Note: The equation of a straight line passing through \displaystyle (x_1, y_1) \text{ and making an angle } \theta with the positive x-axis is given by \displaystyle \frac{x-x_1}{\cos \theta} + \frac{y-y_1}{\sin \theta} = r , where \displaystyle r is the distance of the point \displaystyle (x, y) on the line from the point \displaystyle (x_1, y_1) . 

The coordinate of any point on the line at a distance \displaystyle r from the point \displaystyle ( x_1, y_1) are \displaystyle ( x_1 \pm r \cos \theta, y_1 \pm r \sin \theta) .

Question 1: A line passes through a point \displaystyle A(1,2) and makes an angle of \displaystyle 60^{\circ} with the x-axis and intersects the line \displaystyle x + y =6 at the point \displaystyle P . Find \displaystyle AP .

Answer:

\displaystyle \text{Given } (x_1, y_1) = A ( 1, 2) \text{ and } \theta = 60^{\circ}

\displaystyle \text{The equation passing through } A ( 1, 2) \text{ and making an angle } 60^{\circ} \text{ is }  

 \displaystyle \frac{x-1}{\cos 60^{\circ}} = \frac{y-2}{\sin 60^{\circ}} = r

 \displaystyle \Rightarrow \frac{x-1}{(1/2) } = \frac{y-2}{ (\sqrt{3}/2)} = r … … … … … i)

From equation i)

\displaystyle \text{The coordinates of any point } P(x, y) \text{ on this line are } \Big( 1+ \frac{r}{2} , 2+ \frac{\sqrt{3}}{2} r \Big)

\displaystyle \text{Since } P \text{ lies on } x+y = 6 , \text{ substituting we get }  

 \displaystyle \Big( 1+ \frac{r}{2} \Big) + \Big( 2+ \frac{\sqrt{3}}{2} \Big) r = 6

 \displaystyle \Rightarrow \frac{r}{2} + \frac{\sqrt{3}}{2} r = 3

 \displaystyle \Rightarrow ( 1 + \sqrt{3})r = 6

 \displaystyle \Rightarrow r = \frac{6}{\sqrt{3}+1} \times \frac{\sqrt{3}-1}{\sqrt{3}-1}  

 \displaystyle \Rightarrow r = 3(\sqrt{3}-1)

 \displaystyle \therefore AP = 3(\sqrt{3}-1)

 \displaystyle \\

Question 2: If the straight line through the point \displaystyle P(3,4) makes an angle \displaystyle \pi / 6 with the x-axis and meets the line \displaystyle 12x + 5y + 10 = 0 at \displaystyle Q , find the length \displaystyle PQ .

Answer:

\displaystyle \text{Given } (x_1, y_1) = P ( 3,4) \text{ and } \theta = 30^{\circ}

\displaystyle \text{The equation passing through } P ( 3,4) \text{ and making an angle } 30^{\circ} \text{ is }  

 \displaystyle \frac{x-3}{\cos 30^{\circ}} = \frac{y-4}{\sin 30^{\circ}} = r

 \displaystyle \Rightarrow \frac{x-3}{(\sqrt{3}/2) } = \frac{y-4}{ (1/2)} = r … … … … … i)

From equation i)

\displaystyle \text{The coordinates of any point } Q(x, y) \text{ on this line are } \Big( 3+ \frac{\sqrt{3}r}{2} , 4+ \frac{r}{2} \Big)

\displaystyle \text{Since } Q \text{ lies on } 12x+5y +10=0 , \text{ substituting we get }  

 \displaystyle 12 \Big( 3+ \frac{\sqrt{3}r}{2} \Big) + 5\Big( 4+ \frac{r}{2} \Big) +10 = 0

 \displaystyle \Rightarrow 36 + 6\sqrt{3}r +20+ \frac{5}{2} r +10 = 0

 \displaystyle \Rightarrow ( 6\sqrt{3}+ \frac{5}{2} )r = -66

 \displaystyle \Rightarrow r = \frac{-66 \times 2}{12\sqrt{3}+5}  

 \displaystyle \Rightarrow r = \frac{-132}{12\sqrt{3}+5}  

 \displaystyle \therefore PQ = |r| = \frac{132}{12\sqrt{3}+5}  

 \displaystyle \\

Question 3: A straight line drawn through the point, \displaystyle A (2, 1) making an angle \displaystyle \pi / 4 with positive x-axis intersects another line \displaystyle x + 2y + 1 = 0 in the point \displaystyle B . Find length \displaystyle AB .

