TS EAMCET · Maths · Straight Lines
If \(3 x+6 y+2=0, x+y+1=0,2 x-y+3=0\) are three given lines then the point \(\left(\frac{-4}{3}, \frac{1}{3}\right)\) is
- A the orthocentre of the triangle formed by the lines
- B the point of concurrence of the lines
- C the circumcentre of the triangle formed by the lines
- D the incentre of the triangle formed by the lines
Answer & Solution
Correct Answer
(B) the point of concurrence of the lines
Step-by-step Solution
Detailed explanation
Here we are given that \[ \begin{aligned} & 3 x+6 y+2=0 \\ & x+y+1=0 \\ & 2 x-y+3=0 \end{aligned} \] \(\because \quad\) Here \(\left(\frac{-4}{3}, \frac{1}{3}\right)\) satisfied all of these three lines \(\Rightarrow\left(\frac{-4}{3}, \frac{1}{3}\right)\) is the concurrent…
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