Class 11: The Straight Line – Exercise 23.13 – ICSE / ISC / CBSE Mathematics Portal for K12 Students

Question 1: Find the angles between each of the following pairs of straight lines:

$\displaystyle \text{(i) } 3 x+y+12 = 0 \text{ and } x+2y-1=0$

$\displaystyle \text{(ii) } 3x-y+5=0 \text{ and } x -3y+1=0$

$\displaystyle \text{(iii) } 3x+4y-7 =0 \text{ and } 4x-3y+5=0$

$\displaystyle \text{(iv) } x-4y=3 \text{ and } 6x-y=11$

$\displaystyle \text{(v) } (m^2-mn)y (mn+n^2)x + n^3 \text{ and } (mn+m^2)y = (mn-n^2)x + m^3$

Answer:

$\displaystyle \text{(i) } \text{Let } m_1 \text{ and } m_2$ be the slope of the straight lines $\displaystyle 3 x+y+12 = 0 \text{ and } x+2y-1=0$ respectively.

$\displaystyle \text{Then } m_1 = -3 \text{ and } m_2 = \frac{-1}{2}$

$\displaystyle \text{Let } \theta$ be the angle between the lines. Then,

$\displaystyle \tan \theta = \Big| \frac{m_1 - m_2}{1 + m_1 m_2} \Big| = \Big| \frac{(-3) - (\frac{-1}{2}) }{1 + (-3) (\frac{-1}{2})} \Big| = \Big| \frac{-6+1}{2+3} \Big| = 1$

$\displaystyle \Rightarrow \theta = 45^{\circ}$

$\displaystyle \text{Thus, the acute angle between the lines is of } 45^{\circ}$

$\displaystyle \text{(ii) } \text{Let } m_1 \text{ and } m_2$ be the slope of the straight lines $\displaystyle 3x-y+5=0 \text{ and } x -3y+1=0$ respectively.

$\displaystyle \text{Then } m_1 = 3 \text{ and } m_2 = \frac{1}{3}$

$\displaystyle \text{Let } \theta$ be the angle between the lines. Then,

$\displaystyle \tan \theta = \Big| \frac{m_1 - m_2}{1 + m_1 m_2} \Big| = \Big| \frac{(3) - (\frac{1}{3}) }{1 + (3) (\frac{1}{3})} \Big| = \Big| \frac{9-1}{3+3} \Big| = \frac{4}{3}$

$\displaystyle \Rightarrow \theta = \tan^{-1} \frac{4}{3}$

$\displaystyle \text{Thus, the acute angle between the lines is of } \tan^{-1} \frac{4}{3}$

$\displaystyle \text{(iii) } \text{Let } m_1 \text{ and } m_2$ be the slope of the straight lines $\displaystyle 3x+4y-7 =0 \text{ and } 4x-3y+5=0$ respectively.

$\displaystyle \text{Then } m_1 = \frac{-3}{4} \text{ and } m_2 = \frac{4}{3}$

Since $\displaystyle m_1 \times m_2 = \frac{-3}{4} \times \frac{4}{3} = -1$

Therefore the given lines are perpendicular.

$\displaystyle \text{Thus, the acute angle between the lines is of } 90^{\circ}$

$\displaystyle \text{(iv) } \text{Let } m_1 \text{ and } m_2$ be the slope of the straight lines $\displaystyle x-4y=3 \text{ and } 6x-y=11$ respectively.

$\displaystyle \text{Then } m_1 = \frac{1}{4} \text{ and } m_2 = 6$

$\displaystyle \text{Let } \theta$ be the angle between the lines. Then,

$\displaystyle \tan \theta = \Big| \frac{m_1 - m_2}{1 + m_1 m_2} \Big| = \Big| \frac{(\frac{1}{4}) - 6 }{1 + (\frac{1}{4}) (6)} \Big| = \Big| \frac{1-24}{4+6} \Big| = \frac{23}{10}$

$\displaystyle \Rightarrow \theta = \tan^{-1} \frac{23}{10}$

$\displaystyle \text{Thus, the acute angle between the lines is of } \tan^{-1} \frac{23}{10}$

$\displaystyle \text{(v) } (m^2-mn)y (mn+n^2)x + n^3 \text{ and } (mn+m^2)y = (mn-n^2)x + m^3$

$\displaystyle \text{Then } m_1 = \frac{mn+n^2}{m^2-mn} \text{ and } m_2 = \frac{mn-n^2}{mn+m^2}$

$\displaystyle \text{Let } \theta$ be the angle between the lines. Then,

$\displaystyle \tan \theta = \Bigg| \frac{m_1 - m_2}{1 + m_1 m_2} \Bigg| = \Bigg| \frac{\frac{mn+n^2}{m^2-mn} - \frac{mn-n^2}{mn+m^2} }{1 + \frac{mn+n^2}{m^2-mn} \cdot \frac{mn-n^2}{mn+m^2} } \Bigg|$