Answer:

\displaystyle \text{Given } (x_1, y_1) = A ( 2,1) \text{ and } \theta = 45^{\circ}

\displaystyle \text{The equation passing through } A ( 2,1) \text{ and making an angle } 45^{\circ} \text{ is }  

 \displaystyle \frac{x-2}{\cos 45^{\circ}} = \frac{y-1}{\sin 45^{\circ}} = r

 \displaystyle \Rightarrow \frac{x-2}{(1/\sqrt{2}) } = \frac{y-1}{ (1/\sqrt{2})} = r … … … … … i)

From equation i)

\displaystyle \text{The coordinates of any point } B(x, y) \text{ on this line are } \Big( 2+ \frac{r}{\sqrt{2}} , 1+ \frac{r}{\sqrt{2}} \Big)

\displaystyle \text{Since } B \text{ lies on } x+2y+1=0 , \text{ substituting we get }  

 \displaystyle \Big( 2+ \frac{r}{\sqrt{2}} \Big) + 2 \Big( 1+ \frac{r}{\sqrt{2}} \Big) +1 = 0

 \displaystyle \Rightarrow \frac{3}{\sqrt{2}} r + 5=0

 \displaystyle \Rightarrow r = \frac{-5\sqrt{2}}{3}  

 \displaystyle \therefore AB = |r| = \frac{5\sqrt{2}}{3}  

 \displaystyle \\

Question 4: A line drawn through \displaystyle A (4,-1) parallel to the line \displaystyle 3x-4y +1=0 . Find the coordinates of the two points on this line which are at a distance of \displaystyle 5 units from \displaystyle A .

Answer:

\displaystyle \text{Given } 3x-4y + 1 = 0

 \displaystyle \Rightarrow y = \frac{3}{4} x - \frac{1}{4} ( comparing with \displaystyle y = mx +c )

 \displaystyle \Rightarrow \text{Slope} = \frac{3}{4} \Rightarrow \tan \theta = \frac{3}{4} \Rightarrow \cos \theta = \frac{4}{5} \text{ and } \sin \theta = \frac{4}{5}  

\displaystyle \text{The equation passing through } A ( 4,-1) \text{ and and having slope of } \frac{3}{4} \text{ is }  

 \displaystyle \frac{x-4}{(4/5)} = \frac{y-(-1)}{(3/5)} = \pm 5

Considering \displaystyle \text{+ve} sign we get:

 \displaystyle \frac{x-4}{(4/5)} = \frac{y-(-1)}{(3/5)} = 5

 \displaystyle \Rightarrow x = 4 + \frac{4}{5} \times 5 = 8 \text{ and } y = -1+ \frac{3}{5} \times 5 = 2

Hence the point is \displaystyle ( 8, 2)

Considering \displaystyle \text{-ve} sign we get:

 \displaystyle \frac{x-4}{(4/5)} = \frac{y-(-1)}{(3/5)} = -5

 \displaystyle \Rightarrow x = 4 + \frac{4}{5} \times (-5) = 0 \text{ and } y = -1+ \frac{3}{5} \times (-5) = -4

Hence the point is \displaystyle ( 0, -4 )

 \displaystyle \\

Question 5: The straight line through \displaystyle P (x_1, y_1) inclined at an angle \theta with the x-axis meets the line \displaystyle ax + by + c = 0 in \displaystyle Q . Find the length of \displaystyle PQ .

