Question

The intercepts a, b of a straight line on the co-ordinate axes are given by a + b = 5, ab = 6 then the equation of the line is​

Answer 1

Given a+b=5 __ (1)

ab=6 __ (2)

Substituting b from (2) in (1)

$\bold{⇒a+ \frac{6}{a} = 5⇒a^2 −5a+6=0}$

$\bold{⇒a^2−2a−3a+6=0}$

⇒a(a−2)−3(a−2)=0

⇒(a−3)(a−2)=0

∴a=2,3

substituting a=2,3 in (2)

$\bold{⇒b= \frac{b}{2} =3(a=2)}$

$\bold{b= \frac{b}{3} =2(a=3)}$

since a and b are +ve , the line intercept the

axes in first quadrant.

From right △AOB

$\bold{tanα= \frac{a}{b} \: or \: \frac{b}{a}}$

$\bold{ \frac{ - 2}{3} \: or \: \frac{ - 3}{2}}$

$\bold{ \frac{2}{3} \: or \: \frac{3}{2}}$

slope of the line = tanθ

=tan(180−α)=−tanα

$\small \bold{Equation \: of \: the \: line \: for \: slope \: \frac{ - 2}{3} \: is }$

$\small \bold{y = \frac{ - 2}{3}x+2⇒3y=−2x+6}$

$\small \bold{Equation \: of \: the \: line \: for \: slope \: \frac{ - 3}{2} \: is }$

$\small \bold{y = \frac{ - 3}{2}x+3⇒2y=−3x+6}$

∴ Required equation are 2x+3y−6=0 and 3x+2y−6=0

(or) 2x+3y−6=0 or 3x+2y−6=0