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Solution for Exercise 4: Relative position and angle between two lines. Distance from a point to a line (C7 – Math 10 Kite)
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Solving exercises Lesson 1, page 86, Math textbook 10 Kite, episode 2
Consider the relative positions of each of the following pairs of lines
a) \({d_1}:3x + 2y–5 = 0\) and \({d_2}:x – 4y + 1 = 0\) ;
b) \({d_3}:x – 2y + 3 = 0\) and \({d_4}: – {\rm{ }}2x + 4y + 10 = 0\) ;
c) \({d_5}:4x + 2y – 3 = 0\) and \({d_6}:\left\{ \begin{array}{l}x = – \frac{1}{2} + t\ \y = \frac{5}{2} – 2t\end{array} \right.\)
Solution method
Solving the system of intersection equations:
System of equations with solutions \( \Rightarrow \) intersecting each other
Parallel system of equations with no solutions \( \Rightarrow \)
System of equations with an infinite number of solutions \( \Rightarrow \) coincide
Solution guide
a) The coordinates of the intersection of two lines \({d_1},{d_2}\) is the solution of the system of equations:
\(\left\{ \begin{array}{l}3x + 2y – 5 = 0\\x – 4y + 1 = 0\end{array} \right.\Leftrightarrow \left\{ \begin{array}{ l}x = \frac{9}{7}\\y = \frac{4}{7}\end{array} \right.\)
The system of equations has only one solution, so the two lines intersect.
b) The coordinates of the intersection of two lines \({d_3},{d_4}\) is the solution of the system of equations:
\(\left\{ \begin{array}{l}x – 2y + 3 = 0\\ – 2x + 4y + 10 = 0\end{array} \right.\) .
The system of equations has no solution. So 2 lines are parallel to each other
c) The coordinates of the intersection of two lines \({d_5},{d_6}\) corresponding to t satisfy the equation:
\(4\left( { – \frac{1}{2} + t} \right) + 2\left( {\frac{5}{2} – 2t} \right) – 3 = 0 \Leftrightarrow 0t = 0\) .
This equation has a solution for all t. Hence \({d_5} \equiv {d_6}\).
Solve the exercise Exercise 2 page 86 Math textbook 10 Kite episode 2
Calculate the measure of the angle between two lines \({d_1}:2x–y + 5 = 0\) and \({d_2}:x – 3y + 3 = 0\).
Solution method
In the coordinate plane, for two lines \({\Delta _1}\) and \({\Delta _2}\) whose direction vectors are \(\overrightarrow {{n_1}} = {\rm{ }}\left( {{a_1};{\rm{ }}{b_1}} \right),{\rm{ }}\overrightarrow {{n_2}} {\rm{ }} = {\rm{ }} \left( {{a_2};{b_2}} \right)\) we have:
\(\cos \left( {{\Delta _1},{\Delta _2}} \right) = \left| {\cos \left( {\overrightarrow {{n_1}} ;\overrightarrow {{n_2}} } \right)} \right| = \frac{{\left| {{a_1}{a_2} + {b_1}{b_2}} \right|}}{{\sqrt {a_1^2 + b_1^2} .\ sqrt {a_2^2 + b_2^2} }}.\)
Solution guide
The normal vector of the line \({d_1}\) is: \(\overrightarrow {{n_1}} = \left( {2; – 1} \right)\)
The normal vector of the line \({d_2}\) is: \(\overrightarrow {{n_2}} = \left( {1; – 3} \right)\)
We have: \(\cos \left( {{d_1},{d_2}} \right) = \left| {\cos \left( {\overrightarrow {{n_1}} ;\overrightarrow {{n_2}} } \ right)} \right| = \frac{{\left| {2.1 + \left( { – 1} \right).\left( { – 3} \right)} \right|}}{{\sqrt {{ {\left( 2 \right)}^2} + {{\left( { – 1} \right)}^2}} .\sqrt {{1^2} + {{\left( { – 3} \ right)}^2}} }} = \frac{{\sqrt 2 }}{2}\)
So \(\left( {{d_1},{d_2}} \right) = {45^o}\)
Solve exercises Exercise 3 page 86 Math textbook 10 Kite episode 2
Calculate the distance from a point to a line in each of the following cases:
a)\(A\left( {1; – 2} \right){\rm{ }}v\`a {\rm{ }}{\Delta _1}:{\rm{ }}3x – y + { \rm{ }}4{\rm{ }} = {\rm{ }}0\) ;
