Question

Problem-Solving: ( Show your solutions )

The midpoint of CS has coordinates (2,-1). If the coordinates of C are (11,2), What are the coordinates of S? Explain your answer.


Please help me i need it now
Please answer it correctly
Thank you ^_^​

Answer 1

Answer:-

Let the co - ordinates of S be (x , y).

Given:-

(2 , - 1) is the midpoint of C(11 , 2) and S(x , y).

We know that,

Midpoint of the line segment joining the points (x₁ , y₁) & (x₂ , y₂) is:

$\boxed{\sf \: p(x \:, \: y) = \bigg( \dfrac{x_1 + x_2}{2} \: \:, \: \: \dfrac{y_1 + y_2}{2} \bigg)}$

Let,

• x₁ = 11
• y₁ = 2
• x₂ = x
• y₂ = y

So,

According to the question,

$\implies \sf \: ( 2\: ,\: - 1) = \bigg( \dfrac{11 + x}{2} \: \: ,\: \: \dfrac{2 + y}{2} \bigg)$

On comparing both sides we get,

★ 2 = (11 + x)/2

⟹ 4 = 11 + x

⟹ 4 - 11 = x

⟹ x = - 7

★ - 1 = (2 + y)/2

⟹ - 2 = 2 + y

⟹ - 2 - 2 = y

⟹ y = - 4

∴ The coordinates of the point S are ( - 7 , - 4).

Answer 2

Given :-

The midpoint of CS has coordinates (2,-1). If the coordinates of C are (11,2)

To Find :-

Coordinate of S

Solution :-

By using mid-point formula

$\sf (x,y) =\bigg(\dfrac{x_2 + x_1}{2} + \dfrac{y_2+y_1}{2}\bigg)$

Let the coordinate be a and b

$\sf (2,-1) = \bigg(\dfrac{a + 11}{2},\dfrac{b + 2}{2}\bigg)$

2 = a + 11/2 (Eq 1)

-1 = b + 2/2 (Eq 2)

Solving both

Eq 1

$\sf 2 = \dfrac{a+11}{2}$

$\sf 2\times2 = a+11$

$\sf 4=a+11$

$\sf 4-11=a$

$\sf -7=a$

Eq 2

$\sf -1=\dfrac{b+2}{2}$

$\sf -1 \times 2 =b+2$

$\sf -2 =b+2$

$\sf -2 - 2 = b$

$\sf -4 = b$

Hence

Coordinates = (a,b)

Coordinates = (-7, -4)