Answer:

\displaystyle \text{Given } P(x_1, y_1) \text{ and } \text{angle} = \theta

\displaystyle \text{The equation passing through } P(x_1, y_1) \text{ and making an angle } \text{angle} = \theta \text{ is }  

 \displaystyle \frac{x-x_1}{\cos \theta} = \frac{y-y_1}{\sin \theta } = r … … … … … i)

From equation i)

\displaystyle \text{The coordinates of any point } Q(x, y) \text{ on this line are } (x_1 + r \cos \theta, y_1+r \sin \theta )

\displaystyle \text{Since } Q \text{ lies on } ax+by+c=0 , \text{ substituting we get }  

 \displaystyle a(x_1 + r \cos \theta)+b(y_1+r \sin \theta)+c=0

 \displaystyle ( a \cos \theta + b \sin \theta) r + (ax_1+by_1 +c) = 0

 \displaystyle r = \frac{-(ax_1+by_1 +c)}{( a \cos \theta + b \sin \theta)}  

 \displaystyle \therefore PQ = |r| = \Big| \frac{(ax_1+by_1 +c)}{( a \cos \theta + b \sin \theta)} \Big|

 \displaystyle \\

Question 6: Find the distance of the point \displaystyle (2, 3) from the line \displaystyle 2x-3y+9=0 measured along a line making an angle of \displaystyle 45^{\circ} with the x-axis.

Answer:

\displaystyle \text{Given } (x_1, y_1) = A ( 2,3) \text{ and } \theta = 45^{\circ}

\displaystyle \text{The equation passing through } A ( 2,3) \text{ and making an angle } 45^{\circ} \text{ is }  

 \displaystyle \frac{x-2}{\cos 45^{\circ}} = \frac{y-3}{\sin 45^{\circ}} = r

 \displaystyle \Rightarrow \frac{x-2}{(1/\sqrt{2}) } = \frac{y-3}{ (1/\sqrt{2})} = r … … … … … i)

From equation i)

\displaystyle \text{The coordinates of any point } B(x, y) \text{ on this line are } \Big( 2+ \frac{r}{\sqrt{2}} , 3+ \frac{r}{\sqrt{2}} \Big)

\displaystyle \text{Since } B \text{ lies on } 2x-3+9=0 , \text{ substituting we get }  

 \displaystyle 2\Big( 2+ \frac{r}{\sqrt{2}} \Big) -3 \Big( 3+ \frac{r}{\sqrt{2}} \Big) +9 = 0

 \displaystyle \Rightarrow - \frac{r}{\sqrt{2}} + 4 = 0

 \displaystyle \Rightarrow r = 4\sqrt{2}

 \displaystyle \therefore AB = 4\sqrt{2}

 \displaystyle \\

Question 7: Find the distance of the point \displaystyle (3,5) from the line \displaystyle 2 x + 3 y =14 measured parallel to a line having slope \displaystyle 1/2 .

Answer:

\displaystyle \text{Given } \tan \theta = \frac{1}{2} \Rightarrow \sin \theta = \frac{1}{\sqrt{1^1+ 2^2}} = \frac{1}{\sqrt{5}} \text{ and } \cos \theta = \frac{2}{\sqrt{1^1+ 2^2}} = \frac{2}{\sqrt{5}}  

\displaystyle \text{The equation passing through } A ( 3, 5) \text{ and and having slope of } \frac{1}{2} \text{ is }  

 \displaystyle \frac{x-3}{(2/\sqrt{5})} = \frac{y-5}{(1/\sqrt{5})} = r … … … … … i)

From equation i)

\displaystyle \text{The coordinates of any point } B(x, y) \text{ on this line are } \Big( 3+ \frac{2r}{\sqrt{5}} , 5+ \frac{r}{\sqrt{5}} \Big)

\displaystyle \text{Since } P \text{ lies on } 2x+3y=14 , \text{ substituting we get }  

 \displaystyle 2 \Big( 3+ \frac{2r}{\sqrt{5}} \Big) + 3 \Big( 5+ \frac{r}{\sqrt{5}} \Big) = 14

 \displaystyle \Rightarrow \frac{7}{\sqrt{5}} + 21 = 14

 \displaystyle \Rightarrow r = - 5

 \displaystyle \therefore AB = |r| = 5

 \displaystyle \\

Question 8: Find the distance of the point \displaystyle (2,5) from the line \displaystyle 3x+y +4=0 measured parallel to a line having slope \displaystyle 3/4 .

