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## Solution of Exercise 3: Equation of a line (C7 – Math 10 Kite)

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### Solve the exercises Lesson 1, page 79, Math textbook 10 Kite, episode 2

Write the general equation of the line \(\Delta \) passing through the point A(-1; 2) and

a) There is a normal vector \(\overrightarrow n = \left( {3{\rm{ }};{\rm{ }}2} \right).\)

b) The direction vector is \(\overrightarrow u = \left( { – 2{\rm{ }};{\rm{ 3}}} \right).\)

**Solution method**

The general equation of the line\(\Delta \) passing through the point \({M_o}\left( {{x_o};{y_o}} \right)\) and getting \(\overrightarrow n = \left( { {\rm{a }};{\rm{ b}}} \right)\left( {\overrightarrow n \ne 0} \right)\) as the normal vector is: \(a\left( {x – {x_o}} \right) + b\left( {y – {y_o}} \right) = 0\)

**Solution guide**

a) The general equation of the line\(\Delta \) passing through the point \(A\left( { – 1;{\rm{ }}2} \right)\) and having the normal vector \(\overrightarrow n = \left( {3{\rm{ }};{\rm{ }}2} \right).\)is: \(3\left( {x + 1} \right) + 2\left( { { y – 2} \right) = 0 \Leftrightarrow 3x + 2y – 1 = 0\)

b) Since \(\Delta \) has a direction vector \(\overrightarrow u = \left( { – 2{\rm{ }};{\rm{ 3}}} \right).\) so the legal vector is the route of \(\Delta \) is \(\overrightarrow n = \left( {3{\rm{ }};{\rm{ }}2} \right).\)

The general equation of the line\(\Delta \) passes through the point \(A\left( { – 1;{\rm{ }}2} \right)\) and has the normal vector \(\overrightarrow n = \left( {3{\rm{ }};{\rm{ }}2} \right).\)is: \(3\left( {x + 1} \right) + 2\left( {y – 2} \right) = 0 \Leftrightarrow 3x + 2y – 1 = 0\)

### Solve the exercises Lesson 2, page 79, Math textbook 10 Kite episode 2

Make the equation of the line in Figures 34,35,36,37:

+) Equation of intercept of line d passing through two points \(A\left( {a;0} \right),B\left( {0;b} \right)\left( {ab \ne 0} \right)\) has the equation \(\frac{x}{a} + \frac{y}{b} = 1\)

+) The equation of the line d passing through two points \(A\left( {{x_o};{y_o}} \right);B\left( {{x_1};{y_1}} \right)\) is: \(\frac{{x – {x_o}}}{{{x_1} – {x_o}}} = \frac{{y – {y_o}}}{{{y_1} – {y_o}}}\)

+) The general equation of the line\(\Delta \) passing through the point \({M_o}\left( {{x_o};{y_o}} \right)\) and getting \(\overrightarrow n = \left ( {{\rm{a }};{\rm{ b}}} \right)\left( {\overrightarrow n \ne 0} \right)\) as the normal vector is: \(a\left( { x – {x_o}} \right) + b\left( {y – {y_o}} \right) = 0\)

**Solution guide**

a) Equation of intercept of the line \({\Delta _1}\) passing through 2 points \(\left( {0;4} \right)\) and \(\left( {3;0} \right )\) is: \(\frac{x}{3} + \frac{y}{4} = 1\)

b) Equation of the line \({\Delta _2}\) passing through 2 points \(\left( {2;4} \right)\) and \(\left( { – 2; – 2} \right) \) to be:

\(\frac{{x – 2}}{{ – 2 – 2}} = \frac{{y – 4}}{{ – 2 – 4}} \Leftrightarrow \frac{{x – 2}}{{) – 4}} = \frac{{y – 4}}{{ – 6}} \Leftrightarrow 3x – 2y + 2 = 0\)

c) Since the line \({\Delta _3}\) is perpendicular to \({\rm{O}}x\), the normal vector of \({\Delta _3}\) is: \(\overrightarrow { {n_3}} = \left( {1;0} \right)\)

