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

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Solution

The correct option is A

We have to use the equation $T_{1} = 0$
for getting the tangent at the point $(x_{1}, y_{1})$ situated on the curve.

The required tangent is,

$T_{1} = 0$

i.e., $\frac{y + y_{1}}{2} = x x_{1} - \frac{- 5 (x + x_{1})}{2} + 6$

For (2,0)

$T_{1} = 0$

$\frac{y}{2} = 2 x \frac{- 5}{2} (x + 2) + 6$

$y = 4 x - 5 x - 10 + 6$

$y + x + 4 = 0$

$\Rightarrow x + y + 4 = 0$ .....(1)

For (3,0)

$T_{1} = 0$

$\frac{y}{2} = 3 x - \frac{5}{2} (x + 3) + 6$

$y = 6 x - 5 x - 15 + 6$

x-y-9=0 .............(2)

from equation (1) &(2)

$m_{1} \times m_{2} = - 1$

$∴ A n g l e b e t w e e n t a n g e n t = 90^{\circ}$

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