OP Malhotra Class 11 Maths Solutions Chapter 16 The Straight Line Exercise 16 (G)

S Chand Class 11 ICSE Maths Solutions Chapter 16 The Straight Line Ex 16(g)

 

Question 1. Find the equations of the lines bisecting the angles between the following pairs of straight lines writing first at the bisector of the angle in which the origin lines :
(i) 3x – 4y + 10 = 0, 5x – 12y – 10 = 0;
(ii) 12x – 5y + 3 = 0, 4x + 3y – 2 = 0.

Answer:
(i) Given lines are:
\( 3x - 4y + 10 = 0 \) ...(1)
To ensure the constant term is positive for the second line, we rewrite \( 5x - 12y - 10 = 0 \) as \( -5x + 12y + 10 = 0 \) ...(2)
The equations of the angular bisectors are given by:
\( \frac{3x - 4y + 10}{\sqrt{3^2 + (-4)^2}} = \pm \frac{-5x + 12y + 10}{\sqrt{(-5)^2 + 12^2}} \)
\( \implies \frac{3x - 4y + 10}{\sqrt{9 + 16}} = \pm \frac{-5x + 12y + 10}{\sqrt{25 + 144}} \)
\( \implies \frac{3x - 4y + 10}{5} = \pm \frac{-5x + 12y + 10}{13} \)
For the bisector of the angle in which the origin lies, we take the positive sign (assuming positive constant terms):
\( \frac{3x - 4y + 10}{5} = \frac{-5x + 12y + 10}{13} \)
\( \implies 13(3x - 4y + 10) = 5(-5x + 12y + 10) \)
\( \implies 39x - 52y + 130 = -25x + 60y + 50 \)
\( \implies 39x + 25x - 52y - 60y + 130 - 50 = 0 \)
\( \implies 64x - 112y + 80 = 0 \)
\( \implies 4x - 7y + 5 = 0 \)
This is the equation of the bisector where the origin lies.
The equation of the other bisector is found by taking the negative sign:
\( \frac{3x - 4y + 10}{5} = - \left( \frac{-5x + 12y + 10}{13} \right) \)
\( \implies 13(3x - 4y + 10) = -5(-5x + 12y + 10) \)
\( \implies 39x - 52y + 130 = 25x - 60y - 50 \)
\( \implies 39x - 25x - 52y + 60y + 130 + 50 = 0 \)
\( \implies 14x + 8y + 180 = 0 \)
\( \implies 7x + 4y + 90 = 0 \)
(ii) Given lines are:
\( 12x - 5y + 3 = 0 \) ...(1)
To ensure the constant term is positive for the second line, we rewrite \( 4x + 3y - 2 = 0 \) as \( -4x - 3y + 2 = 0 \) ...(2)
The equations of the angular bisectors are given by:
\( \frac{12x - 5y + 3}{\sqrt{12^2 + (-5)^2}} = \pm \frac{-4x - 3y + 2}{\sqrt{(-4)^2 + (-3)^2}} \)
\( \implies \frac{12x - 5y + 3}{\sqrt{144 + 25}} = \pm \frac{-4x - 3y + 2}{\sqrt{16 + 9}} \)
\( \implies \frac{12x - 5y + 3}{13} = \pm \frac{-4x - 3y + 2}{5} \)
For the bisector of the angle in which the origin lies, we take the positive sign (assuming positive constant terms):
\( \frac{12x - 5y + 3}{13} = \frac{-4x - 3y + 2}{5} \)
\( \implies 5(12x - 5y + 3) = 13(-4x - 3y + 2) \)
\( \implies 60x - 25y + 15 = -52x - 39y + 26 \)
\( \implies 60x + 52x - 25y + 39y + 15 - 26 = 0 \)
\( \implies 112x + 14y - 11 = 0 \)
This is the equation of the bisector where the origin lies.
The equation of the other bisector is found by taking the negative sign:
\( \frac{12x - 5y + 3}{13} = - \left( \frac{-4x - 3y + 2}{5} \right) \)
\( \implies 5(12x - 5y + 3) = -13(-4x - 3y + 2) \)
\( \implies 60x - 25y + 15 = 52x + 39y - 26 \)
\( \implies 60x - 52x - 25y - 39y + 15 + 26 = 0 \)
\( \implies 8x - 64y + 41 = 0 \)
In simple words: We find the equations for two lines that cut the angle between the given lines exactly in half. For part (i), these bisector lines are \( 4x - 7y + 5 = 0 \) and \( 7x + 4y + 90 = 0 \). For part (ii), the bisector lines are \( 112x + 14y - 11 = 0 \) and \( 8x - 64y + 41 = 0 \). One of these bisectors will always pass through the origin.

🎯 Exam Tip: When finding angle bisectors, ensure the constant terms (C) in both line equations are positive before applying the formula. If they are not, multiply the entire equation by -1 to make them positive.

 

Question 2. Find the equations of the bisectors of the angles between 4x + 3y - 4 = 0 and 12x + 5y - 3 = 0. Show that these bisectors are at right angles to each other.

