第 10 类 RD Sharma 解决方案 - 第 14 章坐标几何 - 练习 14.3 |设置 3

问题 41. 平行四边形的三个连续顶点是 (-2, -1), (1, 0) 和 (4, 3)。找到第四个顶点。

解决方案：

Let the coordinates of three vertices are A (-2, -1), B (1, 0), and C (4, 3)

And let the diagonals AC and BD bisect each other at O

As O is the mid-point of AC

Therefore, 

Vertices of O will be

$=\left(\frac{x_1+x_2}{2},\ \frac{y_1+y_2}{2}\right)\\ =\left(\frac{-2+4}{2},\ \frac{-1+3}{2}\right)\\ =\left(\frac{2}{2},\ \frac{2}{2}\right)\\ =(1,\ 1)$

Assume coordinates of the forth vertex D be (x, y)

As O is the mid-point of BD

Thus, coordinates of O will be 

$\left(\frac{1+x}{2},\ \frac{0+y}{2}\right)\\ =\left(\frac{1+x}{2},\ \frac{y}{2}\right)$

Therefore(1 + x)/2 = 1

1 + x = 2

x = 1

and

y/2 = 1

y = 2

Hence, the co-ordinates of D will be (1, 2).

问题 42. 点 (3, -4) 和 (-6, 2) 是平行四边形的对角线的端点。如果第三个顶点是 (-1, -3)。找到第四个顶点的坐标。

解决方案：

Assume the edges of a diagonal AC of a parallelogram ABCD are A (3, -4) and C (-6, 2)

Consider AC and BD bisect each other at O.

Therefore,

Mid-point of AC will be

$=\left(\frac{x_1+x_2}{2},\ \frac{y_1+y_2}{2}\right)\\ =\left(\frac{3-6}{2},\ \frac{-4+2}{2}\right)\\ =\left(\frac{-3}{2},\ \frac{-2}{2}\right)\\ =(\frac{-3}{2},\ -1)$

Assume the fourth vertex of the parallelogram be (x, y)

Therefore,

Mid-point of BD will be 

$=\left(\frac{x_1+x_2}{2},\ \frac{y_1+y_2}{2}\right)\\ =\left(\frac{-1+x}{2},\ \frac{-3+y}{2}\right)\\ =\frac{-1+x}{2}=\frac{-3}{2}$

-1 + x = -3 

x = -2

and

(-3 + y)/2 = -1

-3 + y = -2 

y = -2 + 3 = 1

Hence, the coordinates of D are (-2, 1).

问题 43. 如果三角形三边中点的坐标是 (1, 1), (2, -3)和 (3, 4)，求三角形的顶点。

解决方案：

Assume A (x₁, y₁), B (x₂, y₂) and C (x₃, y₃) be the vertices of the ∆ABC

D, E and F are the mid-points of BC, CA and AB respectively such that their co-ordinates are D (1, 1), E (2, -3) and F (3, 4)

D is mid-point of BC

Therefore,

$\frac{x_2+x_3}{2}=1$

x₂ + x₃ = 2

and

$\frac{y_2+y_3}{2}=1$

y₂ + y₃ = 2

Similarly, E is the mid-point of AC

Therefore,

$\frac{x_3+x_1}{2}=2$

x₃ + x₁ = 4

and

$\frac{y_3+y_1}{2}=-3$

y₃ + y₁ = -6

And

F is the mid-point of AB

Therefore,

$\frac{x_2+x_1}{2}=3$

x₂ + x₁ = 6

and

$\frac{y_2+y_1}{2}=4$

y₂ + y₁ = 8

Now,

x₁ + x₂ = 6  …………..(i)

x₂ + x₃ = 2  ……………(ii)

x₃ + x₁ = 4  …………….(iii)

On adding we will get

2(x₁ + x₂ + x₃) = 12

x₁ + x₂ + x₃ = 6  …………(iv)

On subtracting (ii), (iii) and (i) from (iv), we get

 x₁ = 4, x₂ = 2, x₃ = 0

Similarly

y₁ + y₂ = 8  ……….(v)

y₂ + y₃ = 2  ……….(vi)

y₃ + y₁ = -6  ………(vii)

On adding we will get 

2(y₁ + y₂ + y₃) = 4

y₁ + y₂ + y₃ = 2  ………(viii)

