## Solving Exercise 1: Coordinates of vectors (C7 – Math 10 Kite)——————

### Solve the exercise Exercise 1, page 65, Math textbook 10 Kite, episode 2

Find the coordinates of the vectors in Figure 16 and represent each of them as two vectors $$\overrightarrow i , \overrightarrow j$$

Solution method

Draw the vectors $$\overrightarrow {OA} = \overrightarrow a ,\overrightarrow {OB} = \overrightarrow b ,\overrightarrow {OC} = \overrightarrow c ,\overrightarrow {OD} = \overrightarrow d$$. The coordinates of 4 points A, B, C, D are the coordinates of 4 vectors.

To represent vectors over unit vectors: $$\overrightarrow u {\rm{ }} = {\rm{ }}\left( {a;{\rm{ }}b} \right) \Leftrightarrow \overrightarrow u = a\overrightarrow i + b\overrightarrow j$$

Solution guide

a) Draw the vectors $$\overrightarrow {OA} = \overrightarrow a ,\overrightarrow {OB} = \overrightarrow b ,\overrightarrow {OC} = \overrightarrow c ,\overrightarrow {OD} = \overrightarrow d$$

Based on the figure, we see that the coordinates of 4 points A, B, C, D are:

$$A\left( { – 5; – 3} \right),B\left( {3; – 4} \right),C\left( { – 1;3} \right),D\left( { 2;5} \right)$$

Thus $$\overrightarrow a = \overrightarrow {OA} = \left( { – 5; – 3} \right),\overrightarrow b = \overrightarrow {OB} = \left( {3; – 4} \right) ,\overrightarrow c = \overrightarrow {OC} = \left( { – 1;3} \right),\overrightarrow d = \overrightarrow {OD} = \left( {2;5} \right)$$

b) Since $$\overrightarrow a = \overrightarrow {OA} = \left( { – 5; – 3} \right)$$$$\overrightarrow a = \left( { – 5} \right)\overrightarrow i + \left( { – 3} \right)\overrightarrow j = – 5\overrightarrow i – 3\overrightarrow j$$

Since $$\overrightarrow b = \overrightarrow {OB} = \left( {3; – 4} \right)$$ $$\overrightarrow b = 3\overrightarrow i + \left( { – 4} \right)\ overrightarrow j = 3\overrightarrow i – 4\overrightarrow j$$

Since $$\overrightarrow c = \overrightarrow {OC} = \left( { – 1;3} \right)$$ $$\overrightarrow c = \left( { – 1} \right)\overrightarrow i + \left ( 3 \right)\overrightarrow j = – \overrightarrow i + 3\overrightarrow j$$

Since $$\overrightarrow d = \overrightarrow {OD} = \left( {2;5} \right)$$ $$\overrightarrow d = 2\overrightarrow i + 5\overrightarrow j$$

### Solve the exercise Exercise 2, page 65, Math textbook 10 Kite, episode 2

Find the coordinates of the following vectors:

a) $$\overrightarrow a = 3\overrightarrow i$$ b) $$\overrightarrow b = – \overrightarrow j$$

c) $$\overrightarrow c = \overrightarrow i – 4\overrightarrow j$$ d) $$\overrightarrow d = 0.5\overrightarrow i + \sqrt 6 \overrightarrow j$$

Solution method

Use: $$\overrightarrow u {\rm{ }} = {\rm{ }}\left( {a;{\rm{ }}b} \right) \Leftrightarrow \overrightarrow u = a\overrightarrow i + b \overrightarrow j$$

Solution guide

a) Since $$\overrightarrow a = 3\overrightarrow i$$ so $$\overrightarrow a = \left( {3;0} \right)$$

b) Since $$\overrightarrow b = – \overrightarrow j$$ so $$\overrightarrow b = \left( {0; – 1} \right)$$

c) Since $$\overrightarrow c = \overrightarrow i – 4\overrightarrow j$$ so $$\overrightarrow c = \left( {1; – 4} \right)$$

d) Since $$\overrightarrow d = 0.5\overrightarrow i + \sqrt 6 \overrightarrow j$$ so $$\overrightarrow d = \left( {0.5;\sqrt 6 } \right)$$

### Solve the exercises Lesson 3, page 65, Math textbook 10 Kite, episode 2

Find pairs of real numbers a and b such that each of the following vector pairs are equal:

a) $$\overrightarrow u = \left( {2a – 1; – 3} \right)$$ and $$\overrightarrow v = \left( {3;4b + 1} \right)$$

b) $$\overrightarrow x = \left( {a + b; – 2a + 3b} \right)$$ and $$\overrightarrow y = \left( {2a – 3;4b} \right)$$

