The point of intersection of the lines 2x + 3y – 5 = 0 and 3x – 4y

1.	The point of intersection of the lines 2x + 3y – 5 = 0 and 3x – 4y + 1=0 lies in which quadrant ? A) I B) II C) III D) IV
Answer» 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. Correct option is A) I

The point of intersection of the lines 2x + 3y – 5 = 0 and 3x – 4y + 1=0 lies in which quadrant ? A) I B) II C) III D) IV

Answer»

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.

Correct option is A) I