b) B(-3; 2) and \({\Delta _2}:\left\{ \begin{array}{l}x = – 2 + t\\y = 1 – 2t\end{array} \right .\)
Solution method
In the coordinate plane Oxy, for the line \(\Delta \) there is the equation \({\rm{a}}x + by + c = 0\left( {{a^2} + {b^2} ) > 0} \right)\) and the point \(M\left( {{x_o};{y_0}} \right)\). The distance from the point M to the line \(\Delta \), denoted by \(d\left( {M,\Delta } \right)\) is calculated by the formula: \(d\left( {M, \Delta } \right) = \frac{{\left| {{\rm{a}}{x_o} + b{y_o} + c} \right|}}{{\sqrt {{a^2} + { b^2}} }}\)
Solution guide
a) The distance from point A to \({\Delta _1}\) is: \(d\left( {A,{\Delta _1}} \right) = \frac{{\left| {3.1 – 1. \left( { – 2} \right) + 4} \right|}}{{\sqrt {{3^2} + {{\left( { – 1} \right)}^2}} }} = \ frac{9}{{\sqrt {10} }}\)
b) The general equation of the line \({\Delta _2}\) is: \(2x + y + 3 = 0\)
The distance from point B to \({\Delta _2}\) is: \(d\left( {A,{\Delta _2}} \right) = \frac{{\left| {2.\left( { – 3} \right) + 1.2 + 3} \right|}}{{\sqrt {{2^2} + {1^2}} }} = \frac{1}{{\sqrt 5 }}\)
Solving exercises Lesson 4, page 86, Math textbook 10 Kite, episode 2
For what value of parameter m are the following two lines perpendicular?
\({\Delta _1}:mx – y + 1 = 0\) and \({\Delta _2}:2x – y + 3 = 0\).
Solution method
The two lines \({\Delta _1}\),\({\Delta _2}\) are perpendicular if and only \(\overrightarrow {{n_1}} ;\overrightarrow {{n_2}} \) are square. angle to each other
Solution guide
The normal vector of is: \(\overrightarrow {{n_1}} = \left( {m; – 1} \right)\)
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The normal vector of is: \(\overrightarrow {{n_2}} = \left( {2; – 1} \right)\)
So who the lines \({\Delta _1}\),\({\Delta _2}\) are perpendicular if and only \(\overrightarrow {{n_1}} ;\overrightarrow {{n_2}} \) perpendicular i.e. \(\overrightarrow {{n_1}} .\overrightarrow {{n_2}} = 0 \Leftrightarrow 2m + 1 = 0 \Leftrightarrow m = \frac{{ – 1}}{2}\)
Solve the exercises Exercise 5 page 86 Math textbook 10 Kite episode 2
Given three points A(2;- 1), B(1 ; 2) and C(4;- 2). Calculate the measure of angle BAC and the angle between two lines AB and AC.
Solution method
\(\cos \widehat {BAC} = \cos \left( {\overrightarrow {AB} ,\overrightarrow {AC} } \right)\)
\(\cos \left( {AB,AC} \right) = \left| {\cos \left( {\overrightarrow {AB} ,\overrightarrow {AC} } \right)} \right|\)
Solution guide
We have: \(\overrightarrow {AB} = \left( { – 1;3} \right);\overrightarrow {AC} = \left( {2; – 1} \right)\)
So \(\cos \widehat {BAC} = \cos \left( {\overrightarrow {AB} ,\overrightarrow {AC} } \right) = \frac{{ – 1.2 + 3.\left( { – 1} \ right)}}{{\sqrt {{{\left( { – 1} \right)}^2} + {3^2}} .\sqrt {{2^2} + {{\left( { – 1 } \right)}^2}} }} = \frac{{ – 1}}{{\sqrt 2 }} \Rightarrow \widehat {BAC} = {135^o}\)
So\(\cos \left( {AB,AC} \right) = \left| {\cos \left( {\overrightarrow {AB} ,\overrightarrow {AC} } \right)} \right| = \frac{ {\left| { – 1.2 + 3.\left( { – 1} \right)} \right|}}{{\sqrt {{{\left( { – 1} \right)}^2} + {3 ^2}} .\sqrt {{2^2} + {{\left( { – 1} \right)}^2}} }} = \frac{1}{{\sqrt 2 }} \Rightarrow \widehat {BAC} = {45^o}\)
Solution of Exercises Lesson 6, page 86, Math Textbook 10, Kite episode 2
Given three points A(2;4), B(-1; 2) and C(3;-1). Write the equation of the line that passes through B and is equidistant from A and C.
