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.`

 

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