Question

#Only maths expert's can answer..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:

Distance between two points  =  

(x₂  − x₁ ) ^2  + (y₂  − y₁)^2

 

Given, one point =(2,−3)

Let, other point =(10,b)

absicca is x cordinate

∴  √ (10 − 2) ^2 + (b + 3) ^2  = 10

√8^ 2  + (b^2 + 9 + 6b)  = 10

Squaring both the sides, we get

64 + b^2    + 9 + 6b = 100

(b + 3)^2  = 36

∴ b+3=6 and b+3=−6      

∴ b = 3 and b = −9

Hope it helps you!!

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,

$\begin{gathered} \begin{gathered} : \implies\sf 10 = \sqrt{(x - ( - 2)) ^{2} + (9 - 3) ^{2} } \\ \\ \frak {\color{navy}{ Squaring \: both \: sides \: we \: get,}} \: \\\\ \sf : \implies \sf \: {(10)}^{2} = {(x + 2)}^{2} + {6}^{2} \end{gathered}\end{gathered}$

$\begin{gathered}\\\\\end{gathered} \boxed{ \frak \color{navy}{using (a + b)² = a² + b² + 2ab  we \: get,}}$$\begin{gathered}\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}\end{gathered}$

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

Thank you!!

@itzshivani