A line passes through the point (10,-2) and forms with the axes a triangle of area of 9 sq units. Find the equation of the line.

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Denote the slope of this line as m. The vertical line (which has an undefined slope) doesn't suit us, so we'll not miss a solution. Horizontal line with m=0 doesn't suit also, so we can divide by m.
The equation of such a line is y = m*(x-10) - 2. The triangle formed with this line and the axes is a right one (because the axes are perpendicular to each other). So its area is 1/2 * |OX| * |OY|, where O is the origin, X is the x-intercept of the line and Y is the y-intercept.
The y-intercept is y(0) = -10m-2. The x-intercept is the x for which y= m*(x-10) - 2=0, so it is 2/m+10.
Thus our equation for m becomes
1/2 |(-10m-2)*(2/m+10)| = 9.
It is the same as |(5m+1)*(1/m+5)| = 9/2, or |1/m (5m+1)^2| = 9/2.
If we suppose m is positive, then it becomes
(5m+1)^2 = 9/2 m, or 25m^2+10m+1=9/2 m, or 25m^2+11/2 m +1=0, or 50m^2+11m+2=0. This equation has no solutions.
Well, what about negative m's? The equation becomes (5m+1)^2 = -9/2 m, or 25m^2+10m+1=-9/2 m, or 25m^2+29/2 m +1=0, or 50m^2+29m+2=0.
The discriminant is D = 29^2-4*50*2 = 29^2 - 20^2 = 9*49, so sqrt(D)=3*7=21. The solutions are (-29+-21)/100, m_1 = -50/100=-1/2, m_2 = -8/100 = -2/25. And both are negative as supposed.
Uff. There are two possible equations, y=-1/2(x-10)-2=-1/2x+3 and y=-2/25(x-10)-2=-2/25 x-6/5.