On subtracting (vi), (vii) and (v) from (viii), we get

y₁ = 0

y₂ = 8

y₃ = -6

Hence, the vertices of ∆ABC are A (4, 0), B(2, 8), C(0, -6)

问题 44. 确定直线 x – y – 2 = 0 分割连接 (3, -1) 和 (8, 9) 的线段的比率。

解决方案：

Assume the straight line x – y – 2 = 0 divides the line segment joining the points (3, -1), (8, 9) in the ratio m : n

Co-ordinates of the point will be

$x=\frac{mx_2+nx_1}{m+n}\\ =\frac{m\times8+n\times3}{m+n}\\ =\frac{8m+3n}{m+n}$

and

$y=\frac{my_2+ny_1}{m+n}\\ =\frac{m\times9+n(-1)}{m+n}\\ =\frac{9m-n}{m+n}$

This point (x, y) lies on the line on the line x – y – 2 = 0

⇒ (8m + 3n) – (9m – n) – 2(m + n) = 0

⇒ 8m + 3n – 9m + n – 2m – 2n = 0

⇒ -3m + 2n = 0

⇒ 2n = 3m

$\frac{m}{n}=\frac{2}{3}$

Hence, the ratio = 2 : 3 internally

问题 45. 平行四边形的三个顶点是 (a + b, a – b), (2 a + b, 2a – b), (a – b, a + b)。找到第四个顶点。

解决方案：

In parallelogram ABCD co-ordinates are of A (a + b, a – b), B (2a + b, 2a – b), C (a – b, a + b)

Assume coordinates of D be (x, y)

Join diagonal AC and BD

Which bisect each other at O

O is the mid-point of AC as well as BD

If O is the mid-point of AC, Then its coordinates will be

$=\left(\frac{x_1+x_2}{2},\ \frac{y_1+y_2}{2}\right)\\ =\left(\frac{a+b+a-b}{2},\ \frac{a-b+a+b}{2}\right)\\ =\left(\frac{2a}{a},\ \frac{2a}{a}\right)\\ =(a,\ a)$

and 

If O is mid-point of BD, then coordinates will be

$\left(\frac{x+2a+b}{2},\ \frac{y+2a-b}{2}\right)\\ \therefore\frac{x+2a+b}{2}=a$

x + 2a + b = 2a

x = 2a – 2a – b = -b

and

$\frac{y+2a-b}{2}=a$

y + 2a – b = 2a

y = 2a – 2a + b = b

Hence, the coordinates of D will be (-b, b).

问题 46. 如果一个平行四边形的两个顶点是 (3, 2), (-1, 0) 并且对角线在 (2, -5) 处切割，求平行四边形的其他顶点。

解决方案：

Two vertices of a parallelogram ABCD are A (3,2), and B (-1, 0) and its diagonals bisect each other at O (2, -5)

Assume the coordinates of C be (x₁, y₁) and of D be (x₂, y₂)

If O is the mid-point of AC, then

$x=\frac{x_1+x_2}{2}$

3 + x₁ = 4

x₁ = 4 – 3 = 1

and 

$y=\frac{y_1+y_2}{2}$

$2=\frac{3+x_1}{2}$

and

$-5=\frac{2+y_1}{2}$

2 + y₁ = -10

y₁ = -10 – 2 = -12

Therefore, the coordinates of C will be (1, -12)

Again if O is mid-point of BD then

$2=\frac{-1+x_2}{2}$

-1 + x₂ = 4

x₂ = 4 + 1 = 5

and

$-5=\frac{0+y_2}{2}$

y₂ = -10

Hence, the coordinates of D will be (5, -10).

问题 47. 如果三角形 ar6 (3, 4), (4, 6) 和 (5, 7) 的三边中点坐标，求其顶点。

解决方案：

The coordinates of the mid-points of the sides BC, CA and AB are D (3, 4), E (4, 6) and F (5, 7) of the ∆ABC.

Assume the coordinates of the vertices of the triangle be A (x,₁ y₁), B (x₂, y₂), and C (x₃, y₃).

