\gamma}\right)\)
- B \(\left(\frac{\gamma}{\alpha}, \frac{\beta}{\alpha}\right)\)
- C \(\left(\frac{\alpha}{\gamma}, \frac{\gamma}{\beta}\right)\)
- D \(\left(\frac{\alpha}{\gamma}, \frac{\beta}{\gamma}\right)\)
Answer & Solution
Correct Answer
(D) \(\left(\frac{\alpha}{\gamma}, \frac{\beta}{\gamma}\right)\)
Step-by-step Solution
Detailed explanation
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 \[ c=-a x-b y \]…
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