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

### 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: $$\overrightarrow u = \left( {2;1} \right)$$.

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.