$\displaystyle = \Big| \frac{m^2n^2+m^3n + mn^3 + m^2n^2-m^3n+m^2n^2+m^2n^2-mn^3}{m^3n+m^4-m^2n^2-m^3n+m^2n^2-mn^3+mn^3-n^4} \Big|$

$\displaystyle = \Big| \frac{4m^2n^2}{m^4-n^4} \Big|$

$\displaystyle \Rightarrow \theta = \tan^{-1} \Big( \frac{4m^2n^2}{m^4-n^4} \Big)$

$\displaystyle \text{Thus, the acute angle between the lines is of } \tan^{-1} \Big( \frac{4m^2n^2}{m^4-n^4} \Big)$

$\displaystyle \\$

Question 2: Find the acute angle between the lines $\displaystyle 2 x - y + 3 = 0 \text{ and } x + y + 2 = 0$ .

Answer:

$\displaystyle \text{Let } m_1 \text{ and } m_2$ be the slope of the straight lines $\displaystyle 2 x - y + 3 = 0 \text{ and } x + y + 2 = 0$ respectively.

$\displaystyle \text{Then } m_1 = 2 \text{ and } m_2 = -1$

$\displaystyle \text{Let } \theta$ be the angle between the lines. Then,

$\displaystyle \tan \theta = \Big| \frac{m_1 - m_2}{1 + m_1 m_2} \Big| = \Big| \frac{(2) - (-1) }{1 + (2) (-1)} \Big| = \Big| \frac{2+1}{1-2} \Big| = 3$

$\displaystyle \Rightarrow \theta = \tan^{-1} (3)$

$\displaystyle \text{Thus, the acute angle between the lines is of } \tan^{-1} (3)$

$\displaystyle \\$

Question 3: Prove that the points $\displaystyle (2, - 1), (0,2), (2,3) \text{ and } (4, 0)$ are the coordinates of the vertices of a parallelogram and find the angle between its diagonals.

Answer:

Let the vertices are $\displaystyle A(2, - 1), B(0,2), C(2,3) \text{ and } D(4, 0)$

$\displaystyle \text{Slope of } AB = \frac{2-(-1)}{0-2} = \frac{-3}{2}$

$\displaystyle \text{Slope of } BC = \frac{3-2}{0-2} = \frac{1}{2}$

$\displaystyle \text{Slope of } CD = \frac{0-3}{4-2} = \frac{-3}{2}$

$\displaystyle \text{Slope of } DA = \frac{-1-0}{2-4} = \frac{1}{2}$

Therefore $\displaystyle AB \parallel CD \text{ and } BC \parallel DA$

Hence $\displaystyle ABCD$ is a parallelogram.

$\displaystyle \text{Let } m_1 \text{ and } m_2$ be the slope of the straight lines $\displaystyle AC \text{ and } BD$ respectively.

$\displaystyle \text{Then } m_1 = \frac{3-(-1)}{2-2} = \infty (AC$ is $\displaystyle \parallel$ to y-axis $\displaystyle )$

$\displaystyle m_2 = \frac{0-2}{4-0} = \frac{-1}{2}$

$\displaystyle \tan \theta = \Big| \frac{m_1 - m_2}{1 + m_1 m_2} \Big| = \Big| \frac{1 - \frac{m_2}{m_1} }{\frac{1}{m_1} + m_2} \Big| = \Big| \frac{1-0}{0-\frac{1}{2}} \Big| = 2$

$\displaystyle \Rightarrow \theta = \tan^{-1} (2)$

$\displaystyle \text{Thus, the acute angle between the diagonals is of } \tan^{-1} (2)$

$\displaystyle \\$

Question 4: Find the angle between the line joining the points $\displaystyle (2,0), (0,3)$ and the line $\displaystyle x + y =1$ .

Answer:

Let the points be $\displaystyle A(2,0), B(0,3)$

$\displaystyle \text{Slope of } AB ( m_1) = \frac{3-0}{0-2} = \frac{-3}{2}$

$\displaystyle \text{Slope of } x + y =1 m_2 = -1$

$\displaystyle \text{Let } \theta$ be the angle between the lines. Then,

$\displaystyle \tan \theta = \Big| \frac{m_1 - m_2}{1 + m_1 m_2} \Big| = \Big| \frac{(\frac{-3}{2}) - (-1) }{1 + (\frac{-3}{2}) (-1)} \Big| = \Big| \frac{-3+2}{2+3} \Big| = \frac{1}{5}$