Answer:
The given lines are:
\( 4x + 3y - 4 = 0 \) ...(1)
\( 12x + 5y - 3 = 0 \) ...(2)
To make constant terms positive, we can rewrite them as:
\( -4x - 3y + 4 = 0 \) (Let's call this \( L_1' \))
\( -12x - 5y + 3 = 0 \) (Let's call this \( L_2' \))
The equations of the angular bisectors are given by:
\( \frac{4x + 3y - 4}{\sqrt{4^2 + 3^2}} = \pm \frac{12x + 5y - 3}{\sqrt{12^2 + 5^2}} \)
\( \implies \frac{4x + 3y - 4}{\sqrt{16 + 9}} = \pm \frac{12x + 5y - 3}{\sqrt{144 + 25}} \)
\( \implies \frac{4x + 3y - 4}{5} = \pm \frac{12x + 5y - 3}{13} \)
Taking the positive sign for the first bisector:
\( 13(4x + 3y - 4) = 5(12x + 5y - 3) \)
\( \implies 52x + 39y - 52 = 60x + 25y - 15 \)
\( \implies 60x - 52x + 25y - 39y - 15 + 52 = 0 \)
\( \implies 8x - 14y + 37 = 0 \) ...(3)
The slope of line (3), \( m_1 = - \frac{8}{-14} = \frac{4}{7} \).
Taking the negative sign for the second bisector:
\( 13(4x + 3y - 4) = -5(12x + 5y - 3) \)
\( \implies 52x + 39y - 52 = -60x - 25y + 15 \)
\( \implies 52x + 60x + 39y + 25y - 52 - 15 = 0 \)
\( \implies 112x + 64y - 67 = 0 \) ...(4)
The slope of line (4), \( m_2 = - \frac{112}{64} = - \frac{7}{4} \).
Now, we check if the bisectors are at right angles to each other by multiplying their slopes:
\( m_1 m_2 = \left( \frac{4}{7} \right) \times \left( - \frac{7}{4} \right) \)
\( \implies m_1 m_2 = -1 \)
Since the product of their slopes is -1, the two bisectors are perpendicular to each other. This is a property of angle bisectors that they are always perpendicular.
In simple words: We found two lines that divide the angles between the given lines in half. These new lines are \( 8x - 14y + 37 = 0 \) and \( 112x + 64y - 67 = 0 \). When we multiply the slopes of these two bisector lines, the answer is -1. This means the lines cross each other at a perfect 90-degree angle, or they are perpendicular.

🎯 Exam Tip: Remember that the angle bisectors of two lines are always perpendicular to each other. This fact can be used as a quick check for your calculations.

 

Question 3. Find the locus of a point which moves so that the perpendiculars drawn from it to the two straight lines 3x + 4y = 5, 12x – 5y = 13 are equal.

Answer:
Let the given lines be:
\( 3x + 4y - 5 = 0 \) ...(1)
\( 12x - 5y - 13 = 0 \) ...(2)
Let \( P(x, y) \) be any point whose locus we need to find.
The condition states that the perpendicular distance from \( P(x, y) \) to line (1) is equal to the perpendicular distance from \( P(x, y) \) to line (2).
Using the formula for perpendicular distance from a point \( (x_1, y_1) \) to a line \( Ax + By + C = 0 \), which is \( \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}} \):
\( \frac{|3x + 4y - 5|}{\sqrt{3^2 + 4^2}} = \frac{|12x - 5y - 13|}{\sqrt{12^2 + (-5)^2}} \)
\( \implies \frac{|3x + 4y - 5|}{\sqrt{9 + 16}} = \frac{|12x - 5y - 13|}{\sqrt{144 + 25}} \)
\( \implies \frac{|3x + 4y - 5|}{5} = \frac{|12x - 5y - 13|}{13} \)
To remove the absolute values, we consider both positive and negative cases:
\( \frac{3x + 4y - 5}{5} = \pm \frac{12x - 5y - 13}{13} \)
Taking the positive sign:
\( 13(3x + 4y - 5) = 5(12x - 5y - 13) \)
\( \implies 39x + 52y - 65 = 60x - 25y - 65 \)
\( \implies 60x - 39x - 25y - 52y - 65 + 65 = 0 \)
\( \implies 21x - 77y = 0 \)
\( \implies 3x - 11y = 0 \)
This is one equation for the locus.
Taking the negative sign:
\( 13(3x + 4y - 5) = -5(12x - 5y - 13) \)
\( \implies 39x + 52y - 65 = -60x + 25y + 65 \)
\( \implies 39x + 60x + 52y - 25y - 65 - 65 = 0 \)
\( \implies 99x + 27y - 130 = 0 \)
This is the second equation for the locus. The locus of the point consists of two straight lines, which are the angle bisectors of the given lines.
In simple words: We are looking for all the points that are the same distance away from two given straight lines. To find these points, we use the distance formula and set the distances equal to each other. This gives us two possible equations: \( 3x - 11y = 0 \) and \( 99x + 27y - 130 = 0 \). These two equations represent the paths (locus) of all such points.

🎯 Exam Tip: The locus of a point equidistant from two lines is always a pair of straight lines, which are the bisectors of the angles between the two given lines. Remember to consider both positive and negative signs when equating the distance formulas to get both bisectors.

 

Question 4. Find the equations of the lines bisecting the angles between the lines 4x – 3y + 12 = 0 and 12x + 5y = 20. Find, without using the tables, the tangent of an angle between one of these bisectors and one of the origin lines.

