Correct option is (A) I
Given lines are 2x + 3y – 5 = 0 _________(1)
and 3x – 4y + 1 = 0 _________(2)
Multiply equation (1) by 4 & equation (2) by 3, we get
8x + 12y - 20 = 0 _________(3)
and 9x - 12y + 3 = 0 _________(4)
By adding equations (3) & (4), we obtain
17x - 20 + 3 = 0
\(\Rightarrow\) 17x = 20 - 3 = 17
\(\Rightarrow\) x = \(\frac{17}{17}\) = 1
Then from (1), we obtain
2 + 3y - 5 = 0
\(\Rightarrow\) 3y = 5 - 2 = 3
\(\Rightarrow\) y = \(\frac33\) = 1
Hence, point of intersection of both given lines is (1, 1) which lies in first quadrant.
