Angle between the tangents to the curve $y = x^{2} - 5 x + 6$ at points $(2, 0)$ and $(3, 0)$ is?

\frac{π}{6}

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\frac{π}{4}

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\frac{π}{3}

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\frac{π}{2}

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Solution

The correct option is C $\frac{π}{2}$
$y = x^{2} - 5 x + 6 \Rightarrow \frac{d y}{d x} = (2 x - 5)$
$∴ m_{1} = \frac{d y}{d x} |_{p (2, 0)} = 4 - 5 = - 1$
and $m_{2} = \frac{d y}{d x} |_{Q (3, 0)} = 6 - 5 = 1$
As, $m_{1} \times m_{2} = - 1 \Rightarrow θ = \frac{π}{2}$ .

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