Solution method

With $$\overrightarrow a = \left( {{x_1};{y_1}} \right)$$ and $$\overrightarrow b = \left( {{x_2},{y_2}} \right)$$ , we yes: $$\overrightarrow a = \overrightarrow b \Leftrightarrow \left\{ \begin{array}{l}{x_1} = {x_2}\\{y_1} = {y_2}\end{array} \right.\ ) Solution guide a) Let \(\overrightarrow u = \overrightarrow v \Leftrightarrow \left\{ \begin{array}{l}2a – 1 = 3\\ – 3 = 4b + 1\end{array} \right. \Leftrightarrow \ left\{ \begin{array}{l}a = 2\\b = – 1\end{array} \right.$$

So $$\left\{ \begin{array}{l}a = 2\\b = – 1\end{array} \right.$$ then $$\overrightarrow u = \overrightarrow v$$

b) $$\overrightarrow x = \overrightarrow y \Leftrightarrow \left\{ \begin{array}{l}a + b = 2a – 3\\ – 2a + 3b = 4b\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}a = 1\\b = – 2\end{array} \right.$$

So $$\left\{ \begin{array}{l}a = 1\\b = – 2\end{array} \right.$$ then $$\overrightarrow x = \overrightarrow y$$

### Solution of Exercises Lesson 4, page 66, Math Textbook 10, Kite episode 2

In the coordinate plane Oxy, give three points A(2;3), B(-1; 1), C(3;- 1).

a) Find the coordinates of point M such that $$\overrightarrow {AM{\rm{ }}} = {\rm{ }}\overrightarrow {BC}$$ .

b) Find the coordinates of the midpoint N of the line segment AC. Prove $$\overrightarrow {BN} {\rm{ }} = {\rm{ }}\overrightarrow {NM}$$ .

Solution method

With $$\overrightarrow a = \left( {{x_1};{y_1}} \right)$$ and $$\overrightarrow b = \left( {{x_2},{y_2}} \right)$$ , we yes: $$\overrightarrow a = \overrightarrow b \Leftrightarrow \left\{ \begin{array}{l}{x_1} = {x_2}\\{y_1} = {y_2}\end{array} \right.\ ) Solution guide a) Call \(M\left( {a;b} \right) \Rightarrow \overrightarrow {AM} = \left( {a – 2;b – 3} \right)$$

Vector coordinates $$\overrightarrow {BC} = \left( {4; – 2} \right)$$

Let $$\overrightarrow {AM{\rm{ }}} = {\rm{ }}\overrightarrow {BC} \Leftrightarrow \left\{ \begin{array}{l}a – 2 = 4\\b – 3 = – 2\end{array} \right \Leftrightarrow \left\{ \begin{array}{l}a = 6\\b = 1\end{array} \right.$$

So for $$\overrightarrow {AM{\rm{ }}} = {\rm{ }}\overrightarrow {BC}$$ then the coordinates of point M are:$$M\left( {6;1} \right)$$

b) Call $$N\left( {x,y} \right) \Rightarrow \overrightarrow {NC} = \left( {3 – x, – 1 – y} \right)$$and $$\overrightarrow {AC) } = \left( {x – 2,y – 3} \right)$$

Since N is the midpoint of AC, $$\overrightarrow {AC} = \overrightarrow {NC} \Leftrightarrow \left\{ \begin{array}{l}x – 2 = 3 – x\\y – 3 = – 1 – y\end{array} \right \Leftrightarrow \left\{ \begin{array}{l}x = \frac{5}{2}\\y = 1\end{array} \right.$$ . So $$N\left( {\frac{5}{2},1} \right)$$

We have: $$\overrightarrow {BN} {\rm{ }} = \left( { – \frac{7}{2};0} \right)$$ and $$\overrightarrow {NM} = \left( {\frac{{ – 7}}{2};0} \right)$$. So $$\overrightarrow {BN} {\rm{ }} = {\rm{ }}\overrightarrow {NM}$$

### Solve exercises Exercise 5 page 66 Math textbook 10 Kite episode 2

In the coordinate plane Oxy, give the point M(-1; 3).

a) Find the coordinates of point A symmetrical to point M through the origin O.

b) Find the coordinates of point B symmetrical to point M about the Ox axis.

c) Find the coordinates of point C symmetrical to point M about the axis Oy.

Solution method

Draw a picture and then based on the figure determine the points

Solution guide

a) Based on the figure, we see that $$A\left( {1; – 3} \right)$$

b) Based on the figure, we see that $$B\left( { – 1; – 3} \right)$$

c) Based on the figure, we see that $$C\left( {1;3} \right)$$

### Solution of Exercises Lesson 6, page 66, Math Textbook 10, Kite episode 2

In the coordinate plane Oxy, give three noncollinear points A(- 3 ; 1), B(-1; 3), I(4;2). Find the coordinates of two points C and D such that quadrilateral ABCD is a parallelogram with I as the center of symmetry.

