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4.4 Q-4

Question Statement

Find the condition that the lines y=m1x+c1y = m_1 x + c_1, y=m2x+c2y = m_2 x + c_2, and y=m3x+c3y = m_3 x + c_3 are concurrent.

That is, determine the relationship between the slopes and intercepts of these lines for them to intersect at a single point.


Background and Explanation

In geometry, three lines are said to be concurrent if they all meet at a single point. To solve for the condition of concurrency, we need to find the relationship between the constants m1,m2,m3m_1, m_2, m_3 (slopes) and c1,c2,c3c_1, c_2, c_3 (intercepts) of the three given lines.

The approach involves solving a system of equations for xx and yy, the coordinates of the intersection point. The lines are represented by their general equations, and we aim to derive a condition under which all three equations are satisfied simultaneously.


Solution

We are given three equations of straight lines:

  1. m1xβˆ’y+c1=0m_1 x - y + c_1 = 0
  2. m2xβˆ’y+c2=0m_2 x - y + c_2 = 0
  3. m3xβˆ’y+c3=0m_3 x - y + c_3 = 0

Step 1: Solve for the intersection of the first two lines

From the first two equations, we can solve for xx and yy.

Starting with the first and second equations:

m1xβˆ’y+c1=0(1)m_1 x - y + c_1 = 0 \tag{1} m2xβˆ’y+c2=0(2)m_2 x - y + c_2 = 0 \tag{2}

Subtract equation (2) from equation (1):

(m1xβˆ’y+c1)βˆ’(m2xβˆ’y+c2)=0(m_1 x - y + c_1) - (m_2 x - y + c_2) = 0

Simplifying:

(m1βˆ’m2)x+(c1βˆ’c2)=0(m_1 - m_2) x + (c_1 - c_2) = 0

Solving for xx:

x=c2βˆ’c1m1βˆ’m2(3)x = \frac{c_2 - c_1}{m_1 - m_2} \tag{3}

Step 2: Substitute into one of the original equations to find yy

Now substitute the value of xx from equation (3) into either of the original equations (let’s use equation (1)):

m1(c2βˆ’c1m1βˆ’m2)βˆ’y+c1=0m_1 \left( \frac{c_2 - c_1}{m_1 - m_2} \right) - y + c_1 = 0

Solving for yy:

y=m1(c2βˆ’c1)m1βˆ’m2+c1y = \frac{m_1 (c_2 - c_1)}{m_1 - m_2} + c_1

Thus, the coordinates of the intersection of the first two lines are:

x=c2βˆ’c1m1βˆ’m2,y=m1(c2βˆ’c1)m1βˆ’m2+c1x = \frac{c_2 - c_1}{m_1 - m_2}, \quad y = \frac{m_1 (c_2 - c_1)}{m_1 - m_2} + c_1

Step 3: Substitute these into the third equation

Now, substitute xx and yy into the third equation m3xβˆ’y+c3=0m_3 x - y + c_3 = 0:

m3(c2βˆ’c1m1βˆ’m2)βˆ’(m1(c2βˆ’c1)m1βˆ’m2+c1)+c3=0m_3 \left( \frac{c_2 - c_1}{m_1 - m_2} \right) - \left( \frac{m_1 (c_2 - c_1)}{m_1 - m_2} + c_1 \right) + c_3 = 0

Simplify the equation to find the condition for concurrency:

m3(c2βˆ’c1)βˆ’m2c1βˆ’m1c2+m2c3βˆ’m1c3=0m_3 \left( c_2 - c_1 \right) - m_2 c_1 - m_1 c_2 + m_2 c_3 - m_1 c_3 = 0

Rearranging terms:

m3(c2βˆ’c1)+m2(c3βˆ’c1)+m1(c1βˆ’c3)=0m_3 (c_2 - c_1) + m_2 (c_3 - c_1) + m_1 (c_1 - c_3) = 0

Step 4: Final condition for concurrency

Thus, the condition for the three lines to be concurrent is:

(m2βˆ’m1)(c3βˆ’c1)=(m3βˆ’m1)(c2βˆ’c1)(m_2 - m_1)(c_3 - c_1) = (m_3 - m_1)(c_2 - c_1)

Key Formulas or Methods Used

  • Equation of a line: y=mx+cy = mx + c
  • Solving a system of linear equations
  • Condition for three lines to be concurrent

Summary of Steps

  1. Write the equations of the three lines.
  2. Solve for xx and yy using the first two lines.
  3. Substitute xx and yy into the third line’s equation.
  4. Simplify to find the condition for concurrency:
(m2βˆ’m1)(c3βˆ’c1)=(m3βˆ’m1)(c2βˆ’c1) (m_2 - m_1)(c_3 - c_1) = (m_3 - m_1)(c_2 - c_1)