Question

The points of intersection of the lines 2x+3y-5=0 and 3x-4y+1=0 lies in which quadrant ?

• A. III
• B. IV
• C. I
• D. II

Answer 1

The Correct Option is C

To solve the problem, we need to find the point of intersection of the two lines: \( 2x + 3y - 5 = 0 \) and \( 3x - 4y + 1 = 0 \), and determine in which quadrant this point lies.

1. Write the given equations:
Equation (1): \( 2x + 3y = 5 \)
Equation (2): \( 3x - 4y = -1 \)

2. Solve the system of equations:
We'll use the method of elimination. First, multiply both equations to make the coefficients of \(x\) equal:

Multiply Equation (1) by 3: \( 6x + 9y = 15 \)
Multiply Equation (2) by 2: \( 6x - 8y = -2 \)

3. Subtract the equations:
\( (6x + 9y) - (6x - 8y) = 15 - (-2) \)
\( 6x + 9y - 6x + 8y = 17 \)
\( 17y = 17 \Rightarrow y = 1 \)

4. Substitute \(y = 1\) back to find \(x\):
From Equation (1):
\( 2x + 3(1) = 5 \Rightarrow 2x + 3 = 5 \Rightarrow 2x = 2 \Rightarrow x = 1 \)

5. Coordinates of the point of intersection:
The intersection point is \( (1, 1) \)

6. Determine the quadrant:
Since both \(x\) and \(y\) are positive, the point lies in the first quadrant.

Final Answer:
The point of intersection lies in Quadrant I.