Question

The equation of the line passing through origin which is parallel to the tangent of the curve \(y=\dfrac{x-2}{x-3}\) at \(x=4\) is 

• A. \(y=2x \)
• B. \(y=-2x+1\)
• C. \( y=-x\)
• D. \( y=x+2 \)
• E. \( y=4x\)

Answer 1

The Correct Option is C

Given: \(y=\dfrac{x-3}{x-2}\)

\(⇒\dfrac{dy}{dx}=\dfrac{(x-3).1-(x-2).1}{(x-3)^2}\)

\(⇒\dfrac{dy}{dx}=\dfrac{-1}{(x-3)^2}\)

Then  ⇒\(\dfrac{dy}{dx}\) at \(x=4\) \(=\dfrac{-1}{(4-3)^2}\)

                                    \(=-1\)

means the slope\(=-1\)

Therefore, the equation of the line passing through origin which is parallel to the tangent of the curve \(y=\dfrac{x-2}{x-3}\) at \(x=4\) is 

\(y=-x\) 

Answer 2

Step 1: Differentiate the curve equation

We are given the curve equation:

\[ y = \frac{x - 2}{x - 3} \]

To find the slope of the tangent at \( x = 4 \), we need to differentiate \( y \) with respect to \( x \). We will use the quotient rule for differentiation:

\[ \text{If } y = \frac{u(x)}{v(x)}, \text{ then } y' = \frac{u'v - uv'}{v^2} \]

Here, \( u(x) = x - 2 \) and \( v(x) = x - 3 \). Therefore:

• \( u'(x) = 1 \)
• \( v'(x) = 1 \)

Now, applying the quotient rule:

\[ y' = \frac{(1)(x - 3) - (x - 2)(1)}{(x - 3)^2} \] \[ y' = \frac{(x - 3) - (x - 2)}{(x - 3)^2} \] \[ y' = \frac{x - 3 - x + 2}{(x - 3)^2} \] \[ y' = \frac{-1}{(x - 3)^2} \]

Step 2: Find the slope at \( x = 4 \)

Substitute \( x = 4 \) into the derivative to find the slope of the tangent at this point:

\[ y'(4) = \frac{-1}{(4 - 3)^2} = \frac{-1}{1^2} = -1 \]

Step 3: Equation of the line passing through the origin and parallel to the tangent

The line that passes through the origin and is parallel to the tangent will have the same slope as the tangent at \( x = 4 \). So, the slope of the required line is \( -1 \).

The equation of a line with slope \( m \) passing through the origin is:

\[ y = mx \]

Since the slope is \( -1 \), the equation of the line is:

\[ y = -x \]