Answer:
The given lines are:
\( 4x - 3y + 12 = 0 \) ...(1)
\( 12x + 5y - 20 = 0 \) ...(2)
The equations of the angular bisectors are given by:
\( \frac{4x - 3y + 12}{\sqrt{4^2 + (-3)^2}} = \pm \frac{12x + 5y - 20}{\sqrt{12^2 + 5^2}} \)
\( \implies \frac{4x - 3y + 12}{\sqrt{16 + 9}} = \pm \frac{12x + 5y - 20}{\sqrt{144 + 25}} \)
\( \implies \frac{4x - 3y + 12}{5} = \pm \frac{12x + 5y - 20}{13} \)
Taking the positive sign for the first bisector:
\( 13(4x - 3y + 12) = 5(12x + 5y - 20) \)
\( \implies 52x - 39y + 156 = 60x + 25y - 100 \)
\( \implies 60x - 52x + 25y + 39y - 100 - 156 = 0 \)
\( \implies 8x + 64y - 256 = 0 \)
\( \implies x + 8y - 32 = 0 \) ...(3)
Taking the negative sign for the second bisector:
\( 13(4x - 3y + 12) = -5(12x + 5y - 20) \)
\( \implies 52x - 39y + 156 = -60x - 25y + 100 \)
\( \implies 52x + 60x - 39y + 25y + 156 - 100 = 0 \)
\( \implies 112x - 14y + 56 = 0 \)
\( \implies 8x - y + 4 = 0 \) ...(4)
Now, we find the tangent of an angle between one of these bisectors and one of the original lines. Let's choose bisector (3) and line (1).
Slope of bisector (3) \( x + 8y - 32 = 0 \), \( m_1 = - \frac{1}{8} \).
Slope of line (1) \( 4x - 3y + 12 = 0 \), \( m_2 = - \frac{4}{-3} = \frac{4}{3} \).
Let \( \theta \) be the angle between them. The tangent of the angle is given by:
\( \tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right| \)
\( \implies \tan \theta = \left| \frac{- \frac{1}{8} - \frac{4}{3}}{1 + \left( - \frac{1}{8} \right) \left( \frac{4}{3} \right)} \right| \)
\( \implies \tan \theta = \left| \frac{\frac{-3 - 32}{24}}{1 - \frac{4}{24}} \right| \)
\( \implies \tan \theta = \left| \frac{- \frac{35}{24}}{\frac{24 - 4}{24}} \right| \)
\( \implies \tan \theta = \left| \frac{- \frac{35}{24}}{\frac{20}{24}} \right| \)
\( \implies \tan \theta = \left| - \frac{35}{20} \right| \)
\( \implies \tan \theta = \frac{7}{4} \)
Alternatively, let's choose bisector (4) and line (2).
Slope of bisector (4) \( 8x - y + 4 = 0 \), \( m_3 = - \frac{8}{-1} = 8 \).
Slope of line (2) \( 12x + 5y - 20 = 0 \), \( m_4 = - \frac{12}{5} \).
Let \( \phi \) be the angle between them. The tangent of the angle is given by:
\( \tan \phi = \left| \frac{m_3 - m_4}{1 + m_3 m_4} \right| \)
\( \implies \tan \phi = \left| \frac{8 - \left( - \frac{12}{5} \right)}{1 + 8 \left( - \frac{12}{5} \right)} \right| \)
\( \implies \tan \phi = \left| \frac{8 + \frac{12}{5}}{1 - \frac{96}{5}} \right| \)
\( \implies \tan \phi = \left| \frac{\frac{40 + 12}{5}}{\frac{5 - 96}{5}} \right| \)
\( \implies \tan \phi = \left| \frac{52}{-91} \right| \)
\( \implies \tan \phi = \frac{52}{91} = \frac{4}{7} \)
In simple words: First, we find the two lines that cut the angles between the given lines in half. These bisector lines are \( x + 8y - 32 = 0 \) and \( 8x - y + 4 = 0 \). Then, we picked one bisector and one of the original lines and calculated the "tangent" of the angle between them. The tangent of the angle between bisector \( x + 8y - 32 = 0 \) and original line \( 4x - 3y + 12 = 0 \) is \( \frac{7}{4} \). For the other pair, bisector \( 8x - y + 4 = 0 \) and original line \( 12x + 5y - 20 = 0 \), the tangent of the angle is \( \frac{4}{7} \).

🎯 Exam Tip: When calculating the tangent of an angle between lines, be careful with the signs when using the slope formula. Also, remember to take the absolute value for the final tangent to represent the acute angle.

 

Question 5. Find the equations of the bisectors of the acute angle between the lines :
(i) 3x + 4y = 11 and 12x – 5y = 2;
(ii) 5x = 12y + 24 and 12x = 5y + 10.