So the equation of the line \({\Delta _3}\) passing through the point \(\left( { – \frac{5}{2};0} \right)\) has the normal vector \(\overrightarrow {{ n_3}} = \left( {1;0} \right)\)is: \(1\left( {x + \frac{5}{2}} \right) + 0\left( {y – 0} \right) = 0 \Leftrightarrow x = – \frac{5}{2}\)

d) Since the line \({\Delta _4}\) is perpendicular to \({\rm{O}}x\) the normal vector of \({\Delta _4}\) is: \(\overrightarrow { {n_4}} = \left( {0;1} \right)\)

So the equation of the line \({\Delta _4}\) passing through the point \(\left( {0;3} \right)\) has the normal vector \(\overrightarrow {{n_4}} = \left( { 0;1} \right)\)is: \(0\left( {x – 0} \right) + 1\left( {y – 3} \right) = 0 \Leftrightarrow y = 3\)

### Solve the exercises Lesson 3 page 80 Math textbook 10 Kite episode 2

Given a line d whose parametric equation is: \(\left\{ \begin{array}{l}x = – 1 – 3t\\y = 2 + 2t\end{array} \right.\)

a) Write the general equation of the line d.

b) Find the coordinates of the intersection of the line d with the axes Ox, Oy, respectively.

c) Does the line d pass through the point M(-7; 5)?

**Solution method**

a) Eliminate \(t\) to get the relationship between \(x\) and \(y\)( which is also the PTTQ of the line d )

b) Solve the system of equations consisting of 2 equations of the intersecting line

c) Try the coordinates of the point M into the PTTQ of d to draw conclusions.

**Solution guide**

a) Consider the parametric equation of d: \(\left\{ \begin{array}{l}x = – 1 – 3t\left( 1 \right)\\y = 2 + 2t\left( 2 \right )\end{array} \right.\).

Take \(\left( 1 \right) + \frac{3}{2}.\left( 2 \right) \Rightarrow x + \frac{3}{2}y = 2 \Rightarrow 2x + 3y – 4 = 0\)

So the general equation of the line d is: \(2x + 3y – 4 = 0\)

b) Consider the system of equations: \(\left\{ \begin{array}{l}2x + 3y – 4 = 0\\x = 0\end{array} \right \Leftrightarrow \left\{ \begin{ array}{l}y = \frac{4}{3}\\x = 0\end{array} \right.\) . So the intersection of d with the axis Oy is: \(A\left( {0;\frac{4}{3}} \right)\)

Consider the system of equations: \(\left\{ \begin{array}{l}2x + 3y – 4 = 0\\y = 0\end{array} \right \Leftrightarrow \left\{ \begin{array} {l} = 0\\x = 2\end{array} \right.\) . So the intersection of d with the Ox axis is: \(B\left( {2;0} \right)\)

c) Substituting the coordinates of the point \(M\left( { – 7;{\rm{ }}5} \right)\) into the equation of the line d we have: \(2.\left( { – 7} \) right) + 3.5 – 4 \ne 0\)

So \(M\left( { – 7;{\rm{ }}5} \right)\) is not on the line d.

### Solve the exercises Lesson 4, page 80, Math textbook 10 Kite, episode 2

Given a straight line d whose general equation is: x – 2y – 5 = 0.

a) Make a parametric equation for the line d.

b) Find the coordinates of the point M in d such that OM = 5 where O is the origin.

c) Find the coordinates of point N in d such that the distance from N to the horizontal axis Ox is 3.

