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 ordinate be x

then, by distance formula,

d=

(x

2

−x

1

)

2

+(y

2

−y

1

)

2

⇒10=

(10−2)

2

+(x+3)

2

⇒10

2

=8

2

+(x+3)

2

⇒100=64+x

2

+6x+9

⇒x

2

+6x−27=0

⇒(x−3)(x+9)=0

⇒x=3,−9

Hence the ordinate of the points is 3 or −9.

Answer 2

Given:

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

To determine:

• The absicca of the other end

Solution:

We know that,

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

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

Let,

x₁ = - 2

y₁ = 3

x₂ = x

y₂ = 9

Hence,

$\begin{gathered} : \implies\sf 10 = \sqrt{(x - ( - 2)) ^{2} + (9 - 3) ^{2} } \\ \\ \frak {\color{navy}{ Squaring \: both \: sides \: we \: get,}} \: \\\\ \sf : \implies \sf \: {(10)}^{2} = {(x + 2)}^{2} + {6}^{2} \end{gathered}$$\boxed{ \frak \color{navy}{using (a + b)² = a² + b² + 2ab  we \: get,}} \red\bigstar$

$\begin{gathered} : \implies \sf \:100 = {x}^{2} + 4 + 4x + 36\\ \\ : \implies \sf \:0 = {x}^{2} + 4x + 40 - 100 \\ \\ : \implies \sf \: {x}^{2} + 4x - 60 = 0\\\\ \end{gathered} \\ \begin{gathered} : \implies \sf \: {x}^{2} + 10x - 6x - 60 = 0 \\ \\ : \implies \sf \:x(x + 10) - 6(x + 10) = 0 \\ \\ : \implies \sf \:(x + 10)(x - 6) = 0 \\ \\ : \implies \underline{\boxed{\frak{(- 10 \: ,\: 6)}}} \pink\bigstar\end{gathered}$

$\frak{The abscissa of the other end is}\textsf {\textbf{( -  10) or 6.}}$

Thank you!

@itzshivani