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 $\displaystyle AP$.

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

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

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

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

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

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

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

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

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

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

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

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

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

$\displaystyle \Rightarrow y = \frac{3}{4} x - \frac{1}{4} ($ 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 $\displaystyle PQ$.

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

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

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

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

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

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

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

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

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

$\displaystyle \frac{x-3}{(2/\sqrt{5})} = \frac{y-5}{(1/\sqrt{5})} = r$ … … … … … 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 $\displaystyle 3/4$.

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

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

$\displaystyle \frac{x-2}{(4/5)} = \frac{y-5}{(3/5)} = r$ … … … … … 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 $\displaystyle x -2y=1$.

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

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

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

$\displaystyle \frac{x-3}{(2/\sqrt{5})} = \frac{y-5}{(1/\sqrt{5})} = r$ … … … … … 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 $\displaystyle 3x-4y+8=0$.

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

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

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

$\displaystyle \frac{x-2}{(4/5)} = \frac{y-5}{(3/5)} = r$ … … … … … 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 $\displaystyle 1$.

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

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

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

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

$\displaystyle \text{Let } P_1 P_2$ 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 $\displaystyle 4x + 3y = 12 \text{ and } 4x + 3y = 3$.

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

$\displaystyle \text{Let } P_1 P_2$ 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 :

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

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

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