Solution method
In the coordinate plane Oxy, for the line \(\Delta \) there is the equation \({\rm{a}}x + by + c = 0\left( {{a^2} + {b^2} ) > 0} \right)\) and the point \(M\left( {{x_o};{y_0}} \right)\). The distance from the point M to the line \(\Delta \), denoted by \(d\left( {M,\Delta } \right)\) is calculated by the formula: \(d\left( {M, \Delta } \right) = \frac{{\left| {{\rm{a}}{x_o} + b{y_o} + c} \right|}}{{\sqrt {{a^2} + { b^2}} }}\)
Solution guide
Let \(\Delta \) be the line that passes through B and whose normal vector is \(\overrightarrow n = \left( {a;b} \right)\)
So the equation \(\Delta \) is: \(a\left( {x + 1} \right) + b\left( {y – 2} \right) = 0 \Leftrightarrow {\rm{a}}x + by + \left( {a – 2b} \right) = 0\)
We have: \(d\left( {A,\Delta } \right) = d\left( {C,\Delta } \right) \Leftrightarrow \frac{{\left| {3a + 2b} \right|} }{{\sqrt {{a^2} + {b^2}} }} = \frac{{\left| {4a – 3b} \right|}}{{\sqrt {{a^2} + { b^2}} }} \Leftrightarrow \left[\begin{array}{l}3a+2b=4a–3b\\3a+2b=–4a+3b\end{array}\right\Leftrightarrow\left[\begin{array}{l}a=5b\left(1\right)\\7a=b\left(2\right)\end{array}\right\)[\begin{array}{l}3a+2b=4a–3b\\3a+2b= –4a+3b\end{array}\right\Leftrightarrow\left[\begin{array}{l}a=5b\left(1\right)\\7a=b\left(2\right)\end{array}\right\)
From (1) we can choose a normal vector: \(\overrightarrow n = \left( {5;1} \right)\). So the equation for the line \(\Delta \) is: \(5x + y + 3 = 0\)
From (2) we can choose a normal vector: \(\overrightarrow n = \left( {1;7} \right)\). So the equation for the line \(\Delta \) is: \(x + 7y – 13 = 0\)
Solve the exercises Lesson 7 page 86 Math textbook 10 Kite episode 2
There are two ships A and B both starting from two berths, moving in a straight line at sea. On the radar display of the control station (referred to as the coordinate plane Oxy with units on the axes in kilometers), after starting t (hours) (\(t \ge 0\) ), position
of ship A whose coordinates are determined by the formula \(\left\{ \begin{array}{l}x = 3 – 35t\\y = – 4 + 25t\end{array} \right.\) , The position of ship B has coordinates (4 – 30t; 3 – 40t).
a) Calculate the cosine of the angle between the paths of two trains A and B.
b) How long after departure are the two closest trains?
c) If train A is at rest in the initial position and train B is moving, what is the shortest distance between the two trains?
