Note: We know that the equations of two lines passing through $( x_1, y_1)$  and making and angle $\alpha$  with the given line $y = mx + c$  are

$y - y_1 = \Big($ $\frac{m \pm \tan \alpha }{1 \mp m \tan \alpha}$ $\Big) ( x - x_1)$

Question 1: Find the equation of the straight lines passing through the origin and making an angle of $45^{\circ}$ with the straight line $\sqrt{3} x + y =11$ .

We know that the equations of two lines passing through $( x_1, y_1)$  and making and angle $\alpha$  with the given line $y = mx + c$  are

$y - y_1 = \Big($ $\frac{m \pm \tan \alpha }{1 \mp m \tan \alpha}$ $\Big) ( x - x_1)$

Here $x_1 = 0, \hspace{0.5cm} y_1 = 0 , \hspace{0.5cm} \alpha = 45^{\circ} \hspace{0.5cm} m = -\sqrt{3}$

Therefore the equation are:

$y - 0 = \Big($ $\frac{-\sqrt{3} + \tan 45^{\circ} }{1 + \sqrt{3} \tan 45^{\circ}}$ $\Big) ( x - 0)$

$\Rightarrow y = \Big($ $\frac{-\sqrt{3} + 1 }{1 + \sqrt{3} }$ $\Big) x$

$\Rightarrow y = \Big($ $\frac{-\sqrt{3} + 1 }{1 + \sqrt{3} } \times \frac{\sqrt{3} - 1}{\sqrt{3} -1}$ $\Big) x$

$\Rightarrow y = \Big($ $\frac{3 +1 - 2\sqrt{2} }{3+1 }$ $\Big) x$

$\Rightarrow y = ( \sqrt{3} - 2) x$     … … … … … i)

$y - 0 = \Big($ $\frac{-\sqrt{3} - \tan 45^{\circ} }{1 - \sqrt{3} \tan 45^{\circ}}$ $\Big) ( x - 0)$

$\Rightarrow y = \Big($ $\frac{-\sqrt{3} - 1 }{1 - \sqrt{3} }$ $\Big) x$

$\Rightarrow y = \Big($ $\frac{\sqrt{3} + 1 }{\sqrt{3} - 1} \times \frac{\sqrt{3} + 1}{\sqrt{3} +1}$ $\Big) x$

$\Rightarrow y = \Big($ $\frac{3 +1 + 2\sqrt{2} }{3-1 }$ $\Big) x$

$\Rightarrow y = ( \sqrt{3} + 2) x$     … … … … … ii)

Therefore $y = ( \sqrt{3} - 2) x \text{ and } y = ( \sqrt{3} + 2) x$ are the two equations.

$\\$

Question 2: Find the equations to the straight lines which pass through the origin and are inclined at an angle of $75^{\circ}$ to the straight line $x + y +\sqrt{3} (y -x) =a$.

Given equation:

$x+ y + \sqrt{3}( y - x) = a$     $\Rightarrow ( \sqrt{3}+1) y = ( \sqrt{3}-1) x + a$     $\Rightarrow y =$ $\frac{\sqrt{3}-1}{\sqrt{3}+1}$ $x +$ $\frac{a}{\sqrt{3}+1}$

$\therefore \text{ Slope } m =$ $\frac{\sqrt{3}-1}{\sqrt{3}+1}$ $=$ $\frac{\sqrt{3}-1}{\sqrt{3}+1}$ $\times$ $\frac{\sqrt{3}-1}{\sqrt{3}-1}$ $= 2 - \sqrt{3}$

Here $x_1 = 0, \hspace{0.5cm} y_1 = 0 , \hspace{0.5cm} \alpha = 75^{\circ} \hspace{0.5cm} m = 2-\sqrt{3}$

Therefore the equations are:

$y - 0 = \Big($ $\frac{(2-\sqrt{3}) + \tan 75^{\circ} }{1 -( 2 - \sqrt{3}) \tan 75^{\circ}}$ $\Big) ( x - 0)$

$\Rightarrow y = \Big($ $\frac{(2-\sqrt{3}) + ( 2 + \sqrt{3}) }{1 - (2 - \sqrt{3})(2+\sqrt{3} }$ $\Big) x$

$\Rightarrow y = \Big($ $\frac{4 }{0 }$ $\Big) x$

$\Rightarrow x = 0$     … … … … … i)