Now the coordinates of D will be

$\left(\frac{x_2+x_3}{2},\ \frac{y_2+y_3}{2}\right)\\ \therefore\frac{x_2+x_3}{2}=3$

x₂ + x₃ = 6

and

y₂ + y₃ = 4

y₂ + y₃ = 8

Similarly, the coordinates of E will be

$\frac{x_3+x_1}{2}=4$

 x₃ + x₁ = 8

and

$\frac{y_3+y_1}{2}=6\\ \Rightarrow y_3+y_1=12\\ \frac{x_1+x_2}{2}=5\\ \Rightarrow x_1+x_2=10$

and

$\frac{y_1+y_2}{2}=7$

y₁ + y₂ = 14

Now,

x₂ + x₃ = 6  …….(i)

x₃ + x₁ = 8  ……..(ii)

x₁ + x₂ = 10  ……..(iii)

On adding we will get

2(x₁ + x₂ + x₃) = 24

⇒ x₁ + x₂ + x₃ = 24/2 = 12  ………..(iv)

On subtracting each from (iv), 

We will get

x₁ = 6, x₂ = 4 and x₃ = 2

Similarly,

y₂ + y₃ = 8   ……..(v)

y₃ + y₁ = 12  ………(vi)

y₁ + y₂ = 14   ……..(vii)

On Adding, we will get

2(y₁ + y₂ + y₃) = 34

⇒ y₁ + y₂ + y₃ = 34/2  = 17  ……..(viii)

On subtracting each from (viii),

We will get

y₁ = 9

y₂ = 5

y₃ = 3

Hence, the coordinates will be of A (6, 9), B (4, 5) and C (2, 3)

问题 48. 连接点 P (3, 3) 和 Q (6, -6) 的线段在点 A 和 B 处被三等分，使得 A 更靠近 P。如果 A 也位于由 2x+ y 给出的线上+ k =0，求k的值。

解决方案：

Two points A and B trisect the line segment joining the points P (3, 3) and Q (6, -6) and A is nearer to P and A lies also on the line 2x + y + k = 0

Now,

A divides the line segment PQ in the ratio of 1 : 2

i.e.,

PA = AQ = 1 : 2

Assume coordinates of A be (x, y), then

$x=\frac{m_1x_2+m_2x_1}{m_1+m_2}\\ =\frac{1\times6+2\times3}{1+2}\\ =\frac{6+6}{3}\\ =\frac{12}{3}\\ =4$

and

$y=\frac{m_1y_2+m_2y_1}{m_1+m_2}\\ =\frac{1\times(-6)+2\times3}{1+2}\\ =\frac{-6+6}{3}\\ =\frac{0}{3}\\ =0$

Therefore,

Coordinates of A are (4, 0)

As A lies on the line 2x + y + k = 0

Hence,

It will satisfy it

2 × 4 + 0 + k = 0

8 + k = 0

k = -8

问题 49. 如果平行四边形的三个连续顶点分别是 (1, -2)、(3, 6)和 (5, 10)，求它的第四个顶点。

解决方案：

A (1, -2), B (3, 6) and C (5, 10) are the three consecutive vertices of the parallelogram ABCD

Assume (x, y) be its fourth vertex

AC and BD are its diagonals which bisect each other at O

As O is the mid-point of AC

Therefore,

Coordinates of O will be

$\left(\frac{3+x}{2},\ \frac{6+y}{2}\right)$

On comparing, 

3 + x = 3

3 + x = 6

x = 3

and

(6 + y)/2 = 4

6 + y = 8

y = 2

Hence, the coordinates of fourth vertex D are (3, 2).

问题 50. 如果点 A (a, -11), B (5, b), C (2, 15)和 D (1, 1) 是平行四边形 ABCD 的顶点，求 a 和 b 的值.

解决方案：

A (a, -11), B (5, b), C (2, 15) and D (1, 1) are the vertices of a parallelogram ABCD

Diagonals AC and BD bisect each other at O

As O is the mid-point of AC

Therefore,

Coordinates of O will be

$\left(\frac{2+a}{2},\ \frac{-11+15}{2}\right)\\ \left(\frac{2+a}{2},\ \frac{4}{2}\right)\\ \left(\frac{2+a}{2},\ 2\right)$

Similarly, O is the mid-point of BD also

Therefore,

Coordinates of O will be

(2 + a)/2 = 3

b + 1 = 4

b = 3

Hence, 

a = 4 

b = 3

问题 51. 如果三角形三边中点的坐标是 (3, -2), (-3, 1)和 (4, -3)，则求其顶点坐标。

解决方案：

In a ∆ABC,

D, E and F are the mid-points of the sides BC, CA and AB respectively and co-ordinates of D, E and F are (3, -2), (-3, 1) and (4, -3) respectively.