Answer:
(i) Given lines:
\( 3x + 4y - 11 = 0 \) ...(L1)
\( 12x - 5y - 2 = 0 \) ...(L2)
To ensure the constant terms are positive, we rewrite them as:
\( -3x - 4y + 11 = 0 \) (for L1)
\( -12x + 5y + 2 = 0 \) (for L2)
The equations of the angular bisectors are given by:
\( \frac{-3x - 4y + 11}{\sqrt{(-3)^2 + (-4)^2}} = \pm \frac{-12x + 5y + 2}{\sqrt{(-12)^2 + 5^2}} \)
\( \implies \frac{-3x - 4y + 11}{5} = \pm \frac{-12x + 5y + 2}{13} \)
Taking the positive sign (Bisector 1):
\( 13(-3x - 4y + 11) = 5(-12x + 5y + 2) \)
\( \implies -39x - 52y + 143 = -60x + 25y + 10 \)
\( \implies -39x + 60x - 52y - 25y + 143 - 10 = 0 \)
\( \implies 21x - 77y + 133 = 0 \)
\( \implies 3x - 11y + 19 = 0 \) ...(B1)
Taking the negative sign (Bisector 2):
\( 13(-3x - 4y + 11) = -5(-12x + 5y + 2) \)
\( \implies -39x - 52y + 143 = 60x - 25y - 10 \)
\( \implies -39x - 60x - 52y + 25y + 143 + 10 = 0 \)
\( \implies -99x - 27y + 153 = 0 \)
\( \implies 99x + 27y - 153 = 0 \)
\( \implies 11x + 3y - 17 = 0 \) ...(B2)
To find the acute angle bisector, we find the angle between one of the original lines and one of the bisectors. Let's take line L1 (`-3x - 4y + 11 = 0`) and bisector B2 (`11x + 3y - 17 = 0`).
Slope of L1, \( m_1 = - \frac{-3}{-4} = - \frac{3}{4} \).
Slope of B2, \( m_2 = - \frac{11}{3} \).
\( \tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right| = \left| \frac{- \frac{3}{4} - \left( - \frac{11}{3} \right)}{1 + \left( - \frac{3}{4} \right) \left( - \frac{11}{3} \right)} \right| \)
\( \implies \tan \theta = \left| \frac{- \frac{3}{4} + \frac{11}{3}}{1 + \frac{11}{4}} \right| \)
\( \implies \tan \theta = \left| \frac{\frac{-9 + 44}{12}}{\frac{4 + 11}{4}} \right| \)
\( \implies \tan \theta = \left| \frac{\frac{35}{12}}{\frac{15}{4}} \right| \)
\( \implies \tan \theta = \left| \frac{35}{12} \times \frac{4}{15} \right| \)
\( \implies \tan \theta = \left| \frac{7}{3} \times \frac{1}{3} \right| = \frac{7}{9} \)
Since \( \tan \theta = \frac{7}{9} < 1 \), this means \( \theta < 45^\circ \). Therefore, Bisector 2 (`11x + 3y - 17 = 0`) is the equation of the acute angle bisector.
(ii) Given lines:
\( 5x - 12y - 24 = 0 \) (rewritten from \( 5x = 12y + 24 \)) ...(L1)
\( 12x - 5y - 10 = 0 \) (rewritten from \( 12x = 5y + 10 \)) ...(L2)
To ensure constant terms are positive, we rewrite them as:
\( -5x + 12y + 24 = 0 \) (for L1)
\( -12x + 5y + 10 = 0 \) (for L2)
The equations of the angular bisectors are given by:
\( \frac{-5x + 12y + 24}{\sqrt{(-5)^2 + 12^2}} = \pm \frac{-12x + 5y + 10}{\sqrt{(-12)^2 + 5^2}} \)
\( \implies \frac{-5x + 12y + 24}{13} = \pm \frac{-12x + 5y + 10}{13} \)
Taking the positive sign (Bisector 1):
\( -5x + 12y + 24 = -12x + 5y + 10 \)
\( \implies -5x + 12x + 12y - 5y + 24 - 10 = 0 \)
\( \implies 7x + 7y + 14 = 0 \)
\( \implies x + y + 2 = 0 \) ...(B1)
Taking the negative sign (Bisector 2):
\( -5x + 12y + 24 = -(-12x + 5y + 10) \)
\( \implies -5x + 12y + 24 = 12x - 5y - 10 \)
\( \implies -5x - 12x + 12y + 5y + 24 + 10 = 0 \)
\( \implies -17x + 17y + 34 = 0 \)
\( \implies 17x - 17y - 34 = 0 \)
\( \implies x - y - 2 = 0 \) ...(B2)
To find the acute angle bisector, we take line L1 (`-5x + 12y + 24 = 0`) and bisector B2 (`x - y - 2 = 0`).
Slope of L1, \( m_1 = - \frac{-5}{12} = \frac{5}{12} \).
Slope of B2, \( m_2 = - \frac{1}{-1} = 1 \).
\( \tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right| = \left| \frac{\frac{5}{12} - 1}{1 + \left( \frac{5}{12} \right) (1)} \right| \)
\( \implies \tan \theta = \left| \frac{\frac{5 - 12}{12}}{\frac{12 + 5}{12}} \right| \)
\( \implies \tan \theta = \left| \frac{- \frac{7}{12}}{\frac{17}{12}} \right| \)
\( \implies \tan \theta = \left| - \frac{7}{17} \right| = \frac{7}{17} \)
Since \( \tan \theta = \frac{7}{17} < 1 \), this means \( \theta < 45^\circ \). Therefore, Bisector 2 (`x - y - 2 = 0`) is the equation of the acute angle bisector.
In simple words: For each pair of lines, we first found two lines that divide the angles between them. Then, we picked one original line and one bisector, and calculated the tangent of the angle between them. If this tangent was less than 1, it meant the angle was less than 45 degrees, making that bisector the 'acute angle bisector'. For part (i), the acute angle bisector is \( 11x + 3y - 17 = 0 \). For part (ii), it is \( x - y - 2 = 0 \).

🎯 Exam Tip: To identify the acute angle bisector, calculate \( \tan \theta \) between one of the original lines and one of the bisectors. If \( |\tan \theta| < 1 \), that bisector forms an acute angle with the original line, making it the acute angle bisector.

 

Question 6. Prove that the perpendiculars drawn from any point of the line 2x + 11y = 5 to the lines 24x + 7y = 20 and 4x – 3y = 2 are equal in length.

