Question 68 of 70advanced🔧 ApplyLong Answer5 marks

If three lines whose equations are y=m1x+c1y = m_1x + c_1, y=m2x+c2y = m_2x + c_2 and y=m3x+c3y = m_3x + c_3 are concurrent, then show that m1(c2−c3)+m2(c3−c1)+m3(c1−c2)=0m_1(c_2 - c_3) + m_2(c_3 - c_1) + m_3(c_1 - c_2) = 0.

Correct Answer

Given the three lines: y=m1x+c1y = m_1x + c_1, y=m2x+c2y = m_2x + c_2, and y=m3x+c3y = m_3x + c_3. If they are concurrent, then the point of intersection of any two lines must satisfy the third line.

First, we find the intersection point of lines (1) and (2). Solving them simultaneously: subtract (2) from (1) to get (m1−m2)x=c2−c1(m_1 - m_2)x = c_2 - c_1, so x=c2−c1m1−m2x = \frac{c_2 - c_1}{m_1 - m_2}. Substituting back gives y=m1c2−m2c1m1−m2y = \frac{m_1c_2 - m_2c_1}{m_1 - m_2}. Thus, the intersection point is (c2−c1m1−m2,m1c2−m2c1m1−m2)\left( \frac{c_2 - c_1}{m_1 - m_2}, \frac{m_1c_2 - m_2c_1}{m_1 - m_2} \right).

For concurrency, this point must lie on line (3): y=m3x+c3y = m_3x + c_3. Substituting the coordinates:

m1c2−m2c1m1−m2=m3⋅c2−c1m1−m2+c3.\frac{m_1c_2 - m_2c_1}{m_1 - m_2} = m_3 \cdot \frac{c_2 - c_1}{m_1 - m_2} + c_3.

Multiplying both sides by (m1−m2)(m_1 - m_2) clears the denominator:

m1c2−m2c1=m3(c2−c1)+c3(m1−m2).m_1c_2 - m_2c_1 = m_3(c_2 - c_1) + c_3(m_1 - m_2).

Expanding the right side: m3c2−m3c1+c3m1−c3m2m_3c_2 - m_3c_1 + c_3m_1 - c_3m_2.

Bringing all terms to one side:

m1c2−m2c1−m3c2+m3c1−c3m1+c3m2=0.m_1c_2 - m_2c_1 - m_3c_2 + m_3c_1 - c_3m_1 + c_3m_2 = 0.

Rearranging terms:

m1c2−m1c3+m2c3−m2c1+m3c1−m3c2=0.m_1c_2 - m_1c_3 + m_2c_3 - m_2c_1 + m_3c_1 - m_3c_2 = 0.

Finally, factoring m1m_1, m2m_2, m3m_3:

m1(c2−c3)+m2(c3−c1)+m3(c1−c2)=0.m_1(c_2 - c_3) + m_2(c_3 - c_1) + m_3(c_1 - c_2) = 0.

Hence, the required condition is proved.

Exercise: Miscellaneous Exercise on Chapter 9 | Q: 9 | (Chapter: 22)
For More Understanding

Explanation

The solution follows the standard method for proving concurrency of three lines: find the intersection of two lines, then impose that it satisfies the third line. The intersection point formula for lines in slope-intercept form is derived from solving the equations, as shown in the textbook context. The algebraic manipulation leads to the desired expression.

Solution Steps

  1. Step 1: Find intersection of lines (1) and (2) to get x=c2−c1m1−m2x = \frac{c_2 - c_1}{m_1 - m_2} and y=m1c2−m2c1m1−m2y = \frac{m_1c_2 - m_2c_1}{m_1 - m_2}.

  2. Step 2: Substitute this point into line (3): y=m3x+c3y = m_3x + c_3.

  3. Step 3: Multiply by (m1−m2)(m_1 - m_2) to clear denominators.

  4. Step 4: Expand and rearrange all terms to one side.

  5. Step 5: Factor to obtain m1(c2−c3)+m2(c3−c1)+m3(c1−c2)=0m_1(c_2 - c_3) + m_2(c_3 - c_1) + m_3(c_1 - c_2) = 0.

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