Skip to content
🚨 This site is a work in progress. Exciting updates are coming soon!

4.4 Q-8

Question Statement

We are given the following three lines:

4xβˆ’3yβˆ’8=0(1)4x - 3y - 8 = 0 \tag{1} 3xβˆ’4yβˆ’6=0(2)3x - 4y - 6 = 0 \tag{2} xβˆ’yβˆ’x=0(3)x - y - x = 0 \tag{3}

We need to check if these lines are concurrent (i.e., if they intersect at a single point). If they are, we also need to find the point of intersection.


Background and Explanation

To determine if three lines are concurrent, we need to find if their system of equations has a unique solution. This can be achieved by calculating the determinant of the coefficient matrix of the system. If the determinant is zero, the lines are concurrent. We also need to solve the system of equations to find the point of intersection.


Solution

Step 1: Calculate the Determinant

The three lines are given by the equations (1), (2), and (3). The system of equations can be written as:

4βˆ’3βˆ’83βˆ’4βˆ’61βˆ’1βˆ’2\begin{array}{ccc} 4 & -3 & -8 3 & -4 & -6 1 & -1 & -2 \end{array}

To check if the lines are concurrent, we need to calculate the determinant of the coefficient matrix:

Determinant=∣4βˆ’3βˆ’83βˆ’4βˆ’61βˆ’1βˆ’2∣\text{Determinant} = \left|\begin{array}{ccc} 4 & -3 & -8 3 & -4 & -6 1 & -1 & -2 \end{array}\right|

Expanding this determinant:

=4((βˆ’4)(βˆ’2)βˆ’(βˆ’6)(βˆ’1))βˆ’(βˆ’3)((3)(βˆ’2)βˆ’(βˆ’6)(1))+(βˆ’8)((3)(βˆ’1)βˆ’(βˆ’4)(1))= 4\left( (-4)(-2) - (-6)(-1) \right) - (-3)\left( (3)(-2) - (-6)(1) \right) + (-8)\left( (3)(-1) - (-4)(1) \right) =4(8βˆ’6)βˆ’(βˆ’3)(βˆ’6+6)βˆ’8(βˆ’3+4)= 4(8 - 6) - (-3)(-6 + 6) - 8(-3 + 4) =4(2)βˆ’0βˆ’8(1)= 4(2) - 0 - 8(1) =8βˆ’0βˆ’8=0= 8 - 0 - 8 = 0

Since the determinant is zero, the lines are concurrent.


Step 2: Find the Point of Intersection

To find the point of intersection, we can solve the system of two equations (1) and (2) for xx and yy.

Start by solving equations (1) and (2) simultaneously. From the equations:

4xβˆ’3yβˆ’8=0(1)4x - 3y - 8 = 0 \tag{1} 3xβˆ’4yβˆ’6=0(2)3x - 4y - 6 = 0 \tag{2}

We use the method of elimination or substitution to solve for xx and yy. However, for simplicity, we directly substitute into the formula to find the point of intersection.

Solving the system of equations yields:

xβˆ’14=y0=1βˆ’7\frac{x}{-14} = \frac{y}{0} = \frac{1}{-7}

From this, we get:

x=βˆ’14βˆ’7=2,y=07=0x = \frac{-14}{-7} = 2, \quad y = \frac{0}{7} = 0

Thus, the point of concurrency is (2, 0).


Key Formulas or Methods Used

  1. Determinant of a 3x3 Matrix: To check if the lines are concurrent, we calculate the determinant of the matrix of their coefficients. A determinant of 0 indicates that the lines intersect at a common point.

  2. Simultaneous Equations: To find the point of intersection, we solve the system of equations by substitution or elimination.


Summary of Steps

  1. Write the system of equations for the three lines.
  2. Calculate the determinant of the coefficient matrix.
  3. If the determinant is 0, the lines are concurrent.
  4. Solve the system of equations to find the point of intersection.
  5. The point of concurrency is (2, 0).