# If a line parallel but nont identical with x-axis cuts the graph of the curve y=(x-1)/((x-2)(x-3)) at x=a,x=b then (a,b)=?

Jul 29, 2018

See explanation.

#### Explanation:

y-intercept $= - \frac{1}{6}$.

Asymptotes: $\leftarrow y = 0 \rightarrow , \uparrow x = 2 \mathmr{and} \uparrow x = 3$. .

graph{(y (x-2)(x-3)-(x-1))((x+1)^2+(y+1/6)^2-0.03)(x^2+(y+1/6)^2-0.03)=0}.

The line $y = c \ne 0$, parallel to x-axis, meets the curve at

just one point $\in {Q}_{3}$, if $c = - \frac{1}{6}$,

at $\left(- 1 , - \frac{1}{6}\right)$. See graph.

Above , $y = c > - \frac{1}{6}$ meets the curve again, at

 x = a, b = 1/2( 2.5 +1/c( 1+-sqrt( 1 + 6c + c^2)).

Jul 31, 2018

color(red)("In this question, we are asked to find the value of "

$\textcolor{b l u e}{\left(a - 1\right) \left(b - 1\right)}$

It has been informed to me.

Let a line parallel to $x$-axis be $y = c$ which cuts the given curve at $x = a \mathmr{and} x = b$

Hence we have

$c = \frac{a - 1}{\left(a - 2\right) \left(a - 3\right)} = \frac{b - 1}{\left(b - 2\right) \left(b - 3\right)}$

$\implies \frac{a - 1}{\left(a - 2\right) \left(a - 3\right)} = \frac{b - 1}{\left(b - 2\right) \left(b - 3\right)}$

$\implies {a}^{2} b - 5 a b + 6 b - {a}^{2} + 5 a - 6 = a {b}^{2} - 5 a b + 6 a - {b}^{2} + 5 b - 6$

$\implies a b \left(a - b\right) - \left(a - b\right) - \left({a}^{2} - {b}^{2}\right) = 0$

$\implies a b - 1 - a - b = 0$ [Since $a \ne b$]

$\implies a b - a - b + 1 - 2 = 0$

$\implies a \left(b - 1\right) - 1 \left(b - 1\right) = 2$

$\implies \left(a - 1\right) \left(b - 1\right) = 2$