A straight line L passing through the point (- 5, - 4) intersects two coordinate axes and is tangent to the triangle formed by the two axes. The area of the triangle is 5. The equation of L is obtained . urgent

A straight line L passing through the point (- 5, - 4) intersects two coordinate axes and is tangent to the triangle formed by the two axes. The area of the triangle is 5. The equation of L is obtained . urgent

Let the slope of the line l be K, and the equation is y = KX + B. the intersection of the line L and the coordinate axis is a (- B / K, 0), B (0, b),
Because the point (- 5, - 4) is on the line L, so, - 4 = - 5K + B, B = 5k-4,
Because the area of the line L, which intersects the coordinate axis and forms a triangle with it, is 5 square units,
So, 1 / 2 * | - B / K | * | - B | = 5, (5k-4) ^ 2 = 10 ||
The square of both sides is (5k-4) ^ 4 = 100k ^ 2
[(5k-4)^2-10k]*[(5k-4)^2+10k]=0
(5k-2)(5k-8)(25k^2-30k+16)=0
So, k = 2 / 5, k = 8 / 5
b=-2,b=4
So the equation of line L is: y = 2 / 5x-2, y = 8 / 5x + 4