Assume the coordinates of A are (x₁, y₁), of B are (x₂, y₂) and of C are (x₃, y₃)

Now,

As D is the mid-point BC

Therefore,

$\frac{x_2+x_3}{2}=3\\ \Rightarrow x_3+x_3=6$

and 

$\frac{y_2+y_3}{2}=-2\\ \Rightarrow y_2+y_3=-4$

As E is the mid-point of CA

Therefore,

$-3=\frac{x_3+x_1}{2}\\ \Rightarrow x_3+x_1=-6$

and

$1=\frac{y_3+y_1}{2}\\ \Rightarrow y_3+y_1=2$

and F is the mid-point of AB

Therefore,

$\frac{x_1+x_2}{2}=4\\ \Rightarrow x_1+x_2=8$

and

$\frac{y_1+y_2}{2}=-3\\ \Rightarrow y_1+y_2=-6$

Now,

x₂ + x₃ = 6  ……..(i)

x₃ + x₁ = -6  ……..(ii)

x₁ + x₂ = 8   ………(iii)

On adding we get

2(x₁ + x₂ + x₃) = 8

⇒ x₁ + x₂ + x₃ = 4   …….(iv)

On subtracting from (iv) we get

x₁ = -2, 

x₂ = 10,

x₃ = -4

and

y₂ + y₃ = -4   …….(v)

y₃ + y₁ = 2   …….(vi)

y₁ + y₂ = -6  ………(vii)

On Adding, We will get

2(y₁ + y₂ + y₃) = -8

⇒ y₁ + y₂ + y₃ = -4   ……..(viii)

On subtracting from (viii) we get

y₁ = 0,

y₂ = -6,

y₃ = 2

Hence, the coordinates of A are (-2, 0) of B are (10, -6) and of C (-4, 2).

问题 52. 连接点 (3, -4) 和 (1, 2) 的线段在点 P 和 Q 处被三等分。如果 P 和 Q 的坐标是 (p, -2) 和 (53 , q) 分别求 p 和 q 的值。

解决方案：

Assume line AB whose ends points are A (3, -4) and B (1, 2)

Coordinates of P and Q which trisect AB are P (p, -2) and Q $(\frac{5}{3},\ q)$

Now,

As P divides AB in the ratio 1 : 2

Therefore,

Coordinates of P will be

$p=\frac{mx_2+nx_1}{m+n}\\ =\frac{1\times1+2\times3}{1+2}\\ =\frac{1+6}{3}\\ =\frac{7}{3}$

Similarly, Q divides AB in the ratio 2 : 1

$q=\frac{my_2+ny_1}{m+n}\\ =\frac{2\times2+1\times(-4)}{2+1}\\ =\frac{4-4}{3}\\ =0$

Hence,

p = 7/3, q = 0

问题 53. 连接点(2, 1), (5, -8) 的直线在点 P 和 Q 处被三等分。如果点 P 位于直线 2x – y + k = 0 上，求 k 的值。

解决方案：

Points A (2, 1), and B (5, -8) are the ends points of the line segment AB

Points P and Q trisect it and P lies on the line 2x – y + k = 0

As P divides AB in the ratio of 1 : 2

Therefore,

Coordinates of P will be

$\left(\frac{mx_2+nx_1}{m+n},\ \frac{my_2+ny_1}{m+n}\right)\\ =\left(\frac{1\times5+2\times2}{1+2},\ \frac{1\times(-8)+2\times1}{1+2}\right)\\ =\left(\frac{5+4}{3}=\frac{9}{3}=3,\ \frac{-8+2}{3}=\frac{-6}{3}=-2\right)$

Therefore,

Coordinates of P are (3, -2)