Answer:
Let the given lines be:
Line (1): \( 2x + 11y - 5 = 0 \)
Line (2): \( 24x + 7y - 20 = 0 \)
Line (3): \( 4x - 3y - 2 = 0 \)
We need to show that the perpendicular distances from any point on line (1) to lines (2) and (3) are equal.
Let's choose a convenient point on line (1). If we set \( x = 0 \) in line (1):
\( 2(0) + 11y - 5 = 0 \)
\( \implies 11y = 5 \)
\( \implies y = \frac{5}{11} \)
So, a point on line (1) is \( \left( 0, \frac{5}{11} \right) \).
Now, calculate the length of the perpendicular from this point \( \left( 0, \frac{5}{11} \right) \) to line (2) \( 24x + 7y - 20 = 0 \):
\( d_2 = \frac{|24(0) + 7\left(\frac{5}{11}\right) - 20|}{\sqrt{24^2 + 7^2}} \)
\( \implies d_2 = \frac{|\frac{35}{11} - 20|}{\sqrt{576 + 49}} \)
\( \implies d_2 = \frac{|\frac{35 - 220}{11}|}{\sqrt{625}} \)
\( \implies d_2 = \frac{|-\frac{185}{11}|}{25} \)
\( \implies d_2 = \frac{185}{11 \times 25} = \frac{37 \times 5}{11 \times 5 \times 5} = \frac{37}{55} \)
Next, calculate the length of the perpendicular from the same point \( \left( 0, \frac{5}{11} \right) \) to line (3) \( 4x - 3y - 2 = 0 \):
\( d_3 = \frac{|4(0) - 3\left(\frac{5}{11}\right) - 2|}{\sqrt{4^2 + (-3)^2}} \)
\( \implies d_3 = \frac{|-\frac{15}{11} - 2|}{\sqrt{16 + 9}} \)
\( \implies d_3 = \frac{|\frac{-15 - 22}{11}|}{\sqrt{25}} \)
\( \implies d_3 = \frac{|-\frac{37}{11}|}{5} \)
\( \implies d_3 = \frac{37}{11 \times 5} = \frac{37}{55} \)
Since \( d_2 = d_3 = \frac{37}{55} \), the perpendiculars drawn from a point on line (1) to lines (2) and (3) are equal in length. This proves the statement. A key insight is that line (1) is actually one of the angle bisectors of lines (2) and (3).
In simple words: We took a point from the first line, which was \( (0, \frac{5}{11}) \). Then, we measured how far this point was from the second line and the third line. Both distances turned out to be exactly the same, \( \frac{37}{55} \). This shows that any point on the first line is equally far from the other two lines, which proves the statement.

🎯 Exam Tip: To prove a point is equidistant from two lines, select a simple point (like an intercept) on the reference line, then calculate perpendicular distances to the other two lines. If the distances are equal, the proof is complete.

 

Question 7. A triangle is formed by the lines whose equations are AB : x + y - 5 = 0, BC : x + 7y - 7 = 0, and CA : 7x + y + 14 = 0 . Find (i) the bisector of the interior angle at B, and (ii) the bisector of the exterior angle at C.

