Question 186465
One line would be
{{{y = mx + b[1]}}}
The other line would be
{{{y = -(1/m)x + b[2]}}}
{{{b[1]}}} and {{{b[2]}}} are the y-intercepts
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Let the x-intercepts be {{{x[1]}}} and {{{x[2]}}}
{{{y = 0}}} at the x-intercepts, so
{{{0 = m*x[1] + b[1]}}}
{{{x[1] = -(b[1]/m)}}}
and
{{{0 = -(1/m)*x[2] + b[2]}}}
{{{x[2] = m*b[2]}}}
{{{x[1] + x[2] = m*b[2] - b[1]/m}}}
and
{{{y = mx + b[1]}}}
{{{2 = 9m + b[1]}}}
{{{b[1] = 2 - 9m}}}
and
{{{2 = -(1/m)*9 + b[2]}}}
{{{b[2] = 2 + 9/m}}}
Substituting:
{{{x[1] + x[2] = m*(2 + 9/m) - (2 - 9m)/m}}} 
{{{x[1] + x[2] = 2m + 9 - 2/m + 9}}} 
{{{x[1] + x[2] = 2(m - 1/m) + 18}}} this is the answer
Suppose the slopes of the lines are
{{{m = -3}}}
{{{-(1/m) = 1/3}}}
and the solution is (9,2)
the equations are:
{{{y = -3x + b[1]}}}
{{{2 = -3*9 + b[1]}}}
{{{b[1] = 2 + 27}}}
{{{b[1] = 29}}}
{{{y = -3x + 29}}} 1st equation
and
{{{y = (1/3)*x + b[2]}}}
{{{2 = (1/3)*9 + b[2]}}}
{{{b[2] = 2 - 3}}}
{{{b[2] = -1}}}
{{{y = (1/3)*x - 1}}} 2nd equation
Setting {{{y = 0}}} in both will find the x-intercepts
{{{y = -3x + 29}}}
{{{0 = -3*x[1] + 29}}}
{{{x[1] = 29/3}}}
and
{{{0 = (1/3)*x[2] - 1}}}
{{{x[2] = 3}}}
and
{{{x[1] + x[2] = 2(m - 1/m) + 18}}}
{{{29/3 + 3 = 2(-3 - 1/(-3)) + 18}}}
{{{29/3 + 9/3 = 2(-3 + 1/3) + 18}}}
{{{29/3 + 9/3 = -6 + 2/3 + 18}}}
Multiply both sides by {{{3}}}
{{{29 + 9 = -18 + 2 + 54}}}
{{{38 = -16 + 54}}}
{{{38 = 38}}}
OK