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

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

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

Answer:

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

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

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

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

From equation i)

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

Since P lies on x+y = 6 , substituting we get

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

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

\Rightarrow  ( 1 + \sqrt{3})r = 6

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

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

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

\\

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

Answer:

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

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

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

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

From equation i)

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

Since Q lies on 12x+5y +10=0 , substituting we get

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

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

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

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

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

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

\\

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

Answer:

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

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

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

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

From equation i)

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

Since B lies on x+2y+1=0 , substituting we get

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

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

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

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

\\

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

Answer:

Given 3x-4y + 1 = 0

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

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

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

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

Considering \text{+ve} sign we get:

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

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

Hence the point is ( 8, 2)

Considering \text{-ve} sign we get:

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

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

Hence the point is ( 0, -4 )

\\

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

Answer:

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

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

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

From equation i)

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

Since Q lies on ax+by+c=0 , substituting we get

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

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

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

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

\\

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

Answer:

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

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

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

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

From equation i)

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

Since B lies on 2x-3+9=0 , substituting we get

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

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

\Rightarrow r = 4\sqrt{2}

\therefore AB = 4\sqrt{2}

\\

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

Answer:

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

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

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

From equation i)

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

Since P lies on 2x+3y=14 , substituting we get

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

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

\Rightarrow r = - 5

\therefore AB = |r| = 5

\\

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

Answer:

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

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

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

From equation i)

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

Since P lies on 3x+y+4=0  , substituting we get

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

\Rightarrow 3r+15=0

\Rightarrow r = - 5

\therefore AB = |r| = 5

\\

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

Answer:

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

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

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

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

From equation i)

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

Since B lies on 2x+3y=14  , substituting we get

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

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

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

\Rightarrow  r = - \sqrt{5}

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

\\

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

Answer:

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

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

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

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

From equation i)

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

Since P lies on 3x+y+4=0  , substituting we get

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

\Rightarrow 3r+15=0

\Rightarrow r = - 5

\therefore AB = |r| = 5

\\

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

Answer:

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

The equation passing through A ( -1, -3) and and having slope of  1 is 

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

From equation i)

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

Since P lies on 2x+y=3 , substituting we get

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

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

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

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

\\

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

Answer:

Let P_1 P_2 be the intercept  between lines 5x -y-4=0 and 3x+4y -4=0

Let P_1 P_2 makes an angle \theta   with positive x-axis.

Given (x_1, y_1) = A ( 1, 5)     and    \theta = \theta 

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

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

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

Let AP_1 = AP_2 = r

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

We know P_1 lies on 5x - y - 4 = 0

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

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

Also P_2 lies on 3x+4y-4=0

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

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

Since A(1, 5) bisects P_1 and P_2

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

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

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

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

Therefore the equation of the line would be

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

\Rightarrow 35 y - 175 = 83x - 83

\Rightarrow 83x - 35y +92 = 0 

\\

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

Answer:

Let P_1 P_2 be the intercept  between lines 4x + 3y = 12 and 4x + 3y = 3

Let P_1 P_2 makes an angle \theta   with positive x-axis.

Given (x_1, y_1) = A ( -2,-7)     and    \theta = \theta 

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

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

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

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

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

We know P_1 lies on 4x+3y=3 

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

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

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

We know P_2 lies on 4x+3y=12

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

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

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

Since AP_2 -AP_1 = 3

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

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

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

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

Squaring both sides

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

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

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

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

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

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

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

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

When \sin \theta = 1 \Rightarrow \cos \theta = 0

Therefore the equation of line is x + 2 = 0

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

Therefore the equation of line is :

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

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

\Rightarrow 7x+24y+182=0