$y - 0 = \Big($ $\frac{(2-\sqrt{3}) - \tan 75^{\circ} }{1 +( 2 - \sqrt{3}) \tan 75^{\circ}}$ $\Big) ( x - 0)$

$\Rightarrow y = \Big($ $\frac{(2-\sqrt{3}) - ( 2 + \sqrt{3}) }{1 + (2 - \sqrt{3})(2+\sqrt{3} }$ $\Big) x$

$\Rightarrow y = \Big($ $\frac{-2\sqrt{3} }{1+1 }$ $\Big) x$

$\Rightarrow y + \sqrt{3} x = 0$     … … … … … ii)

Therefore $x = 0 \text{ and } y + \sqrt{3} x = 0$ are the two equations.

$\\$

Question 3: Find the equations of the straight lines passing through $(2, - 1)$ and making an angle of $45^{\circ}$ with the line $6x + 5 y - 8 = 0$.

Given equation:

$6x + 5 y - 8 = 0$     $\Rightarrow y =$ $\frac{-6}{5}$ $x +$ $\frac{4}{3}$     $\therefore \text{ Slope } m =$ $\frac{-6}{5}$

Here $x_1 = 2, \hspace{0.5cm} y_1 = 2 , \hspace{0.5cm} \alpha = 45^{\circ} \hspace{0.5cm} m =$ $\frac{-6}{5}$

Therefore the equations are:

$y - (-1) = \Big($ $\frac{ \frac{-6}{5} + \tan 45^{\circ} }{1 - ( \frac{-6}{5}) \tan 45^{\circ}}$ $\Big) ( x - 2)$

$\Rightarrow y +1 = \Big($ $\frac{ \frac{-6}{5} + 1 }{1 + ( \frac{6}{5}) }$ $\Big) ( x - 2)$

$\Rightarrow y + 1 =$ $\frac{-1}{11}$ $( x- 2)$

$\Rightarrow x + 11y + 9 = 0$     … … … … … i)

$y - (-1) = \Big($ $\frac{\frac{-6}{5} - \tan 45^{\circ} }{1 - ( \frac{6}{5}) \tan 45^{\circ}}$ $\Big) ( x - 2)$

$\Rightarrow y +1 = \Big($ $\frac{\frac{-6}{5} - 1 }{1 - ( \frac{6}{5}) }$ $\Big) ( x - 2)$

$\Rightarrow y + 1 =$ $\frac{-11}{-1}$ $( x- 2)$

$\Rightarrow 11x-y-23 = 0$     … … … … … ii)

Therefore $x + 11y + 9 = 0 \text{ and } 11x-y-23 = 0$ are the two equations.

$\\$

Question 4: Find the equations to the straight lines which pass through the point $(h, k)$ and are inclined at angle $\tan^{-1} m$ to the straight line $y = mx + c$.

Here $x_1 = h, \hspace{0.5cm} y_1 = k , \hspace{0.5cm} \alpha = \tan^{-1} m \hspace{0.5cm} m = m$

Therefore the equations are:

$y - k = \Big($ $\frac{ m + m }{1 - m^2 }$ $\Big) ( x - h)$

$\Rightarrow y - k = \Big($ $\frac{ 2m }{1 - m^2 }$ $\Big) ( x - h)$

$\Rightarrow (y-k)(1-m^2) = 2m ( x - h)$

$\Rightarrow (1- m^2)y - 2mx = k( 1-m^2) - 2mh$ … … … … … i)

$y - k = \Big($ $\frac{ m - m }{1 - m^2 }$ $\Big) ( x - h)$

$\Rightarrow y - k = 0 ( x - h)$

$\Rightarrow y = k$     … … … … … ii)

Therefore $(1- m^2)y - 2nx = k( 1-m^2) - 2mh \text{ and } y = k$ are the two equations.

$\\$

Question 5: Find the equations to the straight lines passing through the point $(2, 3)$ and inclined at an angle of $45^{\circ}$ to the line $3 x + y -5 = 0$.