Answer:
The lines forming the triangle are:
AB: \( x + y - 5 = 0 \) ...(1)
BC: \( x + 7y - 7 = 0 \) ...(2)
CA: \( 7x + y + 14 = 0 \) ...(3)
To find the vertices:
Vertex A: Intersection of AB (1) and CA (3).
\( x + y = 5 \)
\( 7x + y = -14 \)
Subtracting the first equation from the second: \( 6x = -19 \implies x = - \frac{19}{6} \).
Substitute \( x \) back into \( x + y = 5 \): \( - \frac{19}{6} + y = 5 \implies y = 5 + \frac{19}{6} = \frac{30 + 19}{6} = \frac{49}{6} \).
So, \( A \left( - \frac{19}{6}, \frac{49}{6} \right) \).
Vertex B: Intersection of AB (1) and BC (2).
\( x + y = 5 \)
\( x + 7y = 7 \)
Subtracting the first equation from the second: \( 6y = 2 \implies y = \frac{1}{3} \).
Substitute \( y \) back into \( x + y = 5 \): \( x + \frac{1}{3} = 5 \implies x = 5 - \frac{1}{3} = \frac{15 - 1}{3} = \frac{14}{3} \).
So, \( B \left( \frac{14}{3}, \frac{1}{3} \right) \).
Vertex C: Intersection of BC (2) and CA (3).
\( x + 7y = 7 \)
\( 7x + y = -14 \)
Multiply the first equation by 7: \( 7x + 49y = 49 \).
Subtract the second equation from this: \( 48y = 63 \implies y = \frac{63}{48} = \frac{21}{16} \).
Substitute \( y \) back into \( x + 7y = 7 \): \( x + 7 \left( \frac{21}{16} \right) = 7 \implies x + \frac{147}{16} = 7 \implies x = 7 - \frac{147}{16} = \frac{112 - 147}{16} = - \frac{35}{16} \).
So, \( C \left( - \frac{35}{16}, \frac{21}{16} \right) \).
(i) Bisector of the interior angle at B:
Angle B is formed by lines AB (\( x + y - 5 = 0 \)) and BC (\( x + 7y - 7 = 0 \)).
To find the interior bisector, we need to ensure the constant terms for the lines are positive. Rewritten lines are:
\( -(x + y - 5) = 0 \implies -x - y + 5 = 0 \)
\( -(x + 7y - 7) = 0 \implies -x - 7y + 7 = 0 \)
The bisector equations are given by:
\( \frac{-x - y + 5}{\sqrt{(-1)^2 + (-1)^2}} = \pm \frac{-x - 7y + 7}{\sqrt{(-1)^2 + (-7)^2}} \)
\( \implies \frac{-x - y + 5}{\sqrt{2}} = \pm \frac{-x - 7y + 7}{\sqrt{50}} \)
\( \implies \frac{-x - y + 5}{\sqrt{2}} = \pm \frac{-x - 7y + 7}{5\sqrt{2}} \)
\( \implies 5(-x - y + 5) = \pm (-x - 7y + 7) \)
Taking the positive sign:
\( -5x - 5y + 25 = -x - 7y + 7 \)
\( \implies -5x + x - 5y + 7y + 25 - 7 = 0 \)
\( \implies -4x + 2y + 18 = 0 \)
\( \implies 2x - y - 9 = 0 \) ...(B1)
Taking the negative sign:
\( -5x - 5y + 25 = -(-x - 7y + 7) \)
\( \implies -5x - 5y + 25 = x + 7y - 7 \)
\( \implies -5x - x - 5y - 7y + 25 + 7 = 0 \)
\( \implies -6x - 12y + 32 = 0 \)
\( \implies 3x + 6y - 16 = 0 \) ...(B2)
To determine the interior bisector of angle B, we check if points A and C lie on opposite sides of the bisector.
Test Bisector (B2): \( 3x + 6y - 16 = 0 \)
For A\( \left( - \frac{19}{6}, \frac{49}{6} \right) \):
\( 3 \left( - \frac{19}{6} \right) + 6 \left( \frac{49}{6} \right) - 16 = - \frac{19}{2} + 49 - 16 = -9.5 + 33 = 23.5 > 0 \)
For C\( \left( - \frac{35}{16}, \frac{21}{16} \right) \):
\( 3 \left( - \frac{35}{16} \right) + 6 \left( \frac{21}{16} \right) - 16 = - \frac{105}{16} + \frac{126}{16} - \frac{256}{16} = \frac{-105 + 126 - 256}{16} = \frac{-235}{16} < 0 \)
Since A and C yield opposite signs, bisector B2 (`3x + 6y - 16 = 0`) is the interior bisector of angle B.
(ii) Bisector of the exterior angle at C:
Angle C is formed by lines BC (\( x + 7y - 7 = 0 \)) and CA (\( 7x + y + 14 = 0 \)).
The constant terms for these lines are \( -7 \) and \( +14 \). We adjust for positive constant terms:
\( -(x + 7y - 7) = 0 \implies -x - 7y + 7 = 0 \)
\( 7x + y + 14 = 0 \)
The bisector equations are given by:
\( \frac{-x - 7y + 7}{\sqrt{(-1)^2 + (-7)^2}} = \pm \frac{7x + y + 14}{\sqrt{7^2 + 1^2}} \)
\( \implies \frac{-x - 7y + 7}{\sqrt{50}} = \pm \frac{7x + y + 14}{\sqrt{50}} \)
\( \implies -x - 7y + 7 = \pm (7x + y + 14) \)
Taking the positive sign:
\( -x - 7y + 7 = 7x + y + 14 \)
\( \implies -x - 7x - 7y - y + 7 - 14 = 0 \)
\( \implies -8x - 8y - 7 = 0 \)
\( \implies 8x + 8y + 7 = 0 \) ...(B3)
Taking the negative sign:
\( -x - 7y + 7 = -(7x + y + 14) \)
\( \implies -x - 7y + 7 = -7x - y - 14 \)
\( \implies -x + 7x - 7y + y + 7 + 14 = 0 \)
\( \implies 6x - 6y + 21 = 0 \)
\( \implies 2x - 2y + 7 = 0 \) ...(B4)
To determine the interior bisector of angle C, we check if points A and B lie on opposite sides of the bisector.
Test Bisector (B4): \( 2x - 2y + 7 = 0 \)
For A\( \left( - \frac{19}{6}, \frac{49}{6} \right) \):
\( 2 \left( - \frac{19}{6} \right) - 2 \left( \frac{49}{6} \right) + 7 = - \frac{19}{3} - \frac{49}{3} + \frac{21}{3} = \frac{-19 - 49 + 21}{3} = \frac{-47}{3} < 0 \)
For B\( \left( \frac{14}{3}, \frac{1}{3} \right) \):
\( 2 \left( \frac{14}{3} \right) - 2 \left( \frac{1}{3} \right) + 7 = \frac{28}{3} - \frac{2}{3} + \frac{21}{3} = \frac{28 - 2 + 21}{3} = \frac{47}{3} > 0 \)
Since A and B yield opposite signs, bisector B4 (`2x - 2y + 7 = 0`) is the interior bisector of angle C.
Therefore, bisector B3 (`8x + 8y + 7 = 0`) is the exterior bisector of angle C.
In simple words: First, we found the corner points (vertices) of the triangle. Then, for part (i), we found the line that divides angle B exactly in half, pointing inwards. This line is \( 3x + 6y - 16 = 0 \). For part (ii), we found the line that divides the angle outside of angle C in half, which is \( 8x + 8y + 7 = 0 \). We checked which bisector was "inside" the triangle by testing if the other two vertices fell on opposite sides of it.

🎯 Exam Tip: To find an interior angle bisector, ensure the constant terms of the lines forming the angle are positive. Then, check if the other two vertices of the triangle lie on opposite sides of the bisector line; if they do, it's the interior bisector.

 

Question 8. Find the centre of the inscribed circle of the triangle the equations of whose sides are y - 15 = 0, 12y + 5x = 0 and 4y - 3x = 0.

