Question

Consider the family of lines, 5x + 3y − 2 + K1 (3x − y − 4) = 0 and x − y + 1 + K2(2x − y − 2)=0. Find the equation of the line belonging to both the families without determining their vertices.

Answer 1

Problem.

Consider the family of lines, $5x + 3y - 2 + K1(3x - y - 4) = 0$ and $x - y + 1 + K2(2x - y - 2) = 0$. Find the equation of the line belonging to both the families without determining their vertices.

Solution

Reduce the given families of lines to the standard form $y = ax + b$:

$$\left\{ \begin{array}{l} 5x + 3y - 2 + k_1(3x - y - 4) = 0 \\ x - y + 1 + k_2(2x - y - 2) = 0 \end{array} \right.$$ $$\left\{ \begin{array}{l} y(3 - k_1) = (-5 - 3k_1)x + 2 + 4k_1 \\ y(1 + k_2) = (1 + 2k_2)x + 1 - 2k_2 \end{array} \right.$$ $$\left\{ \begin{array}{l} y = -\frac{5 + 3k_1}{3 - k_1}x + \frac{2 + 4k_1}{3 - k_1} \\ y = \frac{1 + 2k_2}{1 + k_2}x + \frac{1 - 2k_2}{1 + k_2} \end{array} \right.$$

Both of these equations will describe the same line, if

$$\left\{ \begin{array}{l} \frac{5 + 3k_1}{k_1 - 3} = \frac{1 + 2k_2}{k_2 + 1} \\ \frac{2 + 4k_1}{3 - k_1} = \frac{1 - 2k_2}{k_2 + 1} \end{array} \right.$$

We have

$$k_1 \neq 3, \quad k_2 \neq -1,$$ $$\left\{ \begin{array}{l} (5 + 3k_1)(k_2 + 1) = (1 + 2k_2)(k_1 - 3) \\ (2 + 4k_1)(k_2 + 1) = (1 - 2k_2)(3 - k_1) \end{array} \right.$$ $$\left\{ \begin{array}{l} 5 + 5k_2 + 3k_1 + 3k_1k_2 = k_1 + 2k_1k_2 - 3 - 6k_2 \\ 2k_2 + 4k_1k_2 + 2 + 4k_1 = 3 - 6k_2 - k_1 + 2k_1k_2 \end{array} \right.$$ $$\left\{ \begin{array}{l} k_1k_2 + 11k_2 + 2k_1 = -8 \\ 2k_1k_2 + 8k_2 + 5k_1 = -17 \end{array} \right.$$

Multiply the first equation by $-2$ and add with the second one

$$-14k_2 + k_1 = 17$$ $$k_1 = 14k_2 + 17$$

Substituting this value in the first equation

$$(14k_2 + 17)k_2 + 11k_2 + 2(14k_2 + 17) = -8$$ $$14k_2^2 + 17k_2 + 11k_2 + 28k_2 = -8$$ $$14k_2^2 + 56k_2 + 42 = 0$$ $$\begin{array}{l} k_{2}^{2} + 4k_{2} + 3 = 0 \\ (k_{2})_{1} = -1 \\ (k_{2})_{2} = -3 \\ \end{array}$$

The value $k_{2} = -1$, doesn't satisfy (*).

If $k_{2} = -3$ then

$$\begin{array}{l} k_{1} = 14(-3) + 17 = -25 \\ a = \frac{1 + 2k_{2}}{1 + k_{2}} = \frac{1 + 2(-3)}{(-3) + 1} = \frac{5}{2} \\ b = \frac{1 - 2(-3)}{1 + (-3)} = -\frac{7}{2} \\ \end{array}$$

Hence, the needed line is

$$\begin{array}{l} y = \frac{5}{2}x - \frac{7}{2} \\ 5x - 2y - 7 = 0 \\ \end{array}$$

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