Answer:

\displaystyle \text{Given } \tan \theta = \frac{3}{4} \Rightarrow \sin \theta = \frac{1}{5} \text{ and } \cos \theta = \frac{4}{5}  

\displaystyle \text{The equation passing through } A ( 2, 5) \text{ and and having slope of } \frac{3}{4} \text{ is }  

 \displaystyle \frac{x-2}{(4/5)} = \frac{y-5}{(3/5)} = r … … … … … i)

From equation i)

\displaystyle \text{The coordinates of any point } B(x, y) \text{ on this line are } \Big( 2+ \frac{4r}{5} , 5+ \frac{3r}{5} \Big)

\displaystyle \text{Since } P \text{ lies on } 3x+y+4=0 , \text{ substituting we get }  

 \displaystyle 3 \Big( 2+ \frac{4r}{5} \Big) + \Big( 5+ \frac{3r}{5} \Big) +4 = 0

 \displaystyle \Rightarrow 3r+15=0

 \displaystyle \Rightarrow r = - 5

 \displaystyle \therefore AB = |r| = 5

 \displaystyle \\

Question 9: Find the distance of the point \displaystyle (3, 5) from the line \displaystyle 2x+3y=14 measured parallel to the line \displaystyle x -2y=1 .

Answer:

\displaystyle \text{Given line: } x - 2y = 1 \ \ \ \Rightarrow y = \frac{1}{2} x - 1 \ \ \ \Rightarrow \text{Slope} = \frac{1}{2}  

\displaystyle \text{Therefore } \cos \theta = \frac{2}{\sqrt{5}} \text{ and } \sin \theta = \frac{1}{\sqrt{5}}  

\displaystyle \text{The equation passing through } A ( 3, 5) \text{ and and having slope of } \frac{1}{2} \text{ is }  

 \displaystyle \frac{x-3}{(2/\sqrt{5})} = \frac{y-5}{(1/\sqrt{5})} = r … … … … … i)

From equation i)

\displaystyle \text{The coordinates of any point } B(x, y) \text{ on this line are } \Big( 3+ \frac{2r}{\sqrt{5}} , 5+ \frac{r}{\sqrt{5}} \Big)

\displaystyle \text{Since } B \text{ lies on } 2x+3y=14 , \text{ substituting we get }  

 \displaystyle 2 \Big( 3+ \frac{2r}{\sqrt{5}} \Big) + 3\Big( 5+ \frac{r}{\sqrt{5}} \Big) =14

 \displaystyle \Rightarrow \frac{7}{\sqrt{5}} r + 21 = 14

 \displaystyle \Rightarrow \frac{7}{\sqrt{5}} r = - 7

 \displaystyle \Rightarrow r = - \sqrt{5}

 \displaystyle \therefore AB = |r| = \sqrt{5}

 \displaystyle \\

Question 10: Find the distance of the point \displaystyle (2, 5) from the line \displaystyle 3x + y + 4 = 0 measured parallel to the line \displaystyle 3x-4y+8=0 .

Answer:

\displaystyle \text{Given } 3x-4y+8=0 \ \ \Rightarrow 4y = 3x + 8 \ \ \Rightarrow y = \frac{3}{4} x+ 2 \ \ \Rightarrow \text{Slope} = \frac{3}{4}  

\displaystyle \text{Now, } \tan \theta = \frac{3}{4} \Rightarrow \sin \theta = \frac{1}{5} \text{ and } \cos \theta = \frac{4}{5}  

\displaystyle \text{The equation passing through } A ( 2, 5) \text{ and and having slope of } \frac{3}{4} \text{ is }  

 \displaystyle \frac{x-2}{(4/5)} = \frac{y-5}{(3/5)} = r … … … … … i)

From equation i)

\displaystyle \text{The coordinates of any point } B(x, y) \text{ on this line are } \Big( 2+ \frac{4r}{5} , 5+ \frac{3r}{5} \Big)

\displaystyle \text{Since } P \text{ lies on } 3x+y+4=0 , \text{ substituting we get }  

 \displaystyle 3 \Big( 2+ \frac{4r}{5} \Big) + \Big( 5+ \frac{3r}{5} \Big) +4 = 0

 \displaystyle \Rightarrow 3r+15=0

 \displaystyle \Rightarrow r = - 5

 \displaystyle \therefore AB = |r| = 5

 \displaystyle \\

Question 11: Find the distance of the line \displaystyle 2x + y= 3 from the point \displaystyle (- 1, - 3) in the direction of the line whose slope is \displaystyle 1 .