Given equation:

$3x+y-5=0$     $\Rightarrow y = - 3x + 5$     $\therefore \text{ Slope } m = -3$

Here $x_1 = 2, \hspace{0.5cm} y_1 = 3 , \hspace{0.5cm} \alpha = 45^{\circ} \hspace{0.5cm} m = -3$

Therefore the equations are:

$y - (3) = \Big($ $\frac{ -3 + \tan 45^{\circ} }{1 + 3 \tan 45^{\circ}}$ $\Big) ( x - 2)$

$\Rightarrow y -3 = \Big($ $\frac{ -3 + 1 }{1 + 3 }$ $\Big) ( x - 2)$

$\Rightarrow y -3 =$ $\frac{-1}{2}$ $( x- 2)$

$\Rightarrow x + 2y -8 = 0$     … … … … … i)

$y - (3) = \Big($ $\frac{ -3 - \tan 45^{\circ} }{1 - 3 \tan 45^{\circ}}$ $\Big) ( x - 2)$

$\Rightarrow y -3 = \Big($ $\frac{ -3 - 1 }{1 3 3 }$ $\Big) ( x - 2)$

$\Rightarrow y -3 = 2 ( x- 2)$

$\Rightarrow 2x-y-1= 0$     … … … … … ii)

Therefore $x + 2y -8 = 0 \text{ and } 2x-y-1= 0$ are the two equations.

$\\$

Question 6: Find the equations to the sides of an isosceles right angled triangle the equation of whose hypotenuse is $3 x + 4 y = 4$ and the opposite vertex is the point $(2, 2)$.

Given equation:

$3x+4y=4$     $\Rightarrow y =$ $\frac{-3}{4}$ $x + 4$     $\therefore \text{ Slope } m =$ $\frac{-3}{4}$

Here $x_1 = 2, \hspace{0.5cm} y_1 = 2 , \hspace{0.5cm} \alpha = 45^{\circ} \hspace{0.5cm} m =$ $\frac{-3}{4}$

Therefore the equations are:

$y - 2 = \Big($ $\frac{ \frac{-3}{4} + \tan 45^{\circ} }{1 - (\frac{-3}{4}) \tan 45^{\circ}}$ $\Big) ( x - 2)$

$\Rightarrow y -2 = \Big($ $\frac{ \frac{-3}{4} + 1 }{1 + \frac{3}{4} }$ $\Big) ( x - 2)$

$\Rightarrow y -2 =$ $\frac{1}{7}$ $( x- 2)$

$\Rightarrow x-7y+12=0$     … … … … … i)

$y - 2 = \Big($ $\frac{ \frac{-3}{4} - \tan 45^{\circ} }{1 - \frac{3}{4} \tan 45^{\circ}}$ $\Big) ( x - 2)$

$\Rightarrow y -2 = \Big($ $\frac{ \frac{-3}{4} - 1 }{1 - \frac{-3}{4} }$ $\Big) ( x - 2)$

$\Rightarrow y -2 =$ $\frac{-7}{1}$ $( x- 2)$

$\Rightarrow 7x+y-16=0$     … … … … … ii)

Therefore $x-7y+12=0 \text{ and } 7x+y-16=0$ are the two equations.

$\\$

Question 7: The equation of one side of an equilateral triangle is $x-y =0$ and one vertex is $(2+\sqrt{3},5)$. Prove that a second side is $y + ( 2- \sqrt{3})x =6$ and find the equation of the third side.

Refer to the adjoining figure. Since the triangle is equilateral triangle, hence all angles are $60^{\circ}$

Given equation:

$x-y=0$     $\Rightarrow y = x$     $\therefore \text{ Slope } m = 1$

Here $x_1 = 2+\sqrt{3} , \hspace{0.5cm} y_1 = 5 , \hspace{0.5cm} \alpha = 60^{\circ} \hspace{0.5cm} m = 1$

Therefore the equations are:

$y - 5 = \Big($ $\frac{ 1 + \tan 60^{\circ} }{1 - \tan 60^{\circ}}$ $\Big) ( x - 2-\sqrt{3} )$

$\Rightarrow y -5 = \Big($ $\frac{ 1+\sqrt{3} }{1 - \sqrt{3} }$ $\Big) ( x - 2 - \sqrt{3} )$

$\Rightarrow y -5 = -(2 +\sqrt{3})( x - 2 - \sqrt{3} )$

$\Rightarrow ( 2+\sqrt{3}) x + y = 5 + ( 2 + \sqrt{3})^2$     … … … … … i)

$y - 5 = \Big($ $\frac{ 1 - \tan 60^{\circ} }{1 + \tan 60^{\circ}}$ $\Big) ( x - 2-\sqrt{3} )$

$\Rightarrow y -5 = \Big($ $\frac{ 1-\sqrt{3} }{1 + \sqrt{3} }$ $\Big) ( x - 2 - \sqrt{3} )$

$\Rightarrow y -5 = -(2 -\sqrt{3})( x - 2 - \sqrt{3} )$

$\Rightarrow ( 2-\sqrt{3}) x + y -6=0$     … … … … … ii)

Therefore $( 2+\sqrt{3}) x + y = 5 + ( 2 + \sqrt{3})^2 \text{ and } ( 2-\sqrt{3}) x + y -6=0$ are the two equations.

