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

(i) 4x-3y-9=0 and 4x-3y-24=0      (ii) 8x +15y = 34 = 0 and 8x +15y + 31 = 0

(iii) y=mx+c and y=mx+d      (iv) 4x + 3y -11 = 0 and 8x + 6y =15

Answer:

(i)      Given parallel lines: 4x-3y-9=0 and 4x-3y-24=0

Let d be the distance between the given parallel lines.

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

(ii)     Given parallel lines: 8x +15y = 34 = 0 and 8x +15y + 31 = 0

Let d be the distance between the given parallel lines.

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

(iii)   Given parallel lines: y=mx+c and y=mx+d

Let d be the distance between the given parallel lines.

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

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

Let d be the distance between the given parallel lines.

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

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

Answer:

Given parallel lines: 5x -12y - 65 = 0 and 5x -12y + 26 = 0

Let d be the distance between the given parallel lines.

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

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

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

Answer:

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

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

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

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

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

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

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

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Question 4:  Prove that the lines 2x+3y=19 and 2x+3y+7=0 are equidistant from the line 2x+3y=6 .

Answer:

Let d_1 is the distance between 2x+3y=19 and 2x+3y=6

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

Let d_2 is the distance between 2x+3y+7=0  and 2x+3y=6

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

Therefore the lines 2x+3y=19 and 2x+3y+7=0 are equidistant from the line 2x+3y=6 as d_1 = d_2

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

Answer:

Given lines:

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

3x+2y+6=0      … … … … … ii)

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

3x+2y+ \lambda =0      … … … … … iii)

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

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

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

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

Equation of the required line is

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

\Rightarrow 18x+12y + 11 = 0

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

Answer:

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

Let d_1 is the distance between 3x + 4y + 2 = 0 and 3x+4y+5 = 0

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

Let d_2 is the distance between 3x + 4y + 2 = 0 and 3x+4y -5 = 0

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

Hence the ratio if  \frac{d_1}{d_2} = \frac{\frac{3}{5}}{\frac{7}{5}} = \frac{3}{7}

Therefore the line 3x + 4y + 2 = 0 divides the distance between the lines 3x+4y+5 = 0 and 3x+4y -5 = 0 in the ratio of  3:7