Answer:

\displaystyle \text{Given } \tan \theta = 1 \Rightarrow \sin \theta = \frac{1}{\sqrt{2}} \Rightarrow \cos \theta = \frac{1}{\sqrt{2}}  

\displaystyle \text{The equation passing through } A ( -1, -3) \text{ and and having slope of } 1 \text{ is }  

 \displaystyle \frac{x-(-1)}{(1/\sqrt{2})} = \frac{y-(-3)}{(1/\sqrt{2})} = r … … … … … i)

From equation i)

\displaystyle \text{The coordinates of any point } B(x, y) \text{ on this line are } \Big( -1+ \frac{r}{\sqrt{2}} , -3+ \frac{r}{\sqrt{2}} \Big)

\displaystyle \text{Since } P \text{ lies on } 2x+y=3 , \text{ substituting we get }  

 \displaystyle 2 \Big( -1+ \frac{r}{\sqrt{2}} \Big) + \Big( -3+ \frac{r}{\sqrt{2}} \Big) = 3

 \displaystyle \Rightarrow \frac{3}{\sqrt{2}} r -5 = 3

 \displaystyle \Rightarrow r = \frac{8\sqrt{2}}{3}  

 \displaystyle \therefore AB = |r| = \frac{8\sqrt{2}}{3}  

 \displaystyle \\

Question 12: A line is such that its segment between the straight line \displaystyle 5x -y-4=0 \text{ and } 3x+4y -4=0 is bisected at the point \displaystyle (1,5) . Obtain its equation.

Answer:

\displaystyle \text{Let } P_1 P_2 \text{ be the intercept between lines } 5x -y-4=0 \text{ and } 3x+4y -4=0

\displaystyle \text{Let } P_1 P_2 makes an angle \displaystyle \theta with positive x-axis.

\displaystyle \text{Given } (x_1, y_1) = A ( 1, 5) \text{ and } \theta = \theta

\displaystyle \text{The equation passing through } A ( 1, 5) \text{ and making an angle } \theta \text{ is }  

 \displaystyle \frac{x-1}{\cos \theta} = \frac{y-5}{\sin \theta} = r

 \displaystyle \Rightarrow \tan \theta = \frac{y-5}{x-1}  

\displaystyle \text{Let } AP_1 = AP_2 = r

Therefore the coordinate of \displaystyle P_1 = ( 1+ r \cos \theta, 5 + r \cos \theta) and coordinate of \displaystyle P_2 = ( 1- r \cos \theta, 5 - r \cos \theta)

We know \displaystyle P_1 \text{ lies on } 5x - y - 4 = 0

 \displaystyle \therefore 5 (1+ r \cos \theta) - (5 + r \cos \theta) - 4 = 0

 \displaystyle \Rightarrow r = \frac{4}{5\cos \theta - \sin \theta}  

\displaystyle \text{Also } P_2 \text{ lies on } 3x+4y-4=0

 \displaystyle \therefore 3 (1+ r \cos \theta) +4 (5 + r \cos \theta) - 4 = 0

 \displaystyle \Rightarrow r = \frac{19}{3\cos \theta +4 \sin \theta}  

\displaystyle \text{Since } A(1, 5) bisects \displaystyle P_1 \text{ and } P_2

 \displaystyle \frac{4}{5\cos \theta - \sin \theta} = \frac{19}{3\cos \theta +4 \sin \theta}  

 \displaystyle \Rightarrow 12 \cos \theta + 16 \sin \theta = 45 \cos \theta - 19 \sin \theta

 \displaystyle \Rightarrow 35 \sin \theta= 83 \cos \theta

 \displaystyle \Rightarrow \tan \theta = \frac{83}{85}  

Therefore the equation of the line would be

 \displaystyle \frac{y-5}{x-1} = \frac{83}{85}  

 \displaystyle \Rightarrow 35 y - 175 = 83x - 83

 \displaystyle \Rightarrow 83x - 35y +92 = 0

 \displaystyle \\

Question 13: Find the equation of straight line passing through \displaystyle (-2, -7) and having an intercept of length \displaystyle 3 between the straight lines \displaystyle 4x + 3y = 12 \text{ and } 4x + 3y = 3 .