$\\$

Question 8: Find the equations of the two straight lines through $(1,2)$ forming two sides of a square of which $4x + 7y=12$ is one diagonal.

Given equation:

$4x+7y=12$     $\Rightarrow y =$ $\frac{-4}{7}$ $x +$ $\frac{12}{7}$     $\therefore \text{ Slope } m =$ $\frac{-4}{7}$

Here $x_1 = 1 , \hspace{0.5cm} y_1 = 2 , \hspace{0.5cm} \alpha = 45^{\circ} \hspace{0.5cm} m =$ $\frac{-4}{7}$

Therefore the equations are:

$y - 2 = \Big($ $\frac{ \frac{-4}{7} + \tan 45^{\circ} }{1 - ( \frac{-4}{7}) \tan 45^{\circ}}$ $\Big) ( x - 1 )$

$\Rightarrow y -2 = \Big($ $\frac{ \frac{-4}{7}+1 }{1 + \frac{4}{7} }$ $\Big) ( x - 1 )$

$\Rightarrow y -2 = \frac{3}{11} ( x - 1)$

$\Rightarrow 3x-11y+ 19 = 0$     … … … … … i)

$y - 2 = \Big($ $\frac{ \frac{-4}{7} - \tan 45^{\circ} }{1 + ( \frac{-4}{7}) \tan 45^{\circ}}$ $\Big) ( x - 1 )$

$\Rightarrow y -2 = \Big($ $\frac{ \frac{-4}{7}+1 }{1 + \frac{4}{7} }$ $\Big) ( x - 1 )$

$\Rightarrow y -2 = \frac{-11}{3} ( x - 1)$

$\Rightarrow 11x+3y-17 = 0$     … … … … … ii)

Therefore $3x-11y+ 19 = 0 \text{ and } 11x+3y-17 = 0$ are the two equations.

$\\$

Question 9: Find the equation of two straight line passing through $(1,2)$ and making an angle of $60^{\circ}$. with the line $x + y = 0$. Find also the area of the triangle formed by the three lines.

Given equation:

$x+y=0$     $\Rightarrow y = -x$     $\therefore \text{ Slope } m = -1$

Here $x_1 = 1 , \hspace{0.5cm} y_1 = 2 , \hspace{0.5cm} \alpha = 60^{\circ} \hspace{0.5cm} m = -1$

Therefore the equations are:

$y - 2 = \Big($ $\frac{ -1 + \tan 60^{\circ} }{1 + \tan 60^{\circ}}$ $\Big) ( x - 1 )$

$\Rightarrow y -2 = \Big($ $\frac{ \sqrt{3}-1 }{\sqrt{3}+1 }$ $\Big) ( x - 1 )$

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

$y - 2 = \Big($ $\frac{ -1 - \tan 60^{\circ} }{1 - \tan 60^{\circ}}$ $\Big) ( x - 1 )$

$\Rightarrow y -2 = \Big($ $\frac{ \sqrt{3}+1 }{\sqrt{3}-1 }$ $\Big) ( x - 1 )$

$\Rightarrow y -2 = (2+\sqrt{3}) ( x - 1)$     … … … … … i)

Therefore $y -2 = (2-\sqrt{3}) ( x - 1) \text{ and } y -2 = (2+\sqrt{3}) ( x - 1)$ are the two equations.