As it lies on 2x – y + k = 0

Therefore, 

It will satisfy it

2 × 3 – (-2) + k = 0

⇒ 6 + 2 + k = 0

⇒ 8 + k = 0

⇒ k = -8

问题 54. A (4, 2), B (6, 5)和 C (1, 4) 是 ΔABC 的顶点，

(i) A 的中位数与 D 中的 BC 相交。求 D 点的坐标。

(ii) 求 AD 上点 P 的坐标，使得 AP : PD = 2 : 1。

(iii) 分别求中位数 BE 和 CF 上点 Q 和 R 的坐标，使得 BQ : QE = 2 : 1 和 CR : RF = 2 : 1。

(iv) 你观察到什么？

解决方案：

In ∆ABC, co-ordinates of A (4, 2) of (6, 5) and of (1, 4) and AD is BE and CF are the medians

 such that D, E and F are the mid-points of the sides BC, CA and AB respectively

P is a point on AD such that AP : PD = 2 : 1

(i) Now coordinates of D will be

$\left(\frac{x_1+x_2}{2},\ \frac{y_1+y_2}{2}\right)\\ \left(\frac{6+1}{2},\ \frac{5+4}{2}\right)\\ \left(\frac{7}{2},\ \frac{9}{2}\right)$

(ii) As P divides AD in the ratio of 2 : 3

Therefore,

Coordinates of P will be

$x=\frac{mx_2+nx_1}{m+n}\\ =\frac{2\times\frac{7}{2}+1\times4}{2+1}\\ =\frac{9+2}{3}\\ =\frac{11}{3}$

Thus,

Coordinates of P are $\left(\frac{11}{3},\ \frac{11}{3}\right)$

(iii) As E and F are the mid-point if CA and AB respectively

Coordinates of E will be 

$\left(\frac{1+4}{2},\ \frac{4+2}{2}\right)\\ \left(\frac{5}{2},\ \frac{6}{2}\right)\\ \left(\frac{5}{2},\ 3\right)$

and of F will be

$\left(\frac{4+6}{2},\ \frac{2+5}{2}\right)\\ \left(\frac{10}{2},\ \frac{7}{2}\right)\\ \left(5,\ \frac{7}{2}\right)$

As Q and R divides BE and CF in such a way thet BQ : QE = 2 : 1 and CR : RF = 2 : 1

Therefore,

Coordinates of Q will be

$x=\frac{2\times\frac{5}{2}+1\times6}{2+1}\\ =\frac{5+6}{3}\\ =\frac{11}{3}$

and 

$\frac{2\times3+1\times5}{2+1}\\ \frac{6+5}{3}\\ \frac{11}{3}$

i.e., Q₁ is (11/3, 11/3)

and similarly the coordinates of R will be

$x=\frac{2\times5+1\times1}{2+1}\\ =\frac{10+1}{3}\\ =\frac{11}{3}$

and

$y=\frac{2\times\frac{7}{2}+1\times4}{2+1}\\ =\frac{7+4}{3}\\ =\frac{11}{3}$

R is (11/3, 11/3)

(iv) We can see that coordinates of P, Q and R are same 

i.e., P, Q and R coincides each other. 

Medians of the sides of a triangle pass through the same point which is called the centroid of the triangle.

问题 55. 如果点 A (6, 1), B (8, 2), C (9, 4)和 D (k, p) 是按顺序排列的平行四边形的顶点，则求 k 的值和 p。

解决方案：

The diagonals of a parallelogram bisect each other

O is the mid-point of AC and also of BD

O is the mid-point of AC

Therefore,

Coordinates of O will be

$\left(\frac{6+9}{2},\ \frac{1+4}{2}\right)\\ \left(\frac{15}{2},\ \frac{5}{2}\right)$

As O is also the mid-point of BD

Therefore,

$\frac{15}{2}=\frac{8+k}{2}$

and

$\frac{5}{2}=\frac{2+p}{2}$

⇒ 8 + k = 15

⇒ k = 15 – 8 = 7

and

⇒ 2 + p = 5

⇒ p = 5 – 2 = 3

Hence, 

k = 7, p = 3

问题 56. 点 P 分割连接点 A (3, -5) 和 B (-4, 8) 的线段，使得$\frac{AP}{PB}=\frac{k}{1}$ .如果 P 位于 x + y = 0 线上，则求 k 的值。

解决方案：

Point P divides the line segment by joining the points A (3, -5) and B (-4, 8)