**Solution method**

a) Parametric equation of the line\(\Delta \) passing through the point \({M_o}\left( {{x_o};{y_o}} \right)\) and getting \(\overrightarrow u = \left ( {{\rm{a }};{\rm{ b}}} \right)\left( {\overrightarrow u \ne 0} \right)\) as the direction vector: \(\left\{ \ begin{array}{l}x = {x_o} + at\\y = {y_o} + bt\end{array} \right.\) ( \(t\) is the parameter )

b) Parameterization of the M . point

If \(A\left( {{x_1};{y_1}} \right),B\left( {{x_2};{y_2}} \right)\) then \(AB = \left| {\overrightarrow { AB} } \right| = \sqrt {{{\left( {{x_2} – {x_1}} \right)}^2} + {{\left( {{y_2} – {y_1}} \right)} ^2}} \)

c) Parameterize the point N and then use the distance assumption

**Solution guide**

a) From the general equation of the line, we get a normal vector: \(\overrightarrow n = \left( {1; – 2} \right)\) so we choose the direction vector of the line d is: \(\overrightarrow u = \left( {2;1} \right)\).

adsense

Choose the point \(A\left( {1; – 2} \right) \in d\).So the parametric equation of the line d is: \(\left\{ \begin{array}{l}x = 1 + 2t\\y = – 2 + t\end{array} \right.\) (t is parameter)

b) Since the point M belongs to d, we have: \(M\left( {1 + 2m; – 2 + m} \right);m \in \mathbb{R}\).

We have: \(OM = 5 \Leftrightarrow \sqrt {{{\left( {1 + 2m} \right)}^2} + {{\left( { – 2 + m} \right)}^2}} = 5 \Leftrightarrow {m^2} = 4 \Leftrightarrow m = \pm 2\)

With \(m = 2 \Rightarrow M\left( {5;0} \right)\)

With \(m = – 2 \Rightarrow M\left( { – 3; – 4} \right)\)

So we have 2 points M satisfying the problem condition.

c) Since point N belongs to d, we have: \(N\left( {1 + 2n; – 2 + n} \right)\)

The distance from N to the horizontal axis is equal to the absolute value of the coordinates of the point N. Therefore, the distance from N to the horizontal axis is 3 if and only if: \(\left| { – 2 + n} \right| = 3 \Leftrightarrow \left[\begin{array}{l}n=5\\n=–1\end{array}\right\)[\begin{array}{l}n=5\\n= –1\end{array}\right\)

With \(n = 5 \Rightarrow N\left( {11;3} \right)\)

With \(n = – 1 \Rightarrow N\left( { – 1; – 3} \right)\)

So there are 2 points N satisfying the problem

### Solve the exercises Exercise 5 page 80 Math textbook 10 Kite episode 2

Given triangle ABC, know A(1; 3), B(-1;- 1), C(5 – 3). Set up the general equation for:

a) Three straight lines AB, BC, AC;

b) The midline perpendicular to AB;

c) Altitude AH and median AM of triangle ABC.

**Solution method**

a) The equation of the line d passing through the two points \(A\left( {{x_o};{y_o}} \right);B\left( {{x_1};{y_1}} \right)\) is: \(\frac{{x – {x_o}}}{{{x_1} – {x_o}}} = \frac{{y – {y_o}}}{{{y_1} – {y_o}}}\)

b) and c) The general equation of the line\(\Delta \) passing through the point \({M_o}\left( {{x_o};{y_o}} \right)\) and getting \(\overrightarrow n) = \left( {{\rm{a }};{\rm{ b}}} \right)\left( {\overrightarrow n \ne 0} \right)\) as the normal vector is: \(a\ left( {x – {x_o}} \right) + b\left( {y – {y_o}} \right) = 0\)

**Solution guide**

a) The equation of the line AB passing through 2 points A and B is: \(\frac{{x – 1}}{{ – 1 – 1}} = \frac{{y – 3}}{{ – 1 – 3}} \Leftrightarrow \frac{{x – 1}}{{ – 2}} = \frac{{y – 3}}{{ – 4}} \Leftrightarrow 2x – y + 1 = 0\)

The equation of the line AC passing through 2 points A and C is: \(\frac{{x – 1}}{{5 – 1}} = \frac{{y – 3}}{{ – 3 – 3}} \Leftrightarrow \frac{{x – 1}}{4} = \frac{{y – 3}}{{ – 6}} \Leftrightarrow 3x + 2y – 9 = 0\)

The equation of the line BC passing through 2 points B and C is:

\(\frac{{x + 1}}{{5 + 1}} = \frac{{y + 1}}{{ – 3 + 1}} \Leftrightarrow \frac{{x + 1}}{6} = \frac{{y + 1}}{{ – 2}} \Leftrightarrow x + 3y + 4 = 0\)

b) Let d be the perpendicular bisector of side AB.

