ExamBro
ExamBro
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

  1. A \(\left(\frac{\alpha}{\beta}, \frac{\beta}{\gamma}\right)\)
  2. B \(\left(\frac{\gamma}{\alpha}, \frac{\beta}{\alpha}\right)\)
  3. C \(\left(\frac{\alpha}{\gamma}, \frac{\gamma}{\beta}\right)\)
  4. D \(\left(\frac{\alpha}{\gamma}, \frac{\beta}{\gamma}\right)\)
Verified Solution

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 \[ c=-a x-b y \]…