$\displaystyle \Rightarrow \theta = \tan^{-1} \Big( \frac{1}{5} \Big)$

$\displaystyle \text{Thus, the acute angle between the lines is of } \tan^{-1} \Big( \frac{1}{5} \Big)$

$\displaystyle \\$

Question 5: If $\displaystyle \theta$ is the angle which the straight line joining the points $\displaystyle (x_1, y_1) \text{ and } (x_2, y_2)$ subtends at the origin, prove that $\displaystyle \tan \theta = \frac{x_2y_1 - x_1y_2}{x_1x_2 + y_1y_2} \text{ and } \cos \theta = \frac{x_1x_2 + y_1y_2}{ \sqrt{{x_1}^2 + {y_1}^2} \sqrt{{x_2}^2 + {y_2}^2} }$

Answer:

Let the two points be $\displaystyle A ( x_1, y_1) \text{ and } B ( x_2, y_2)$

$\displaystyle \text{Slope of } OA = m_1 = \frac{y_1-0}{x_1-0} = \frac{y_1}{x_1}$

$\displaystyle \text{Slope of } OB = m_2 = \frac{y_2-0}{x_2-0} = \frac{y_2}{x_2}$

$\displaystyle \therefore \tan \theta = \Bigg| \frac{\frac{y_1}{x_1} - \frac{y_2}{x_2}}{1 + \frac{y_1}{x_1} \cdot \frac{y_2}{x_2}} \Bigg| = \Big| \frac{y_1x_2-y_2x_1}{x_1x_2+y_1y_2} \Big|$

We know $\displaystyle \cos \theta = \sqrt{ \frac{1}{1 + \tan^2 \theta} }$

$\displaystyle = \frac{x_1x_2+y_1y_2}{ \sqrt{( x_1x_2+y_1y_2)^2 + ( x_2y_1-x_1y_2)^2} }$

$\displaystyle = \frac{x_1x_2+y_1y_2}{ \sqrt{ {x_1}^2{x_2}^2 + {y_1}^2 {y_1}^2 + 2 x_1x_2y_1y_2 + {x_2}^2{y_1}^2 + {x_1}^2 {y_2}^2 - 2 x_1x_2y_1y_2 }}$

$\displaystyle = \frac{x_1x_2+y_1y_2}{ \sqrt{ {x_1}^2 ( {x_2}^2+ {y_2}^2) + {y_1}^2 ( {x_2}^2+ {y_2}^2) }}$

$\displaystyle = \frac{x_1x_2+y_1y_2}{ \sqrt{ ({x_1}^2 + {y_1}^2) ( {x_2}^2+ {y_2}^2) }}$

$\displaystyle = \frac{x_1x_2+y_1y_2}{ \sqrt{ ({x_1}^2 + {y_1}^2)} \sqrt{( {x_2}^2+ {y_2}^2) }}$

$\displaystyle \\$

Question 6: Prove that the straight lines $\displaystyle (a+ b)x+(a-b)y=2ab,(a-b)x + (a+b) y=2ab \text{ and } x + y =0$ form an isosceles triangle whose vertical angle is $\displaystyle 2 \tan^{-1} \Big( \frac{a}{b} \Big)$

Answer:

Given Lines:

Line i) : $\displaystyle (a+ b)x+(a-b)y=2ab \Rightarrow \text{Slope } = m_1 = \frac{-(a+b)}{(a-b)}$

Line ii) : $\displaystyle (a-b)x + (a+b) y=2ab \Rightarrow \text{Slope } = m_2 = \frac{-(a-b)}{(a+b)}$

Line iii) : $\displaystyle x + y =0 \Rightarrow \text{Slope} = m_3 = -1$

$\displaystyle \text{Let } \theta_1$ is the angle between line i) and line ii)

$\displaystyle \tan \theta_1 = \Bigg| \frac{\frac{-(a+b)}{(a-b)}+\frac{-(a-b)}{(a+b)} }{1 + \frac{(a+b)}{(a-b)} \cdot \frac{(a-b)}{(a+b)} } \Bigg| = \Big| \frac{2ab}{b^2-a^2} \Big| = \Big| \frac{2( \frac{a}{b})}{ 1 - ( \frac{a}{b})^2} \Big|$

$\displaystyle \tan \theta_1 = \tan \Big( 2 \tan^{-1} \Big( \frac{a}{b} \Big ) \Big)$

$\displaystyle \Rightarrow \theta_1 = 2 \tan^{-1} \Big( \frac{a}{b} \Big )$

$\displaystyle \text{Let } \theta_2$ is the angle between line ii) and line iii)