Answer:
The equations of the sides of the triangle are:
Side 1: \( y - 15 = 0 \) ...(L1)
Side 2: \( 5x + 12y = 0 \) ...(L2)
Side 3: \( 4y - 3x = 0 \) ...(L3)
First, find the vertices of the triangle:
Vertex A: Intersection of L2 and L3.
\( 5x + 12y = 0 \)
\( -3x + 4y = 0 \)
Multiply the second equation by 3: \( -9x + 12y = 0 \). Subtract this from the first equation: \( 14x = 0 \implies x = 0 \).
Substitute \( x = 0 \) into \( -3x + 4y = 0 \): \( 4y = 0 \implies y = 0 \). So, \( A(0, 0) \).
Vertex B: Intersection of L1 and L3.
\( y = 15 \)
\( -3x + 4y = 0 \)
Substitute \( y = 15 \) into the second equation: \( -3x + 4(15) = 0 \implies -3x + 60 = 0 \implies 3x = 60 \implies x = 20 \). So, \( B(20, 15) \).
Vertex C: Intersection of L1 and L2.
\( y = 15 \)
\( 5x + 12y = 0 \)
Substitute \( y = 15 \) into the second equation: \( 5x + 12(15) = 0 \implies 5x + 180 = 0 \implies 5x = -180 \implies x = -36 \). So, \( C(-36, 15) \).
The center of the inscribed circle (incenter) is the intersection of the internal angle bisectors. Let's find the internal bisector for angle A and angle B.
**Bisector of Angle A (between L2 and L3):**
Lines are \( 5x + 12y = 0 \) and \( -3x + 4y = 0 \). Since vertex A is the origin, we don't have constant terms to adjust. We use the formula directly:
\( \frac{5x + 12y}{\sqrt{5^2 + 12^2}} = \pm \frac{-3x + 4y}{\sqrt{(-3)^2 + 4^2}} \)
\( \implies \frac{5x + 12y}{13} = \pm \frac{-3x + 4y}{5} \)
Taking the positive sign:
\( 5(5x + 12y) = 13(-3x + 4y) \)
\( \implies 25x + 60y = -39x + 52y \)
\( \implies 25x + 39x + 60y - 52y = 0 \)
\( \implies 64x + 8y = 0 \)
\( \implies 8x + y = 0 \) ...(B_A1)
Taking the negative sign:
\( 5(5x + 12y) = -13(-3x + 4y) \)
\( \implies 25x + 60y = 39x - 52y \)
\( \implies 25x - 39x + 60y + 52y = 0 \)
\( \implies -14x + 112y = 0 \)
\( \implies x - 8y = 0 \) ...(B_A2)
To determine the internal bisector, check if vertices B(20,15) and C(-36,15) lie on opposite sides.
Test \( 8x + y = 0 \):
For B(20,15): \( 8(20) + 15 = 160 + 15 = 175 > 0 \).
For C(-36,15): \( 8(-36) + 15 = -288 + 15 = -273 < 0 \).
Since they have opposite signs, \( 8x + y = 0 \) is the internal bisector of angle A. Let's call this B_A.
**Bisector of Angle B (between L1 and L3):**
Lines are \( y - 15 = 0 \) and \( -3x + 4y = 0 \). To make constant terms positive for L1, use \( -y + 15 = 0 \). For L3, it's already 0 so we can use \( -3x+4y=0 \).
\( \frac{-y + 15}{\sqrt{0^2 + (-1)^2}} = \pm \frac{-3x + 4y}{\sqrt{(-3)^2 + 4^2}} \)
\( \implies \frac{-y + 15}{1} = \pm \frac{-3x + 4y}{5} \)
Taking the positive sign:
\( 5(-y + 15) = -3x + 4y \)
\( \implies -5y + 75 = -3x + 4y \)
\( \implies 3x - 5y - 4y + 75 = 0 \)
\( \implies 3x - 9y + 75 = 0 \)
\( \implies x - 3y + 25 = 0 \) ...(B_B1)
Taking the negative sign:
\( 5(-y + 15) = -(-3x + 4y) \)
\( \implies -5y + 75 = 3x - 4y \)
\( \implies -3x - 5y + 4y + 75 = 0 \)
\( \implies -3x - y + 75 = 0 \)
\( \implies 3x + y - 75 = 0 \) ...(B_B2)
To determine the internal bisector, check if vertices A(0,0) and C(-36,15) lie on opposite sides.
Test \( x - 3y + 25 = 0 \):
For A(0,0): \( 0 - 3(0) + 25 = 25 > 0 \).
For C(-36,15): \( -36 - 3(15) + 25 = -36 - 45 + 25 = -56 < 0 \).
Since they have opposite signs, \( x - 3y + 25 = 0 \) is the internal bisector of angle B. Let's call this B_B.
**Finding the Incenter (intersection of B_A and B_B):**
Solve B_A: \( 8x + y = 0 \implies y = -8x \)
Solve B_B: \( x - 3y + 25 = 0 \)
Substitute \( y = -8x \) into B_B:
\( x - 3(-8x) + 25 = 0 \)
\( \implies x + 24x + 25 = 0 \)
\( \implies 25x = -25 \)
\( \implies x = -1 \)
Now find \( y \): \( y = -8(-1) = 8 \).
The incenter (center of the inscribed circle) is \( (-1, 8) \).
We can verify this by checking if it lies on the third internal bisector (for angle C).
Bisector of Angle C (between L1 and L2):
Lines are \( y - 15 = 0 \) and \( 5x + 12y = 0 \). To make constant terms positive, use \( -y + 15 = 0 \) and \( 5x+12y=0 \).
\( \frac{-y + 15}{\sqrt{0^2 + (-1)^2}} = \pm \frac{5x + 12y}{\sqrt{5^2 + 12^2}} \)
\( \implies \frac{-y + 15}{1} = \pm \frac{5x + 12y}{13} \)
Taking the positive sign:
\( 13(-y + 15) = 5x + 12y \)
\( \implies -13y + 195 = 5x + 12y \)
\( \implies 5x + 25y - 195 = 0 \)
\( \implies x + 5y - 39 = 0 \) ...(B_C1)
Test \( x + 5y - 39 = 0 \) for A(0,0) and B(20,15).
For A(0,0): \( 0 + 5(0) - 39 = -39 < 0 \).
For B(20,15): \( 20 + 5(15) - 39 = 20 + 75 - 39 = 56 > 0 \).
So, \( x + 5y - 39 = 0 \) is the internal bisector of angle C.
Check if \( (-1, 8) \) lies on \( x + 5y - 39 = 0 \):
\( -1 + 5(8) - 39 = -1 + 40 - 39 = 0 \). Yes, it does.
Thus, all three internal bisectors intersect at \( (-1, 8) \), which is the center of the inscribed circle.
In simple words: First, we found the three corners of the triangle using the given lines. Then, we found the equations for the lines that split each angle of the triangle exactly in half (these are called angle bisectors). We did this for two of the angles. Where these two angle-splitting lines meet, that's the special point we're looking for. By solving their equations, we found this point to be \( (-1, 8) \). This point is the center of the circle that can be drawn perfectly inside the triangle, touching all three sides.

