Question

Find the coordinates of points which trisect the line segment joining the point A(5,-3) and B (2,-9)​

Answer 1

Answer: (4, -5)

A trisection point on a line segment trisects that line segment into three equal parts.

Here,

1 : 2

A •------------•----------------------------• B

C

Given that,

• A = (5, -3)
• B = (2, -9)
• C = (x, y)

Which means, The point c divides the line segment in the ratio 1 : 2 which is m : n

Using the section formula, we can find the coordinate of the point C,

⇒ Cₓ = ( mx₂ + nx₁ ) / (m + n)

⇒ Cₓ = (1 × 2 + 2 × 5) / (2 + 1)

⇒ Cₓ = (2 + 10) / 3

⇒ Cₓ = 12 / 3

⇒ Cₓ = 4

Here, We got the abscissa of the point C to be 4.

Let us find the ordinate now,

⇒ Cᵧ = ( my₂ + ny₁ ) / (m + n)

⇒ Cᵧ = ( 1 × -9 + 2 × -3 ) / (1 + 2)

⇒ Cᵧ = ( -9 - 6 ) / 3

⇒ Cᵧ = -15/3

⇒ Cᵧ = -5

So, we got the ordinate to be -5.

Hence, The point C is (4, -5)

Note:-

• Abscissa = x
• Ordinate = y
• Coordinate = (Abscissa, Ordinate)

Answer 2

Question:-

• Find the coordinates of points which trisect the line segment joining the point A(5,-3) and B (2,-9).

To Find:-

• Find the coordinates.

Solution:-

Given ,

• A = ( 5 , - 3 )

• B = ( 2 , - 9 )

• C = ( x , y )

$\tt\implies \: C_x = \dfrac { mx_2 + nx_1 } { m + n }$

$\tt\implies \: C_x = \dfrac { 1 \times 2 + 2 \times 5 } { 2 + 1 }$

$\tt\implies \: C_x = \cancel\dfrac { 12 } { 3 }$

$\tt\implies \: C_x = 4$

Now ,

$\tt\implies \: C_y = \dfrac { my_2 + ny_1 } { m + n }$

$\tt\implies \: C_y = { 1 \times - 9 + 2 \times - 3 } { 1 + 2 }$

$\tt\implies \: C_y = \cancel\dfrac { - 15 } { 3 }$

$\tt\implies \: C_y = - 5$

Hence ,

• Point C is ( 4 , - 5 )