Solving $x+y=0$ & $y -2 = (2-\sqrt{3}) ( x - 1)$

We get $x = - \Big($ $\frac{\sqrt{3}+1}{2}$ $\Big) , y = \Big($ $\frac{\sqrt{3}+1}{2}$ $\Big)$

$\therefore B = \Big( - ($ $\frac{\sqrt{3}+1}{2}$ $\Big) , \Big($ $\frac{\sqrt{3}+1}{2}$ $) \Big)$

Similarly solving $x+y = 0$ and  $y -2 = (2+\sqrt{3}) ( x - 1)$

We get $x = \Big($ $\frac{\sqrt{3}-1}{2}$ $\Big) , y = - \Big($ $\frac{\sqrt{3}-1}{2}$ $\Big)$

$\therefore C = \Bigg( \Big($ $\frac{\sqrt{3}-1}{2}$ $\Big) , -\Big($ $\frac{\sqrt{3}-1}{2}$ $\Big) \Bigg)$

$AB =\sqrt{ [ 1 + ( \frac{\sqrt{3}+1}{2} )^2 ] + [ 2 - (\frac{\sqrt{3}+1}{2} )^2 ] } = \sqrt{6}$

Similarly, $BC = AD = \sqrt{6}$ unit.

Therefore are of $\triangle ABC =$ $\frac{\sqrt{3} \times ( \sqrt{6})^2}{4}$ $= \frac{3\sqrt{3}}{2}$ sq. units.

$\\$

Question 10: Two sides of an isosceles triangle are given by the equations $7 x-y+ 3=0$ and $x +y - 3 = 0$ and its third side passes through the point $(1, - 10)$. Determine the equation of the third side.

$AB = BC$

$\tan B = \tan C$

Therefore Slope of $AB = 7$

and Slope of $AC = - 1$

Let $m$ be the slope of $BC$

$\therefore \Big|$ $\frac{m-7}{1+7m}$ $\Big | = \Big|$ $\frac{m+1}{1-m}$ $\Big | = \Big|$ $\frac{m+1}{m-1}$ $\Big |$

$\Rightarrow$ $\frac{m-7}{1+7m}$ $= \pm$ $\frac{m+1}{m-1}$

Taking the $+$ sign, we get

$m^2 - 8m+7 = 7m^2 + 8m+1$

$\Rightarrow ( m+3) ( m -$ $\frac{1}{3}$ $) = 0$

$\Rightarrow m = - 3,$ $\frac{1}{3}$

Also now taking $-$ ve sign,

$( m-7)(m-1) = - ( 7m+1)(m+1)$

$\Rightarrow m^2 - 8m + 7 = - 7m^2 - 8m - 1$

$\Rightarrow m^2 = - 1$ ( not possible)

Therefore the equation of sides are

$y + 10 = - 3 ( x-1)$

$\Rightarrow 3x + y + 7 = 0$     … … … … … i)

$y + 10 =$ $\frac{1}{3}$ $( x-1)$

$\Rightarrow x - 3y - 31 = 0$     … … … … … ii)

$\\$

Question 11: Show that the point $(3, - 5)$ lies between the parallel lines $2 x+3y-7=0$ and $2x+ 3y +12=0$ and find the equation of lines through $(3,-5)$ cutting the above lines at an angle of $45^{\circ}$.

Given lines:

$2 x+3y-7=0$     … … … … … i)

$2x+ 3y +12=0$     … … … … … ii)

The distance between line i) and ii)

$d = \Big|$ $\frac{12-(-7)}{\sqrt{2^2 + 3^2}}$ $\Big| =$ $\frac{19}{\sqrt{13}}$

Distance for $(3, -5)$ from line i)

$d_1 = \Big|$ $\frac{2(3)+3(-5)-7}{\sqrt{2^2 + 3^2}}$ $\Big| =$ $\frac{16}{\sqrt{13}}$

Distance for $(3, -5)$ from line ii)

$d_1 = \Big|$ $\frac{2(3)+3(-5)+12}{\sqrt{2^2 + 3^2}}$ $\Big| =$ $\frac{3}{\sqrt{13}}$

Now we see that $d_1 + d_2 =$ $\frac{19}{\sqrt{13}}$ $= d$

Therefore we can say that $(3, -5)$ is between the two given lines.

Let m be the slope of the line passing through $( 3, -5)$. This line makes $45^{\circ}$ with the lines. The slope of the given lines is $\frac{-2}{3}$ . Therefore,

$\tan 45^{\circ} = \Big|$ $\frac{m- ( \frac{-2}{3}) }{1 + m ( \frac{-2}{3})}$ $\Big|$

$\Rightarrow$ $\frac{3m+2}{3-2m}$ $= \pm 1$

Two cases arise: Case 1:  $3m + 2 = 3 - 2m \hspace{0.5cm} \Rightarrow m =$ $\frac{1}{5}$

Case 2:  $3m + 2 = -(3 - 2m) \hspace{0.5cm} \Rightarrow m = -5$

Therefore, the required lines are

$y - ( -5) =$ $\frac{1}{5}$ $(x-3) \hspace{0.5cm} \Rightarrow x - 5y - 28 = 0$

and

$y - ( -5) = -5 (x-3) \hspace{0.5cm} \Rightarrow 5x + y - 10 = 0$

$\\$

Question 12: The equation of the base of an equilateral triangle is $x + y =2$ and its vertex is $(2, - 1)$. Find the length and equations of its sides.

