Question

The points of intersection of three lines \(2X + 3Y - 5 = 0\), \(5X - 7Y + 2 = 0\) and \(9X - 5Y - 4 = 0\).

• A. form a triangle
• B. are on lines perpendicular to each other
• C. are on lines parallel to each other
• D. are coincident

Hint:

When solving system of linear equations, always check if the points of intersection are distinct to determine if they form a triangle.

Answer 1

The Correct Option is A

To check the relationship between the three lines, we first find the points of intersection. Let's solve the system of two equations at a time to find the points. Step 1: Solve the first and second equation \(2X + 3Y - 5 = 0\) and \(5X - 7Y + 2 = 0\). \[ 2X + 3Y = 5 \quad \text{(1)}
5X - 7Y = -2 \quad \text{(2)} \] Multiply (1) by 5 and (2) by 2 to eliminate \(X\): \[ 10X + 15Y = 25 \quad \text{(3)}
10X - 14Y = -4 \quad \text{(4)} \] Subtract (4) from (3): \[ (10X + 15Y) - (10X - 14Y) = 25 - (-4)
29Y = 29 \Rightarrow Y = 1 \] Substitute \(Y = 1\) into equation (1): \[ 2X + 3(1) = 5 \quad \Rightarrow 2X + 3 = 5 \quad \Rightarrow 2X = 2 \quad \Rightarrow X = 1 \] So, the point of intersection of the first two lines is \((1, 1)\). Step 2: Now solve the second and third equation \(5X - 7Y + 2 = 0\) and \(9X - 5Y - 4 = 0\). \[ 5X - 7Y = -2 \quad \text{(5)}
9X - 5Y = 4 \quad \text{(6)} \] Multiply (5) by 9 and (6) by 5: \[ 45X - 63Y = -18 \quad \text{(7)}
45X - 25Y = 20 \quad \text{(8)} \] Subtract (8) from (7): \[ (45X - 63Y) - (45X - 25Y) = -18 - 20
-38Y = -38 \Rightarrow Y = 1 \] Substitute \(Y = 1\) into equation (5): \[ 5X - 7(1) = -2 \quad \Rightarrow 5X - 7 = -2 \quad \Rightarrow 5X = 5 \quad \Rightarrow X = 1 \] So, the point of intersection of the second and third lines is also \((1, 1)\). Step 3: Finally, check if these lines are not coincident. Since the points of intersection of all three lines are the same, the lines are not coincident. They form a triangle. Thus, the answer is: a. form a triangle.