Question 1185850
The lines {{{y=4x-10 }}}and {{{y=x-1 }}}intersect at the point {{{T}}}. 

{{{y=4x-10 }}}....eq.1
 {{{y=x-1 }}}..........eq.2
----------------------------if left sides equal, right sides are equal to

{{{4x-10 =x-1}}}.....solve for {{{x}}}

{{{4x-x =10-1}}}

{{{3x =9}}}

{{{x =3}}}

find {{{y}}}

{{{y=3-1 }}}..........eq.2

{{{y=2 }}}

=> the point {{{T}}} is at ({{{3}}},{{{2}}})


then, the equation of the line with gradient {{{-2/3}}} that passes through the point {{{T}}} is:

{{{y-y[1]=m(x-x[1])}}}

{{{y-2=-(2/3)(x-3)}}}

{{{y-2=-(2/3)x+(2/3)3}}}

{{{y-2=-(2/3)x+2}}}

{{{y=-(2/3)x+2+2}}}

{{{y=-(2/3)x+4}}}




{{{ drawing( 600, 600, -10, 10, -10, 10,
circle(3,2,.12), locate(3,2,T(3,2)),locate(-3,4.5,y=-(2/3)x+4),locate(-3,-3,y=x-1),locate(2.5,5,y=4x-10),
graph( 600, 600, -10, 10, -10, 10, -(2/3)x+4, 4x-10, x-1)) }}}