Class 11: The Straight Line – Exercise 23.18 – ICSE / ISC / CBSE Mathematics Portal for K12 Students

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

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

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

Answer:

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

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

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

Therefore the equation are:

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

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

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

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

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

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

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

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

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

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

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

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

Answer:

Given equation:

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

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

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

Therefore the equations are:

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

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

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

$\displaystyle \Rightarrow x = 0$ … … … … … i)

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

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

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

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

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

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

Answer:

Given equation:

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

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

Therefore the equations are:

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

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

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

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

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

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

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

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

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

$\displaystyle \\$

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

Answer:

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

Therefore the equations are:

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

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

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

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

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

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

$\displaystyle \Rightarrow y = k$ … … … … … ii)

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

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

Answer:

Given equation:

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

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

Therefore the equations are:

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

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

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

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

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

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

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

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

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

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

Answer:

Given equation:

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

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

Therefore the equations are:

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

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

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

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

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

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

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

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

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

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

Answer:

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

Given equation:

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

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

Therefore the equations are:

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

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

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

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

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

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

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

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

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

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

Answer:

Refer to the adjoining figure.

Given equation:

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

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

Therefore the equations are:

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

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

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

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

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

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

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

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

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

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

Answer:

Refer to the adjoining figure.

Given equation:

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

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

Therefore the equations are:

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Answer:

Refer to the adjoining figure.

$\displaystyle AB = BC$

$\displaystyle \tan B = \tan C$

Therefore $\displaystyle \text{Slope of } AB = 7$

and $\displaystyle \text{Slope of } AC = - 1$

Let $\displaystyle m$ be the $\displaystyle \text{Slope of } BC$

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

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

Taking the $\displaystyle +$ sign, we get

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

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

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

Also now taking $\displaystyle -$ ve sign,

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

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

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

Therefore the equation of sides are

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

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

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

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

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

Answer:

Given lines:

$\displaystyle 2 x+3y-7=0$ … … … … … i)

$\displaystyle 2x+ 3y +12=0$ … … … … … ii)

The distance between line i) and ii)

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

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

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

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

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

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

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

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

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

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

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

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

Therefore, the required lines are

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

and

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

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

Answer:

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

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

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

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

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

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

Therefore, the required lines are

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

and

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

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

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

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

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

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

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

Answer:

Please refer to the adjoining figure.

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

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