Since this is an equilateral triangle, hence all the three angles are $60^{\circ}$.

The slope of the given lines is $-1$ . Therefore,

$\tan 60^{\circ} = \Big|$ $\frac{m- ( -1) }{1 + m ( -1)}$ $\Big|$

$\Rightarrow$ $\frac{m+1}{m-1}$ $= \pm \sqrt{3}$

Two cases arise: Case 1:  $m+1 = \sqrt{3} ( m-1) \hspace{0.5cm} \Rightarrow m =$ $\frac{\sqrt{3}+1}{\sqrt{3}-1}$ $= 2+\sqrt{3}$

Case 2:  $m+1 = -\sqrt{3} (m-1) \hspace{0.5cm} \Rightarrow m =$ $\frac{\sqrt{3}-1}{\sqrt{3}+1}$ $= 2-\sqrt{3}$

Therefore, the required lines are

$y - ( -1) = (2+\sqrt{3}) (x-1) \hspace{0.5cm} \Rightarrow y +1 = (2+\sqrt{3}) (x-2)$

and

$y - ( -1) = (2-\sqrt{3}) (x-1) \hspace{0.5cm} \Rightarrow y + 1= (2-\sqrt{3}) (x-2)$

Solving $x + y =2$ and $y +1 = (2+\sqrt{3}) (x-2)$ we get

$B = \Big($ $\frac{15+\sqrt{3}}{6}$ $,$ $\frac{-(3+\sqrt{3})}{6}$ $\Big)$

Similarly, Solving $x + y =2$ and $y + 1= (2-\sqrt{3}) (x-2)$ we get

$C = \Big($ $\frac{15-\sqrt{3}}{6}$ $,$ $\frac{-(3-\sqrt{3})}{6}$ $\Big)$

Therefore $AB = BC = CA =$ $\sqrt{\frac{2}{3}}$

$\\$

Question 13: If two opposite vertices of a square arc $(1,2)$ and $(5, 8)$, find the coordinates of its other two vertices and the equations of its sides.

Slope of $AC =$ $\frac{8-2}{5-1}$ $=$ $\frac{3}{2}$

The sides $AB$ and $AD$ passes through $(1, 2)$  and makes an angle of $45^{\circ}$  with $AC$  whose slope is $\frac{3}{2}$

The equation of $AB$ and $AD$ are given by

$y - 2 =$ $\frac{\frac{3}{2} \pm \tan 45^{\circ} }{1 \mp \frac{3}{2} \tan 45^{\circ}}$ $( x - 1)$

$\Rightarrow y - 2 =$ $\frac{3 \pm 2}{2 \mp 3}$ $( x - 1)$

$\Rightarrow y - 2 = -5( x-1) \text{ and } y - 2 =$ $\frac{1}{5}$ $( x-1)$

$\Rightarrow 5x + y -7 = 0$  and $x - 5y + 9 = 0$

Therefore the equations of $AB$ and $AD$  are $5x + y -7 = 0$  and $x - 5y + 9 = 0$  respectively.

Since $BC$ is parallel to $AD$, the equation of $BC$ is $x - 5y + \lambda_1 = 0$

This line passes through $( 5, 8)$. Therefore,

$5 - 40 + \lambda_1 = 0 \Rightarrow \lambda_1 = 35$

Therefore the equation of $BC$ is $x - 5y + 35 = 0$

Since $CD$ is parallel to $AB$, the equation of $CD$ is $5x + y + \lambda_2 = 0$

This line passes through $( 5, 8)$. Therefore,

$25+8 + \lambda_2 = 0 \Rightarrow \lambda_2 = -33$

Therefore the equation of $CD$ is $5x + y -33 = 0$

Solving equation of $AB$ and $BC$ we get $B ( 0,7)$.

Solving equation of $AD$ and $CD$ we get $D ( 6, 3)$.

Hence the two vertices are $B ( 0,7)$ and $D ( 6, 3)$.