Question 1: Determine the distance between the following pair of parallel lines :

(i) $\displaystyle 4x-3y-9=0 \text{ and } 4x-3y-24=0$

(ii) $\displaystyle 8x +15y = 34 = 0 \text{ and } 8x +15y + 31 = 0$

(iii) $\displaystyle y=mx+c \text{ and } y=mx+d$

(iv) $\displaystyle 4x + 3y -11 = 0 \text{ and } 8x + 6y =15$

(i) Given parallel lines: $\displaystyle 4x-3y-9=0 \text{ and } 4x-3y-24=0$

Let $\displaystyle d$ be the distance between the given parallel lines.

$\displaystyle \therefore d = \Big| \frac{-9+24}{\sqrt{(4)^2 + ( -3)^2}} \Big| = \frac{15}{5} = 3 \text{ units }$

(ii) Given parallel lines: $\displaystyle 8x +15y = 34 = 0 \text{ and } 8x +15y + 31 = 0$

Let d be the distance between the given parallel lines.

$\displaystyle \therefore d = \Big| \frac{-34-31}{\sqrt{(8)^2 + ( 15)^2}} \Big| = \frac{65}{17} \text{ units }$

(iii) Given parallel lines: $\displaystyle y=mx+c \text{ and } y=mx+d$

Let d be the distance between the given parallel lines.

$\displaystyle \therefore d = \Big| \frac{c-d}{\sqrt{m^2+1}} \Big| = \text{ units }$

(iv) Given parallel lines: $\displaystyle 4x + 3y -11 = 0 \text{ and } 8x + 6y =15 \Rightarrow 4x + 3y - \frac{15}{2} = 0$

Let d be the distance between the given parallel lines.

$\displaystyle \therefore d = \Big| \frac{-11+\frac{15}{2}}{\sqrt{(4)^2 + ( 3)^2}} \Big| = \frac{7}{10} \text{ units }$

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Question 2: The equations of two sides of a square are $\displaystyle 5x -12y - 65 = 0 \text{ and } 5x -12y + 26 = 0$ Find the area of the square.

Given parallel lines: $\displaystyle 5x -12y - 65 = 0 \text{ and } 5x -12y + 26 = 0$

Let d be the distance between the given parallel lines.

$\displaystyle \therefore d = \Big| \frac{-65-26}{\sqrt{(15)^2 + ( -12)^2}} \Big| = \frac{91}{13} = 7$units

Therefore the area of the square $\displaystyle = 7 \times 7 = 49$ sq. units.

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Question 3: Find the equation of two straight lines which are parallel to $\displaystyle x+7y +2=0$ and at unit distance from the point $\displaystyle (1, - 1)$.

The given line: $\displaystyle x+7y +2=0$ … … … … … i)

Equation of a like parallel to line i) is $\displaystyle x+7y + \lambda =0$ … … … … … ii)

Line ii) is at a unit distance from point $\displaystyle ( 1, -1)$ . Therefore,

$\displaystyle \Big| \frac{1-7+\lambda}{\sqrt{(1)^2 + ( 7)^2}} \Big| = 1$

$\displaystyle \Rightarrow \lambda - 6 = \pm 5\sqrt{2}$

$\displaystyle \Rightarrow \lambda = 6 + 5\sqrt{2}$ or $\displaystyle 6 - 5\sqrt{2}$

Hence the required lines are $\displaystyle x+7y + 6 + 5\sqrt{2} =0 \text{ and } x+7y + 6 - 5\sqrt{2} =0$

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Question 4: Prove that the lines $\displaystyle 2x+3y=19 \text{ and } 2x+3y+7=0$ are equidistant from the line $\displaystyle 2x+3y=6$.

Let $\displaystyle d_1$ is the distance between $\displaystyle 2x+3y=19 \text{ and } 2x+3y=6$

$\displaystyle \therefore d_1 = \Big| \frac{19-(-6)}{\sqrt{2^2 + 3^2}} \Big| = \Big| \frac{-13}{\sqrt{13}} \Big| = \sqrt{13}$

Let $\displaystyle d_2$ is the distance between $\displaystyle 2x+3y+7=0 \text{ and } 2x+3y=6$

$\displaystyle \therefore d_2 = \Big| \frac{7-(-6)}{\sqrt{2^2 + 3^2}} \Big| = \Big| \frac{13}{\sqrt{13}} \Big| = \sqrt{13}$

Therefore the lines $\displaystyle 2x+3y=19 \text{ and } 2x+3y+7=0$ are equidistant from the line $\displaystyle 2x+3y=6$ as $\displaystyle d_1 = d_2$

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Question 5: Find the equation of the line mid-way between the parallel lines $\displaystyle 9x + 6y -7 = 0 \text{ and } 3x+2y+6=0$.

Given lines:

$\displaystyle 9x + 6y -7 = 0 \Rightarrow 3x + 2y - \frac{7}{3} = 0$ … … … … … i)

$\displaystyle 3x+2y+6=0$ … … … … … ii)

Let the equation of the line midway between the two lines be

$\displaystyle 3x+2y+ \lambda =0$ … … … … … iii)

Distance between [i) and iii)] and [ii) and iii)] is equal.

$\displaystyle \Big| \frac{-\frac{7}{3}-\lambda}{\sqrt{3^2 + 2^2}} \Big| = \Big| \frac{6-\lambda}{\sqrt{3^2 + 2^2}} \Big|$

$\displaystyle \Big|-( \lambda + \frac{7}{3} )\Big| = \Big|6 - \lambda\Big|$

$\displaystyle \Rightarrow \lambda = \frac{11}{6}$

Equation of the required line is

$\displaystyle 3x+2y + \frac{11}{6} = 0$

$\displaystyle \Rightarrow 18x+12y + 11 = 0$

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Question 6: Find the ratio in which the line $\displaystyle 3x + 4y + 2 = 0$ divides the distance between the lines $\displaystyle 3x+4y+5 = 0 \text{ and } 3x+4y -5 = 0$.

The three lines are parallel lines are the slope of all the lines is same $\displaystyle \Big( \text{ slope } = \frac{-3}{4} \Big)$

Let $\displaystyle d_1$ is the distance between $\displaystyle 3x + 4y + 2 = 0 \text{ and } 3x+4y+5 = 0$

$\displaystyle \therefore d_1 = \Big| \frac{2-5}{\sqrt{3^2 + 4^2}} \Big| = \Big| \frac{3}{\sqrt{25}} \Big| = \frac{3}{5}$

Let $\displaystyle d_2$ is the distance between $\displaystyle 3x + 4y + 2 = 0 \text{ and } 3x+4y -5 = 0$

$\displaystyle \therefore d_2 = \Big| \frac{2+5}{\sqrt{3^2 + 4^2}} \Big| = \Big| \frac{7}{\sqrt{25}} \Big| = \frac{7}{5}$

$\displaystyle \text{Hence the ratio of } \frac{d_1}{d_2} = \frac{\frac{3}{5}}{\frac{7}{5}} = \frac{3}{7}$

Therefore the line $\displaystyle 3x + 4y + 2 = 0$ divides the distance between the lines $\displaystyle 3x+4y+5 = 0 \text{ and } 3x+4y -5 = 0$ in the ratio of $\displaystyle 3:7$