Let N be the midpoint of AB, deduce \(N\left( {0;1} \right)\).

Since \(d \bot AB\) we have the normal vector of d: \(\overrightarrow {{n_d}} = \left( {1;2} \right)\)

So the equation of the line d through N with the normal vector \(\overrightarrow {{n_d}} = \left( {1;2} \right)\) is:

\(1\left( {x – 0} \right) + 2\left( {y – 1} \right) = 0 \Leftrightarrow x + 2y – 2 = 0\)

c) Since AH is perpendicular to BC, the normal vector of AH is \(\overrightarrow {{n_{AH}}} = \left( {3; – 1} \right)\)

So the equation of altitude AH passing through point A has normal vector \(\overrightarrow {{n_{AH}}} = \left( {3; – 1} \right)\)is: \(3\left( { {) x – 1} \right) – 1\left( {y – 3} \right) = 0 \Leftrightarrow 3x – y = 0\)

Since M is the midpoint of BC, \(M\left( {2; – 2} \right)\). So we have: \(\overrightarrow {AM} = \left( {1; – 5} \right) \Rightarrow \overrightarrow {{n_{AM}}} = \left( {5;1} \right)\)

The equation of median AM passing through point A with normal vector \(\overrightarrow {{n_{AM}}} = \left( {5;1} \right)\) is:

\(5\left( {x – 1} \right) + 1\left( {y – 3} \right) = 0 \Leftrightarrow 5x + y – 8 = 0\)

### Solve exercises Lesson 6 page 80 Math textbook 10 Kite episode 2

To join a gym, a practitioner must pay an initial joining fee and gym usage fee. The line \(\Delta \) in Figure 38 represents the total cost (unit: million dong) to join a gym per person’s training time (unit: month).

a) Write the equation of the line \(\Delta \).

b) What does the intersection of the line \(\Delta \) with the vertical axis mean in this situation?

c) Calculate the total cost that the person has to pay for joining the gym for a period of 12 months.

a) The equation of the line d passing through the two points \(A\left( {{x_o};{y_o}} \right);B\left( {{x_1};{y_1}} \right)\) is: \(\frac{{x – {x_o}}}{{{x_1} – {x_o}}} = \frac{{y – {y_o}}}{{{y_1} – {y_o}}}\)

c) Substitute the corresponding value into the equation of the line

**Solution guide**

a) The line \(\Delta \) passing through two points has coordinates \(\left( {0;1,5} \right),\left( {7;5} \right)\) so \ (\Delta \) whose equation is:

\(\frac{{x – 0}}{{7 – 0}} = \frac{{y – 1.5}}{{5 – 1.5}} \Leftrightarrow \frac{x}{7} = \frac{{y – 1.5}}{{3,5}} \Leftrightarrow x – 2y + 3 = 0\)

b) The intersection of the line \(\Delta \) with the axis \(Oy\) corresponds to \(x = 0\). The time \(x = 0\) indicates the initial participation fee that the practitioner must pay. When \(x = 0\) then \(y = 1.5\) , so the initial participation fee that the practitioner has to pay is 1 500 000 VND.

c) The first 12 months corresponds to \(x = 12\)

From the equation of the line \(\Delta \) we have: \(x – 2y + 3 = 0 \Leftrightarrow y = \frac{1}{2}x + \frac{3}{2}\)

Substituting \(x = 12\) into the equation of the line we have: \(y = \frac{1}{2}.12 + \frac{3}{2} = 7.5\)

So the total cost that person has to pay when joining the gym for 12 months is 7 million VND.