Question

The equations of tangent at those points where the curve y = x² − 3x + 2 meets x-axis are _______________ .

Options

• x − y + 2 = 0 = x − y − 1

• x + y − 1 = 0 = x − y − 2

• x − y − 1 = 0 = x − y

• x − y = 0 = x + y

Answer 1

`x + y − 1 = 0 = x − y − 2`

 

Let the tangent meet the x-axis at point (x, 0).
Now,

\[y = x^2 - 3x + 2\]

\[ \Rightarrow \frac{dy}{dx} = 2x - 3\]

\[\text { The tangent passes through point (x, 0) }.\]

\[ \therefore 0 = x^2 - 3x + 2\]

\[ \Rightarrow \left( x - 2 \right)\left( x - 1 \right) = 0\]

\[ \Rightarrow x = 2 \ or \ x = 1\]

\[\text { Case  1: When } x=2:\]

\[\text { Slope of the tangent },m= \left( \frac{dy}{dx} \right)_\left( 2, 0 \right) =4-3=1\]

\[ \therefore \left( x_1 , y_1 \right) = \left( 2, 0 \right)\]

\[\text { Equation of the tangent }:\]

\[y - y_1 = m \left( x - x_1 \right)\]

\[ \Rightarrow y - 0 = 1 \left( x - 2 \right)\]

\[ \Rightarrow x - y - 2 = 0\]

\[\text { Case 2: When } x=1:\]

\[\text { Slope of the tangent },m= \left( \frac{dy}{dx} \right)_\left( 2, 0 \right) =2-3=-1\]

\[ \therefore \left( x_1 , y_1 \right) = \left( 1, 0 \right)\]

\[\text { Equation of the tangent }:\]

\[y - y_1 = m \left( x - x_1 \right)\]

\[ \Rightarrow y - 0 = - 1 \left( x - 1 \right)\]

\[ \Rightarrow x + y - 1 = 0\]