$\displaystyle \tan \theta_2 = \Bigg| \frac{\frac{-(a-b)}{(a+b) + 1}}{1 + \frac{(a-b)}{(a+b)}} \Bigg| = \Big| \frac{-a+b+a+b}{a+b+a-b} \Big| = \Big| \frac{b}{a} \Big|$

$\displaystyle \Rightarrow \theta_2 = \tan^{-1} \Big( \frac{b}{a} \Big)$

$\displaystyle \text{Let } \theta_3$ is the angle between line iii) and line i)

$\displaystyle \tan \theta_2 = \Bigg| \frac{\frac{-1 + (a+b)}{(a-b)}}{1 + (-1) \frac{-(a+b)}{(a-b)} } \Bigg| = \Big| \frac{-a+b+a+b}{a-b+a+b} \Big| = \Big| \frac{b}{a} \Big|$

Therefore the triangle is an isosceles as $\displaystyle \theta_2 = \theta_3$

The vertical angle is $\displaystyle \theta_1 = 2 \tan^{-1} \Big( \frac{a}{b} \Big )$

$\displaystyle \\$

Question 7: Find the angle between the lines $\displaystyle x = a \text{ and } by + c =0$ .

Answer:

Given: Line i) : $\displaystyle x = a \Rightarrow \text{Slope } = m_1 = \frac{1}{0}$ . This line is $\displaystyle \parallel$ to y-axis.

Line ii) $\displaystyle by + c =0 \Rightarrow \text{Slope } = m_2 = 0$ . This line is $\displaystyle \parallel$ to x-axis.

$\displaystyle \therefore \tan \theta = \Bigg| \frac{ \frac{1}{0} - 0}{1 + \frac{1}{0} \cdot 0} \Bigg| = \Big| \frac{1}{0} \Big| = \infty$

$\displaystyle \therefore \theta = \tan^{-1} ( \infty) = 90^{\circ}$

$\displaystyle \\$

Question 8: Find the tangent of the angle between the lines which have intercepts $\displaystyle 3,4 \text{ and } 1, 8$ on the axes respectively.

Answer:

Equation of the line with intercepts $\displaystyle 3, 4$ :

$\displaystyle \frac{x}{3} + \frac{y}{4} = 1 \Rightarrow 4x + 3y = 12$ … … … … … i)

Equation of the line with intercepts $\displaystyle 1, 8$ :

$\displaystyle \frac{x}{1} + \frac{y}{8} = 1 \Rightarrow 8x + y = 8$ … … … … … ii)

Slope of line i) $\displaystyle m_1 = \frac{-4}{3}$

Slope of line ii) $\displaystyle m_2 = -8$

$\displaystyle \therefore \tan \theta = \Bigg| \frac{ \frac{-4}{3} - (-8)}{ 1 + ( \frac{-4}{3}) ( -8) } \Bigg| = \Big| \frac{-4+24}{3+32} \Big| = \frac{20}{35} = \frac{4}{7}$

$\displaystyle \\$

Question 9: Show that the line $\displaystyle a^2 x + ay + 1= 0$ is perpendicular to the line $\displaystyle x-ay=1$ for all non-zero real values of $\displaystyle a$ .

Answer:

Line i) : $\displaystyle a^2 x + ay + 1 = 0 \Rightarrow \text{Slope } m_1 = \frac{-a^2}{a} =-a$

Line ii) : $\displaystyle x-ay=1 \Rightarrow \text{Slope } m_2 = \frac{1}{a}$

Now, $\displaystyle m_1 \times m_2 = (-a) \times \Big( \frac{1}{a} \Big) = - 1$

Therefore the two lines are perpendicular to each other.

$\displaystyle \\$

Question 10: Show that the tangent of an angle between the lines $\displaystyle \frac{x}{a} + \frac{y}{b} = 1 \text{ and } \frac{x}{a} - \frac{y}{b} = 1$ is $\displaystyle \frac{2ab}{a^2- b^2}$

Answer:

Given lines:

$\displaystyle \frac{x}{a} + \frac{y}{b} = 1 \Rightarrow bx + ay = ab$ … … … … … i)

$\displaystyle \frac{x}{a} - \frac{y}{b} = 1 \Rightarrow bx -a y = ab$ … … … … … ii)

Slope of line i) $\displaystyle m_1 = \frac{-b}{a}$

Slope of line ii) $\displaystyle m_2 = \frac{b}{a}$

$\displaystyle \therefore \tan \theta = \Bigg| \frac{ \frac{-b}{a} - \frac{b}{a}}{ 1 + ( \frac{b}{a}) ( -8) } \Bigg| = \Big| \frac{-2ab}{a^2-b^2} \Big|= \frac{2ab}{a^2-b^2}$

ICSE / ISC / CBSE Mathematics Portal for K12 Students

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Class 11: The Straight Line – Exercise 23.13

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