Solution method

The center of symmetry of a parallelogram is the midpoint of the two diagonals.

With $$\overrightarrow a = \left( {{x_1};{y_1}} \right)$$ and $$\overrightarrow b = \left( {{x_2},{y_2}} \right)$$ , we yes: $$\overrightarrow a = \overrightarrow b \Leftrightarrow \left\{ \begin{array}{l}{x_1} = {x_2}\\{y_1} = {y_2}\end{array} \right.\ ) Solution guide Call \(C\left( {a;b} \right),D\left( {m,n} \right) \Rightarrow \overrightarrow {IC} = \left( {a – 4,b – 2} \right )$$ and $$\overrightarrow {ID} = \left( {m – 4,n – 2} \right)$$

Since I is the center of parallelogram ABCD, I is the midpoints of AC and BD.

So we have:$$\overrightarrow {AI} = \overrightarrow {IC}$$ and $$\overrightarrow {BI} = \overrightarrow {ID}$$

We have: $$\overrightarrow {AI} = \left( {7;1} \right)$$ and $$\overrightarrow {BI} = \left( {5; – 1} \right)$$

Do $$\overrightarrow {AI} = \overrightarrow {IC} \Leftrightarrow \left\{ \begin{array}{l}7 = a – 4\\1 = b – 2\end{array} \right. \Leftrightarrow. \left\{ \begin{array}{l}a = 11\\b = 3\end{array} \right.$$ .So $$C\left( {11;3} \right)$$

Do $$\overrightarrow {BI} = \overrightarrow {ID} \Leftrightarrow \left\{ \begin{array}{l}5 = m – 4\\ – 1 = n – 2\end{array} \right. \ Leftrightarrow \left\{ \begin{array}{l}m = 9\\n = 1\end{array} \right.$$. So $$D\left( {9;1} \right)$$

### Solve exercises Exercise 7 page 66 Math textbook 10 Kite episode 2

In the coordinate plane Oxy, let ABC be a triangle. The points M(1;- 2), N(4;- 1) and P(6 ; 2) are the midpoints of the sides BC, CA, AB, respectively. Find the coordinates of the points A, B, C.

Solution method

The moving average is parallel to and one-half of the corresponding bottom edge

With $$\overrightarrow a = \left( {{x_1};{y_1}} \right)$$ and $$\overrightarrow b = \left( {{x_2},{y_2}} \right)$$ , we yes: $$\overrightarrow a = \overrightarrow b \Leftrightarrow \left\{ \begin{array}{l}{x_1} = {x_2}\\{y_1} = {y_2}\end{array} \right.\ ) Solution guide By product of the mean lines in a triangle we have: \(\overrightarrow {PN} = \overrightarrow {BM} = \overrightarrow {MC}$$ and $$\overrightarrow {MP} = \overrightarrow {NA}$$

Call $$A\left( {{a_1},{a_2}} \right),B\left( {{b_1};{b_2}} \right),C\left( {{c_1};{c_2}} \right)$$

We have: $$\overrightarrow {PN} = \left( {2;3} \right)$$,$$\overrightarrow {BM} = \left( {1 – {b_1}; – 2 – {b_2}} \right)$$, $$\overrightarrow {MC} = \left( {{c_1} – 1;{c_2} + 2} \right)$$, $$\overrightarrow {MP} = \left( {5; 4} \right)$$, $$\overrightarrow {NA} = \left( {{a_1} – 4;{a_2} + 1} \right)$$

Yes $$\overrightarrow {PN} = \overrightarrow {BM} \Leftrightarrow \left\{ \begin{array}{l}2 = 1 – {b_1}\\3 = – 2 – {b_2}\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}{b_1} = – 1\\{b_2} = – 5\end{array} \right.$$ .So $$B\left( { {) – 1; – 5} \right)$$

Yes $$\overrightarrow {PN} = \overrightarrow {MC} \Leftrightarrow \left\{ \begin{array}{l}2 = {c_1} – 1\\3 = {c_2} + 2\end{array} \ right. \Leftrightarrow \left\{ \begin{array}{l}{c_1} = 3\\{c_2} = 1\end{array} \right.$$ .So $$C\left( {3;1 } \right)$$

Yes $$\overrightarrow {NA} = \overrightarrow {MP} \Leftrightarrow \left\{ \begin{array}{l}5 = {a_1} – 4\\4 = {a_2} + 1\end{array} \ right.\Leftrightarrow \left\{ \begin{array}{l}{a_1} = 9\\{a_2} = 3\end{array} \right.$$ .So $$A\left( {9;3 } \right)$$