🎯 Exam Tip: The incenter of a triangle is the point of concurrence of its three internal angle bisectors. You only need to find the intersection of any two internal angle bisectors to locate the incenter.

 

Question 9. The co-ordinates of A, B, C are respectively (- 4, 0),(0, 2) and (- 3, 2). Find (i) the equation of the straight line which bisects the angle CAB internally; (ii) the co-ordinates of the point where this straight line meets the straight line joining C to the middle point of AB.

Answer:
Given coordinates: \( A(-4, 0) \), \( B(0, 2) \), \( C(-3, 2) \).
(i) Equation of the straight line which bisects the angle CAB internally:
Angle CAB is formed by lines AB and AC.
Equation of line AB (joining \( A(-4, 0) \) and \( B(0, 2) \)):
\( \frac{y - 0}{x - (-4)} = \frac{2 - 0}{0 - (-4)} \)
\( \implies \frac{y}{x + 4} = \frac{2}{4} \)
\( \implies \frac{y}{x + 4} = \frac{1}{2} \)
\( \implies 2y = x + 4 \)
\( \implies x - 2y + 4 = 0 \) ...(L1)
Equation of line AC (joining \( A(-4, 0) \) and \( C(-3, 2) \)):
\( \frac{y - 0}{x - (-4)} = \frac{2 - 0}{-3 - (-4)} \)
\( \implies \frac{y}{x + 4} = \frac{2}{1} \)
\( \implies y = 2(x + 4) \)
\( \implies y = 2x + 8 \)
\( \implies 2x - y + 8 = 0 \) ...(L2)
The equations of the bisectors of angle CAB are:
\( \frac{x - 2y + 4}{\sqrt{1^2 + (-2)^2}} = \pm \frac{2x - y + 8}{\sqrt{2^2 + (-1)^2}} \)
\( \implies \frac{x - 2y + 4}{\sqrt{1 + 4}} = \pm \frac{2x - y + 8}{\sqrt{4 + 1}} \)
\( \implies \frac{x - 2y + 4}{\sqrt{5}} = \pm \frac{2x - y + 8}{\sqrt{5}} \)
\( \implies x - 2y + 4 = \pm (2x - y + 8) \)
Taking the positive sign (Bisector 1):
\( x - 2y + 4 = 2x - y + 8 \)
\( \implies x - 2x - 2y + y + 4 - 8 = 0 \)
\( \implies -x - y - 4 = 0 \)
\( \implies x + y + 4 = 0 \) ...(B1)
Taking the negative sign (Bisector 2):
\( x - 2y + 4 = -(2x - y + 8) \)
\( \implies x - 2y + 4 = -2x + y - 8 \)
\( \implies x + 2x - 2y - y + 4 + 8 = 0 \)
\( \implies 3x - 3y + 12 = 0 \)
\( \implies x - y + 4 = 0 \) ...(B2)
To find the internal bisector of angle CAB, we check if points B(0,2) and C(-3,2) lie on opposite sides of the bisector.
Test Bisector (B2): \( x - y + 4 = 0 \)
For B(0,2): \( 0 - 2 + 4 = 2 > 0 \).
For C(-3,2): \( -3 - 2 + 4 = -1 < 0 \).
Since B and C yield opposite signs, \( x - y + 4 = 0 \) is the internal bisector of angle CAB.
(ii) Coordinates of the point where this straight line meets the straight line joining C to the middle point of AB:
Middle point of AB, let's call it D:
\( D = \left( \frac{-4 + 0}{2}, \frac{0 + 2}{2} \right) = (-2, 1) \).
Now, find the equation of the straight line CD (joining \( C(-3, 2) \) and \( D(-2, 1) \)):
\( \frac{y - 2}{x - (-3)} = \frac{1 - 2}{-2 - (-3)} \)
\( \implies \frac{y - 2}{x + 3} = \frac{-1}{1} \)
\( \implies y - 2 = -(x + 3) \)
\( \implies y - 2 = -x - 3 \)
\( \implies x + y + 1 = 0 \) ...(L_CD)
We need to find the intersection of the internal bisector of angle CAB (B2: \( x - y + 4 = 0 \)) and line CD (\( x + y + 1 = 0 \)).
Add the two equations:
\( (x - y + 4) + (x + y + 1) = 0 \)
\( \implies 2x + 5 = 0 \)
\( \implies x = - \frac{5}{2} \)
Substitute \( x = - \frac{5}{2} \) into \( x + y + 1 = 0 \):
\( - \frac{5}{2} + y + 1 = 0 \)
\( \implies y = \frac{5}{2} - 1 = \frac{3}{2} \)
The coordinates of the intersection point are \( \left( - \frac{5}{2}, \frac{3}{2} \right) \).
In simple words: We are given three points A, B, and C that form a triangle. For part (i), we found the equation of the line that cuts the angle at A (angle CAB) exactly in half, pointing into the triangle. This line is \( x - y + 4 = 0 \). For part (ii), we found the middle point of the line AB, then found the equation of the line connecting this middle point to C. Finally, we found where this new line crosses the angle-splitting line from part (i). This crossing point is \( \left( - \frac{5}{2}, \frac{3}{2} \right) \).

🎯 Exam Tip: When finding internal angle bisectors, always verify by checking if the other vertices of the triangle lie on opposite sides of the bisector line. For coordinate geometry problems, carefully calculate midpoints and line equations to avoid arithmetic errors.

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