Question

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

(a) x − y + 2 = 0 = x − y − 1
(b) x + y − 1 = 0 = x − y − 2
(c) x − y − 1 = 0 = x − y
(d) x − y = 0 = x + y

Answer 1

(b) x + y − 1 = 0 = x − y − 2

Let the tangent meet the x-axis at point (x, 0).
Now,
$$\begin{gathered} y=x^2-3x+2 \\ \Rightarrow \frac{dy}{dx}=2x-3 \\ \text{The tangent passes through point (}\mathit{\text{x}}\text{, 0).} \\ \therefore 0=x^2-3x+2 \\ \Rightarrow \left(x-2\right)\left(x-1\right)=0 \\ \Rightarrow x=2\text{or}x=1 \\ \text{Case 1: When}\mathit{\text{x}}\text{=2:} \\ \text{Slope of the tangent,}\mathit{\text{m}}\text{=}{\left(\frac{dy}{dx}\right)}_{\left(2,0\right)}\text{=4-3=1} \\ \therefore \left(x_1,y_1\right)=\left(2,0\right) \\ \text{Equation of the tangent:} \\ y-y_1=m\left(x-x_1\right) \\ \Rightarrow y-0=1\left(x-2\right) \\ \Rightarrow x-y-2=0 \\ \text{Case 2: When}\mathit{\text{x}}\text{=1:} \\ \text{Slope of the tangent,}\mathit{\text{m}}\text{=}{\left(\frac{dy}{dx}\right)}_{\left(2,0\right)}\text{=2-3=-1} \\ \therefore \left(x_1,y_1\right)=\left(1,0\right) \\ \text{Equation of the tangent:} \\ y-y_1=m\left(x-x_1\right) \\ \Rightarrow y-0=-1\left(x-1\right) \\ \Rightarrow x+y-1=0 \end{gathered}$$