Answer
385.8k+ views
Hint:Here we will use the concepts of the algebra and the arithmetic equations. Also, we will concentrate on the reasoning and the calculus. Remember the slope of the two parallel lines are always equal.
Complete step by step solution:
A line which is parallel to the given line is $x - 2y = 2$ must have the same slope. Comparing the given equation with the standard line equation: $y = mx + c$
Take the given expression: $x - 2y = 2$
Convert the above equation in the form of the standard equation.
$ - 2y = 2 - x$
Taking constant on the opposite side.
$ \Rightarrow y = - 1 + \dfrac{x}{2}$
So, the slope of the given line is equal to $\dfrac{1}{2}$
Now, the slope of the tangent line can be derived using the derivation.
Now, the given expression is $y = \dfrac{{x - 1}}{{x + 1}}$ …(A)
So, here we will apply $\left( {\dfrac{u}{v}} \right)$rule to find the derivation.
$y' = \dfrac{{x(x - 1) - (x - 1).1}}{{{{\left( {x + 1} \right)}^2}}}$
Simplify the above equation, like terms with the same value and the opposite sign cancel each other.
$y' = \dfrac{2}{{{{\left( {x + 1} \right)}^2}}}$
Now, to find the value of “x”, take $y' = \dfrac{1}{2}$ in the above equation.
$\dfrac{1}{2} = \dfrac{2}{{{{\left( {x + 1} \right)}^2}}}$
Take cross multiplication, where the numerator of one side is multiplied with the denominator of the opposite side and vice-versa.
$ \Rightarrow {(x + 1)^2} = 4$
Take the square root on both the sides of the equation.
$ \Rightarrow \sqrt {{{(x + 1)}^2}} = \sqrt 4 $
Square and square root cancel each other on the left hand side of the equation.
$ \Rightarrow x + 1 = \pm 2$
So, $x + 1 = 2$ or $x + 1 = - 2$
When you move any term from one side of the equation to the other side, then the sign of the term changes. Positive terms become negative and vice-versa.
$ \Rightarrow x = 1$ or $x = ( - 3)$
Now, placing the values of “x” in the equation (A)
When $x = 1,$
$ \Rightarrow y = 0$
And when $x = ( - 3)$
$ \Rightarrow y = 2$
Therefore, the equation of the line $(1,0)$ with the slope $m = \dfrac{1}{2}$ and equation of the line through $( - 3,2)$ with the slope $m = \dfrac{1}{2}$ using the $y - {y_0} = m(x - {x_0})$
Which gives –
$y = \dfrac{1}{2}x + \dfrac{7}{2}$ and
$y = \dfrac{1}{2}x - \dfrac{1}{2}$
Note: Be careful about the sign convention and the simplification of the algebraic expressions. Always remember when you move any term from one side to another sign of the term changes. Positive terms become negative and vice-versa.
Complete step by step solution:
A line which is parallel to the given line is $x - 2y = 2$ must have the same slope. Comparing the given equation with the standard line equation: $y = mx + c$
Take the given expression: $x - 2y = 2$
Convert the above equation in the form of the standard equation.
$ - 2y = 2 - x$
Taking constant on the opposite side.
$ \Rightarrow y = - 1 + \dfrac{x}{2}$
So, the slope of the given line is equal to $\dfrac{1}{2}$
Now, the slope of the tangent line can be derived using the derivation.
Now, the given expression is $y = \dfrac{{x - 1}}{{x + 1}}$ …(A)
So, here we will apply $\left( {\dfrac{u}{v}} \right)$rule to find the derivation.
$y' = \dfrac{{x(x - 1) - (x - 1).1}}{{{{\left( {x + 1} \right)}^2}}}$
Simplify the above equation, like terms with the same value and the opposite sign cancel each other.
$y' = \dfrac{2}{{{{\left( {x + 1} \right)}^2}}}$
Now, to find the value of “x”, take $y' = \dfrac{1}{2}$ in the above equation.
$\dfrac{1}{2} = \dfrac{2}{{{{\left( {x + 1} \right)}^2}}}$
Take cross multiplication, where the numerator of one side is multiplied with the denominator of the opposite side and vice-versa.
$ \Rightarrow {(x + 1)^2} = 4$
Take the square root on both the sides of the equation.
$ \Rightarrow \sqrt {{{(x + 1)}^2}} = \sqrt 4 $
Square and square root cancel each other on the left hand side of the equation.
$ \Rightarrow x + 1 = \pm 2$
So, $x + 1 = 2$ or $x + 1 = - 2$
When you move any term from one side of the equation to the other side, then the sign of the term changes. Positive terms become negative and vice-versa.
$ \Rightarrow x = 1$ or $x = ( - 3)$
Now, placing the values of “x” in the equation (A)
When $x = 1,$
$ \Rightarrow y = 0$
And when $x = ( - 3)$
$ \Rightarrow y = 2$
Therefore, the equation of the line $(1,0)$ with the slope $m = \dfrac{1}{2}$ and equation of the line through $( - 3,2)$ with the slope $m = \dfrac{1}{2}$ using the $y - {y_0} = m(x - {x_0})$
Which gives –
$y = \dfrac{1}{2}x + \dfrac{7}{2}$ and
$y = \dfrac{1}{2}x - \dfrac{1}{2}$
Note: Be careful about the sign convention and the simplification of the algebraic expressions. Always remember when you move any term from one side to another sign of the term changes. Positive terms become negative and vice-versa.
Recently Updated Pages
How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Why Are Noble Gases NonReactive class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let X and Y be the sets of all positive divisors of class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x and y be 2 real numbers which satisfy the equations class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x 4log 2sqrt 9k 1 + 7 and y dfrac132log 2sqrt5 class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x22ax+b20 and x22bx+a20 be two equations Then the class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Trending doubts
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
At which age domestication of animals started A Neolithic class 11 social science CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Which are the Top 10 Largest Countries of the World?
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Give 10 examples for herbs , shrubs , climbers , creepers
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference Between Plant Cell and Animal Cell
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Write a letter to the principal requesting him to grant class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Change the following sentences into negative and interrogative class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)