Question

a line segment is of length 10 units and one of its and is (- 2,3) if the ordinate of the other end is 9 then determine the absicca of the other end​

Answer 1

Answer:-

Let the absissca of the other point be x.

Given:-

Distance between (- 2 , 3) & (x , 9) = 10 units. (According to the question).

We know that,

Distance between two points (x₁ , y₁) & (x₂ , y₂) is:

$\red{\sf \: D = \sqrt{ {(x_2 - x_1)}^{2} + {(y_2 - y_1)}^{2} } }$

Let,

• x₁ = - 2
• y₁ = 3
• x₂ = x
• y₂ = 9

Hence,

$\implies \sf \: 10 = \sqrt{(x - ( - 2)) ^{2} + (9 - 3) ^{2} }$

Squaring both sides we get,

$\implies \sf \: {(10)}^{2} = {(x + 2)}^{2} + {6}^{2}$

using (a + b)² = a² + b² + 2ab we get,

$\: \implies \sf \:100 = {x}^{2} + 4 + 4x + 36\\ \\ \\ \implies \sf \:0 = {x}^{2} + 4x + 40 - 100 \\ \\ \\ \implies \sf \: {x}^{2} + 4x - 60 = 0$

$\implies \sf \: {x}^{2} + 10x - 6x - 60 = 0 \\ \\ \\ \implies \sf \:x(x + 10) - 6(x + 10) = 0 \\ \\ \\ \implies \sf \:(x + 10)(x - 6) = 0 \\ \\ \\ \implies \boxed{\sf \:x = (- 10 \: ,\: 6)}$

∴ The abscissa of the other end is - 10 or 6.

Answer 2

Answer:

Let the absissca of the other point be x.

Given:-

Distance between (- 2 , 3) & (x , 9) = 10 units. (According to the question).

We know that,

Distance between two points (x₁ , y₁) & (x₂ , y₂) is:

\red{\sf \: D = \sqrt{ {(x_2 - x_1)}^{2} + {(y_2 - y_1)}^{2} } }D=

(x

2

−x

1

)

2

+(y

2

−y

1

)

2

Let,

x₁ = - 2

y₁ = 3

x₂ = x

y₂ = 9

Hence,

\implies \sf \: 10 = \sqrt{(x - ( - 2)) ^{2} + (9 - 3) ^{2} } ⟹10=

(x−(−2))

2

+(9−3)

2

Squaring both sides we get,

\implies \sf \: {(10)}^{2} = {(x + 2)}^{2} + {6}^{2} ⟹(10)

2

=(x+2)

2

+6

2

using (a + b)² = a² + b² + 2ab we get,

\begin{gathered} \: \implies \sf \:100 = {x}^{2} + 4 + 4x + 36\\ \\ \\ \implies \sf \:0 = {x}^{2} + 4x + 40 - 100 \\ \\ \\ \implies \sf \: {x}^{2} + 4x - 60 = 0\\\\\\ \end{gathered}

⟹100=x

2

+4+4x+36

⟹0=x

2

+4x+40−100

⟹x

2

+4x−60=0

\begin{gathered} \implies \sf \: {x}^{2} + 10x - 6x - 60 = 0 \\ \\ \\ \implies \sf \:x(x + 10) - 6(x + 10) = 0 \\ \\ \\ \implies \sf \:(x + 10)(x - 6) = 0 \\ \\ \\ \implies \boxed{\sf \:x = (- 10 \: ,\: 6)}\end{gathered}

⟹x

2

+10x−6x−60=0

⟹x(x+10)−6(x+10)=0

⟹(x+10)(x−6)=0

⟹

x=(−10,6)