Such that $\frac{AP}{PB}=\frac{k}{1}$ 

⇒ AP : PB = k : 1

Assume coordinates of P be (x, y), then

$x=\frac{m_1x_2+m_2x_1}{m_1+m_2}\\ =\frac{k\times(-4)+1\times3}{k+1}\\ =\frac{-4k+3}{k+1}$

and

$y=\frac{m_1y_2+m_2y_1}{m_1+m_2}\\ =\frac{k\times(8)+1\times(-5)}{k+1}\\ =\frac{8k-5}{k+1}$

As x + y = 0

Therefore,

$\frac{-4k+3}{k+1}+\frac{8k-5}{k+1}=0\\ \Rightarrow \frac{-4k+3+8k-5}{k+1}=0$

4k – 2 = 0

4k = 2

k = 2/4

k = 1/2 

问题 57. 连接点 A (-10, 4) 和 B (-2, 0) 的线段的中点 P 位于连接点C (-9, -4) 和 D (- 4，是的）。求 P 除以 CD 的比率。另外，求 y 的值。

解决方案：

P is the mid-point of line segment joining the points A (-10, 4) and B (-2, 0)

Coordinates of P will be

$=\left(\frac{x_1+x_2}{2},\ \frac{y_1+y_2}{2}\right)\\ =\left(\frac{-10-2}{2},\ \frac{4+0}{2}\right)\\ =\left(\frac{-12}{2},\ \frac{4}{2}\right)\\ =(-6,\ 2)$

P lies on CD also,

Let P divides C (9, 4) and D (-4, y) in the ratio m₁ : m₂

Therefore,

$-6=\frac{m_1(-4)+m_2(-9)}{m_1+m_2}$

⇒ -6m₁ – 6m₂ = -4m₁ – 9m₂

⇒ -6m₁ + 4m₁ = -9m₂ + 6m₂

⇒ -2m₁ = -3m₂

⇒ $\frac{m_1}{m_2}=\frac{-3}{-2}=\frac{3}{2}$

Therefore,

m₁ : m₂ = 3 : 2

and

$2=\frac{3\times y+2\times(-4)}{3+2}\\ ⇒ 2=\frac{3y-8}{5}$

⇒ 10 = 3y – 8x 

⇒ 3y = 10 + 8 = 18

⇒ y = 18/3 = 6

⇒ y = 6

问题 58. 如果点 C (-1, 2) 在内部将连接点 A (2, 5) 和 B (x, y) 的线段以 3 : 4 的比例分割，求 x ² + y ²的值.

解决方案：

As we know that if a point (x, y) divides the line segment joining 

the points (x₁, y₁) and (x,₂ y₂) in the ration m : n, then

$x=\frac{mx_2+nx_1}{m+n}$

and

$y=\frac{my_2+ny_1}{m+n}$

Now here, C (-1, 2) divides the line segment joining A (2, 5) and B (x, y) in the ratio of 3 : 4

Now, 

$-1=\frac{3x+8}{3+4}$

and

$2=\frac{20-3y}{3+4}$

⇒ 3x + 8 = -7

and

⇒ 20 – 3y = 14

⇒ x = -5

and

⇒ y = 2

Now,

x² + y² = (-5)² + (2)²

x² + y² = 25 + 4 = 29

问题 59. ABCD 是具有顶点 A (x ₁ , y ₁ )、B (x ₂ , y ₂ )和 C (x ₃ , y ₃ ) 的平行四边形。根据 x ₁ 、 x ₂ 、 x ₃ 、 y ₁ 、 y ₂和 y ₃求第四个顶点 D 的坐标

解决方案：

Assume the coordinates of D be (x, y). We know that diagonals of a parallelogram bisect each other.