Answer:

\displaystyle \text{Let } P_1 P_2 \text{ be the intercept between lines } 4x + 3y = 12 \text{ and } 4x + 3y = 3

\displaystyle \text{Let } P_1 P_2 makes an angle \displaystyle \theta with positive x-axis.

\displaystyle \text{Given } (x_1, y_1) = A ( -2,-7) \text{ and } \theta = \theta

\displaystyle \text{The equation passing through } A ( -2,-7) \text{ and making an angle } \theta \text{ is }  

 \displaystyle \frac{x-(-2)}{\cos \theta} = \frac{y-(-7)}{\sin \theta}  

 \displaystyle \Rightarrow \tan \theta = \frac{y+7}{x+2}  

\displaystyle \text{Let } AP_1 =r_1 \text{ and } AP_2 = r_2

Therefore the coordinate of \displaystyle P_1 = ( -2+ r_1 \cos \theta, -7 + r_1 \cos \theta) and coordinate of \displaystyle P_2 = ( -2+ r_2 \cos \theta, -7 + r_2 \cos \theta)

We know \displaystyle P_1 \text{ lies on } 4x+3y=3

 \displaystyle \therefore 4 (-2+ r_1 \cos \theta) +3 (-7 + r_1 \cos \theta) = 3

 \displaystyle \Rightarrow r_1 ( 4 \cos \theta + 3 \sin \theta) = 32

 \displaystyle \Rightarrow r_1 = \frac{32}{( 4 \cos \theta + 3 \sin \theta)}  

We know \displaystyle P_2 \text{ lies on } 4x+3y=12

 \displaystyle \therefore 4 (-2+ r_2 \cos \theta) +3 (-7 + r_2 \cos \theta) = 12

 \displaystyle \Rightarrow r_2 ( 4 \cos \theta + 3 \sin \theta) = 41

 \displaystyle \Rightarrow r_2 = \frac{41}{( 4 \cos \theta + 3 \sin \theta)}  

\displaystyle \text{Since } AP_2 -AP_1 = 3

 \displaystyle \frac{41}{( 4 \cos \theta + 3 \sin \theta)} - \frac{32}{( 4 \cos \theta + 3 \sin \theta)} = 3

 \displaystyle \Rightarrow \frac{9}{( 4 \cos \theta + 3 \sin \theta)} = 3

 \displaystyle \Rightarrow ( 4 \cos \theta + 3 \sin \theta) = 3

 \displaystyle \Rightarrow 3( 1 - \sin \theta) = 4 \cos \theta

Squaring both sides

 \displaystyle \Rightarrow 16 \cos^2 \theta = 9( 1 + \sin^2 \theta - 2 \sin \theta)

 \displaystyle \Rightarrow 16 ( 1 - \sin^2 \theta) = 9 + 9 \sin^2 \theta - 18 \sin \theta

 \displaystyle \Rightarrow 25 \sin^2 \theta - 18 \sin \theta - 7 = 0

 \displaystyle \Rightarrow 25 \sin^2 \theta - 25 \sin \theta + 7 \sin \theta - 7 = 0

 \displaystyle \Rightarrow 25 \sin \theta ( \sin \theta - 1) + 7 ( \sin \theta-1)=0

 \displaystyle \Rightarrow (\sin \theta-1)( 25 \sin \theta + 7) = 0 … … … … … ii)

From ii) either \displaystyle (\sin \theta-1) = 0 \text{ or } ( 25 \sin \theta + 7) = 0

 \displaystyle \Rightarrow \sin \theta = 1 \text{ or } \sin \theta = \frac{-7}{25}  

\displaystyle \text{When } \sin \theta = 1 \Rightarrow \cos \theta = 0

Therefore the equation of line is \displaystyle x + 2 = 0

\displaystyle \text{When } \sin \theta = \frac{-7}{25} \Rightarrow \cos \theta = \frac{24}{25}  

Therefore the equation of line is :

 \displaystyle \frac{x-(-2)}{\frac{24}{25}} = \frac{y-(-7)}{\frac{-7}{25}}  

 \displaystyle \Rightarrow - 7x - 14 = 24y + 168

 \displaystyle \Rightarrow 7x+24y+182=0