Question

If ${m_i}x + \dfrac{1}{{{m_i}}}\left( {i = 1,2,3} \right)$ represent three straight lines whose slopes are the roots of the equation$2{m^3} - 3{m^3} - 3m - 2 = 0$, then

Answer 1

We have given a general equation of three lines and slopes of the lines are given by the sports of the equation $2{m^3} - 3{m^2} - 3m + 2 = 0$. We have to tell the lines by solving this. Firstly we have to solve the equation. Degree of the equation is $3$. So the number of roots will be three. Their roots will be the slope of the line. After that we will find the following
(A)The algebraic sum of the intercept mode by the line on x-axis.
(B)Algebraic sum of the intercept mode by the line on y-axis.
(C)The perpendicular distance of line from origin.
(D)Length of line intercepted between the co-ordinate axis.

Complete step-by-step answer:
The equations of lines is given as
$y = {m_i}x + \dfrac{1}{{{m_i}}}\left( {i = 1,2,3} \right)$
The equation of slopes is

$2{m^3} - 3{m^2} - 3m + 2 = 0$

$\Rightarrow$ $2{m^3} + 2 - 3{m^2} - 3 = 0$

$\Rightarrow$ $2({m^3} + 1) - 3m(m + 1) = 0$

Now we have formula ${a^3} + {b^3} = (a + b)({a^2} - ab + {b^2})$

Therefore ${m^3} + {1^3} = (m + 1)({m^2} - m + 1)$

$\Rightarrow$ $2[(m + 1)({m^2} - m + 1)] - 3m(m + 1) = 0$

$\Rightarrow$ $m + 1[2({m^2} - m + 1) - 3m] = 0$

$\Rightarrow$ $(m + 1)[2{m^2} - 2m + 2 - 3m] = 0$

$\Rightarrow$ $(m + 1)[2{m^2} - 5m + 2] = 0$

$\Rightarrow$ $(m + 1)(2m - 1)(m - 2) = 0$

$\Rightarrow$ $m + 1 = 0,2m - 1 = 0,m - 2 = 0$

$\Rightarrow$ $m = - 1,m = \dfrac{1}{2},m = 0$

Slope of the lines are $m = - 1,\dfrac{1}{2},2$
Now (A). Algebraic sum of the intercepts made by lines an x – axis
$\Rightarrow$ $\sum {\dfrac{1}{{{m_i}^2}}} = - \left[ {\dfrac{1}{{{m_1}^2}} + \dfrac{1}{{{m_2}^2}} + \dfrac{1}{{{m_2}^2}}} \right] = - \left[ {1 + \dfrac{1}{4} + 4} \right]$
$= \dfrac{{ - 21}}{4}$
(B) Algebraic sum of the intercept made by the line on y-axis
$\Rightarrow$ $\sum {\dfrac{1}{{{m_i}}}} = - 1 + 2 + \dfrac{1}{2} = \dfrac{3}{2}$
(C) The perpendicular distance of the line from the origin
$\Rightarrow$ $\left| {\dfrac{{ - {\raise0.5ex\hbox{$\scriptstyle 1$} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{$\scriptstyle {{m_i}}$}}}}{{\sqrt {1 + {m_i}^2} }}} \right| = \dfrac{1}{{\sqrt {1 + 1} }} + \dfrac{2}{{\sqrt {1 + {\raise0.5ex\hbox{$\scriptstyle 1$} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{$\scriptstyle 4$}}} }} + \dfrac{{{\raise0.5ex\hbox{$\scriptstyle 1$} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{$\scriptstyle 2$}}}}{{\sqrt {14} }}$
$= \dfrac{1}{{\sqrt 2 }} + \dfrac{4}{{\sqrt 5 }} + \dfrac{1}{{\sqrt {1 + {\raise0.5ex\hbox{$\scriptstyle 1$} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{$\scriptstyle 4$}}} }} + \dfrac{{{\raise0.5ex\hbox{$\scriptstyle 1$} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{$\scriptstyle 2$}}}}{{\sqrt {1 + 4} }}$
(D) Lengths of the line intercepted between co-ordinate axes
$= \sqrt {{{\left( {\dfrac{1}{{{m_i}^2}}} \right)}^2} + {{\left( {\dfrac{1}{{{m_i}}}} \right)}^2}} = \sqrt {1 + 1} + \sqrt {16 + 4} + \sqrt {\dfrac{1}{{16}} + \dfrac{1}{4}}$
$= \dfrac{{4\sqrt 2 + 9\sqrt 5 }}{4}$

Note: Slope of the line: The slope of the line characteristics the direction of the line. To find slope we divide the difference of true points to of y – co-ordinate by the difference of two co-ordinate of x - co-ordinate.
Intercepts of line: The x – intercept is the point where line crosses x – axis. y – Intercept is, point where the line crosses the y-axis.