Solution method
a) In the coordinate plane, for two lines \({\Delta _1}\) and \({\Delta _2}\) whose direction vectors are \(\overrightarrow {{u_1}} = {\, respectively. rm{ }}\left( {{a_1};{\rm{ }}{b_1}} \right),{\rm{ }}\overrightarrow {{u_2}} {\rm{ }} = {\rm{ }}\left( {{a_2};{b_2}} \right)\) we have:
\(\cos \left( {{\Delta _1},{\Delta _2}} \right) = \left| {\cos \left( {\overrightarrow {{u_1}} ;\overrightarrow {{u_2}} } \right)} \right| = \frac{{\left| {{a_1}{a_2} + {b_1}{b_2}} \right|}}{{\sqrt {a_1^2 + b_1^2} .\ sqrt {a_2^2 + b_2^2} }}.\)
b) Step 1: In the coordinate plane Oxy, for the line \(\Delta \) has the equation \({\rm{a}}x + by + c = 0\left( {{a^2} + {b^2} > 0} \right)\) and the point \(M\left( {{x_o};{y_0}} \right)\). The distance from the point M to the line \(\Delta \), denoted by \(d\left( {M,\Delta } \right)\) is calculated by the formula: \(d\left( {M, \Delta } \right) = \frac{{\left| {{\rm{a}}{x_o} + b{y_o} + c} \right|}}{{\sqrt {{a^2} + { b^2}} }}\)
Step 2: Evaluate by parameter t
c) In the coordinate plane Oxy, for the line \(\Delta \) there is the equation \({\rm{a}}x + by + c = 0\left( {{a^2} + {b^) 2} > 0} \right)\) and the point \(M\left( {{x_o};{y_0}} \right)\). The distance from the point M to the line \(\Delta \), denoted by \(d\left( {M,\Delta } \right)\) is calculated by the formula: \(d\left( {M, \Delta } \right) = \frac{{\left| {{\rm{a}}{x_o} + b{y_o} + c} \right|}}{{\sqrt {{a^2} + { b^2}} }}\)
Solution guide
a) Ship A moves in a vector direction \(\overrightarrow {{u_1}} = \left( { – 35;25} \right)\)
Ship B moves in the vector direction \(\overrightarrow {{u_2}} = \left( { – 30; – 40} \right)\)
Let \(\alpha \) be the angle between the two paths of the two ships, we have:
\(\cos \alpha = \left| {\cos \left( {\overrightarrow {{u_1}} ;\overrightarrow {{u_2}} } \right)} \right| = \frac{{\left| {\ left( { – 35} \right).\left( { – 30} \right) + 25.\left( { – 40} \right)} \right|}}{{\sqrt {{{\left( { – 35} \right)}^2} + {{25}^2}} .\sqrt {{{\left( { – 30} \right)}^2} + {{\left( { – 40} \ right)}^2}} }} = \frac{1}{{5\sqrt {74} }}.\)
b) After t hours, the position of train A is point M with coordinates: \(M\left( {3 – 35t; – 4 + 25t} \right)\)
After t hours, the position of train B is point N with coordinates: \(N\left( {4 – 30t;3 – 40t} \right)\)
Thus, \(\overrightarrow {MN} = \sqrt {{{\left( {1 + 5t} \right)}^2} + {{\left( {7 – 65t} \right)}^2}} = \sqrt {4250{t^2} – 900t + 50} = \sqrt {4250{{\left( {t – \frac{9}{{85}}} \right)}^2} + \frac{ {40}}{{17}}} \ge \sqrt {\frac{{40}}{{17}}} \approx 1.53\left( {km} \right)\)
The smallest MN is approximately 1.53km when \(t = \frac{9}{{85}}\)
So after \(\frac{9}{{85}}\) hours from the time of departure, the two trains are closest to each other and are 1.53km apart.
c) The initial position of ship A at \({M_o}\) corresponds to \(t = 0\) , then \({M_o}\left( {3; – 4} \right)\)
Train B moves in a straight line with normal vector \(\overrightarrow n = \left( {40; – 30} \right)\) and passes through the point \(K\left( {4;3} \right)\ ) The general equation of is: \(40\left( {x – 4} \right) – 30\left( {y – 3} \right) = 0 \Leftrightarrow 4x – 3y – 7 = 0\) \( \Delta \)
We have: \(d\left( {{M_o},\Delta } \right) = \frac{{\left| {4.3 – 3.\left( { – 4} \right) – 7} \right|} }{{\sqrt {{4^2} + {{\left( { – 3} \right)}^2}} }} = \frac{{17}}{5} = 3.4\left( { km} \right)\)
So if train A is at rest at the initial position and train B is moving, the shortest distance between the two trains is 3.4km.
