Straight Lines - Distance between two parallel lines

Find the distance between the parallel lines $3x - 4y + 7 = 0$ and $3x - 4y + 5 = 0$.

Given lines are $3x - 4y + 7 = 0$ and $3x - 4y + 5 = 0$. Comparing with $Ax + By + C_1 = 0$ and $Ax + By + C_2 = 0$, we have: $A = 3$, $B = -4$, $C_1 = 7$, and $C_2 = 5$. Using the formula $d = \frac{|C_1 - C_2|}{\sqrt{A^2 + B^2}}$, we get: $d = \frac{|7 - 5|}{\sqrt{3^2 + (-4)^2}} = \frac{2}{\sqrt{9 + 16}} = \frac{2}{\sqrt{25}} = \frac{2}{5}$ units.

Identify the common coefficients $A$ and $B$ and the constants $C_1$ and $C_2$, then substitute them into the distance formula.

Find the distance between the lines $15x + 8y - 34 = 0$ and $30x + 16y + 62 = 0$.

The first line is $15x + 8y - 34 = 0$. The second line is $30x + 16y + 62 = 0$. Dividing the second equation by $2$ to make coefficients of $x$ and $y$ same: $15x + 8y + 31 = 0$. Now, $A = 15$, $B = 8$, $C_1 = -34$, and $C_2 = 31$. Using the formula: $d = \frac{|-34 - 31|}{\sqrt{15^2 + 8^2}} = \frac{|-65|}{\sqrt{225 + 64}} = \frac{65}{\sqrt{289}} = \frac{65}{17}$ units.

Before using the formula, the coefficients of $x$ and $y$ must be identical. We divided the second equation by $2$ to match the first equation.

Calculate the difference between the constant terms of the lines $L_1: 5x + 12y - 20 = 0$ and $L_2: 5x + 12y + 6 = 0$ using vertical subtraction for the numerator calculation.

Here $C_1 = -20$ and $C_2 = 6$. To find $|C_1 - C_2|$, we perform: $-20 - 6 = -26$. In vertical form for absolute values: $\begin{array}{r} 20 \\ + 6 \\ \hline 26 \end{array}$ Thus $|-26| = 26$. The denominator is $\sqrt{5^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13$. The distance is $d = \frac{26}{13} = 2$ units.

The distance is calculated by finding the absolute difference of the constants and dividing by the magnitude of the normal vector.

1. Identify the slope $m$ and intercepts $c_1, c_2$: $m = 2$, $c_1 = -3$, $c_2 = 7$.
2. Use the formula $d = \frac{|c_1 - c_2|}{\sqrt{1 + m^2}}$: $d = \frac{|-3 - 7|}{\sqrt{1 + 2^2}}$ $d = \frac{|-10|}{\sqrt{1 + 4}}$ $d = \frac{10}{\sqrt{5}}$
3. Rationalize the denominator: $d = \frac{10\sqrt{5}}{5} = 2\sqrt{5}$ units.

Since the lines are in $y = mx + c$ form with the same slope $m=2$, we directly subtract the y-intercepts and divide by the magnitude of the normal vector component.

1. Normalize the first equation to match the coefficients of the second: Divide $8x + 6y - 12 = 0$ by 2: $4x + 3y - 6 = 0$
2. Now $A = 4, B = 3, C_1 = -6, C_2 = 4$.
3. Calculate the numerator $|C_1 - C_2|$: $\begin{array}{r} -6 \\ -4 \\ \hline -10 \end{array}$ Absolute value is $10$.
4. Calculate the denominator $\sqrt{A^2 + B^2}$: $\sqrt{4^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5$.
5. Distance $d = \frac{10}{5} = 2$ units.

To use the general distance formula, the coefficients of x and y must be identical in both equations. Dividing the first equation by 2 makes $A=4$ and $B=3$ for both.