Question 1155733
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The solution by the other tutor is incomplete; there are two lines that satisfy the given conditions.<br>
In her solution, she assumes that, since the product of the two intercepts is 3, one of the must be 3 and the other must be 1.  That is not true.<br>
Let the intercepts be (a,0) and (0,b).<br>
The equation of the line is y = mx+b.  We know
(1) {{{m = -b/a}}}
(2) (6,-1) is on the line, so {{{-1 = (-b/a)(6)+b}}}<br>
(3) {{{ab=3}}}<br>
Simplify (2) and substitute (3):<br>
{{{-1 = (-b/a)(6)+b}}}
{{{-a = -6b+ab}}}
{{{-a = -6b+3}}}
{{{a = 6b-3}}} (4)<br>
Substitute (4) in (3):<br>
{{{(6b-3)(b) = 3}}}
{{{6b^2-3b-3 = 0}}}
{{{2b^2-b-1 = 0}}}
{{{(2b+1)(b-1) = 0}}}
{{{b = -1/2}}} or {{{b = 1}}}<br>
ANSWERS:<br>
b=-1/2 and ab=3 means a = -6; the intercepts are (-6,0) and (0,-1/2); the equation is y = (-1/12)x-1/2<br>
b = 1 and ab=3 means a=3; the intercepts are (3,0) and (0,1); the equation is y = (-1/3)x+1<br>
A graph....<br>
{{{graph(400,400,-10,10,-4,4,(-1/12)x-1/2,(-1/3)x+1)}}}<br>