Hence, 

Mid-point of AC = Mid-point of BD 

$\left(\frac{x_1+x_3}{2},\ \frac{y_1+y_3}{2}\right)=\left(\frac{x_2+x}{2},\ \frac{y_2+y}{2}\right)$

i.e., x₁ + x₃ = x₂ + x and y₁ + y₃ = y₂ + y

i.e, x₁ + x₃ – x₂ = x and y₁ + y₃ + y₂ = y

Hence, the coordinates of D are (x₁ + x₃ – x₂, y₁ + y₃ + y₂)

问题 60. 点 A (x ₁ , y ₁ )、B (x ₂ , y ₂ )和 C (x ₃ , y ₃ ) 是 ΔABC 的顶点。

(i) A 的中位数在 D 处与 BC 相交。求 D 点的坐标。

(ii) 求 AD 上点 P 的坐标，使得 AP : PD = 2 : 1。

(iii) 分别在中位数 BE 和 CF 上找到坐标 Q 和 R 的点，使得 BQ : QE = 2 : 1 和 CR : RF = 2 : 1。

(iv) 三角形 ABC 的质心坐标是多少？

解决方案：

Given: The points A (x₁, y₁), B (x₂, y₂) and C (x₃, y₃) are the vertices of ∆ABC.

(i) We know that, the median bisect the line segment into two equal parts i.e., here D is the mid-point of BC.

Therefore,

Coordinates of mid-point of BC = $\left(\frac{x_2+x_3}{2},\ \frac{y_2+y_3}{2}\right)$

D = $\left(\frac{x_2+x_3}{2},\ \frac{y_2+y_3}{2}\right)$

(ii) Assume the coordinates of a point P be (x, y)

Given that, the point P (x, y), divide the line joining A (x_1, y₁) and 

D $\left(\frac{x_2+x_3}{2},\ \frac{y_2+y_3}{2}\right)$ in the ration of 2 : 1,

Then the coordinates of P

$=\left[\frac{2\left(\frac{x_2+x_3}{2}\right)+1.x_1}{2+1},\ \frac{2\left(\frac{y_2+y_3}{2}\right)+1.y_1}{2+1}\right]$

Using internal section formula, we get

$=\left(\frac{x_2+x_3+x_1}{3},\ \frac{y_2+y_3+y_1}{3}\right)$

Therefore,

Required coordinates of points P = $\left(\frac{x_2+x_3+x_1}{3},\ \frac{y_2+y_3+y_1}{3}\right)$

(iii) Assume the coordinates of a point Q be (p, q)

Given: The point Q (p, q) divide the line joining B(x_2, y₂) and E $=\left(\frac{x_1+x_3}{2},\ \frac{y_1+y_3}{2}\right)$

In the ratio of 2 : 1

Then the coordinates of Q

$=\left[\frac{2\left(\frac{x_1+x_3}{2}\right)+1.x_2}{2+1},\ \frac{2\left(\frac{y_1+y_3}{2}\right)+1.y_2}{2+1}\right]\\ \left(\frac{x_2+x_3+x_1}{3},\ \frac{y_2+y_3+y_1}{3}\right)$

[Since, BE is the median of side CA. So, BE divides AC into two equal parts]

Therefore,

Mid-point of AC = Coordinates of E

⇒ E = $=\left(\frac{x_1+x_3}{2},\ \frac{y_1+y_3}{2}\right)$

So, the required coordinate of point Q = $\left(\frac{x_1+x_2+x_3}{3},\ \frac{y_1+y_2+y_3}{3}\right)$

Now, let us considered that the coordinates of a point E be (α, β). 

It is given that, the point R (α, β), divide the line joining C (x₃, y₃) and 

F $\left(\frac{x_1+x_2}{2},\ \frac{y_1+y_2}{2}\right)$ in the ration 2 : 1, 

So, the coordinates of R be

$=\left[\frac{2\left(\frac{x_1+x_2}{2}\right)+1.x_3}{2+1},\ \frac{2\left(\frac{y_1+y_2}{2}\right)+1.y_3}{2+1}\right]\\ \left(\frac{x_2+x_3+x_1}{3},\ \frac{y_2+y_3+y_1}{3}\right)$

Here, CF is the median of side AB, hence CF divides AB into two equal parts

Hence, 

Mid-point of AB = Coordinates of CF

⇒ F $=\left(\frac{x_1+x_2}{2},\ \frac{y_1+y_2}{2}\right)$

So, required coordinate of point R $=\left(\frac{x_1+x_2+x_3}{3},\ \frac{y_1+y_2+y_3}{3}\right)$

(iv) Coordinate of the centroid of the ∆ABC = (sum of abscissa of all vertices/3, sum of all vertices/2)

$=\left(\frac{x_1+x_2+x_3}{3},\ \frac{y_1+y_2+y_3}{3}\right)$