When crossing point a (- 5, - 4) on a straight line L, it intersects two coordinate axes and the area of the triangle enclosed by the two axes is 5, the equation for solving the straight line L is obtained 1. Let the slope of the line be K y+4=k(x+5) x=0,y=5k-4 y=0,x=4/k-5=(4-5k)/k So area = | 5k-4 | * | (4-5k) / K} / 2 = 5 |(5k-4)^2/k|=10 (5k-4)^2=±10k 25k^2-40k+16=±10k -No solution at 10K 25k^2-50k+16=0 (5k-8)(5k-2)=0 k=8/5,k=2/5 8x-5y+20=0 2x-5y-10=0 Or? 2. Let y = KX + 5k-4. (k is not equal to 0) Let y = 0, x = (4-5k) / K Let x = 0.y = 5k-4 S = 1 / 2 * {5k-4} * {(4-5k) / K} ({} is absolute value.) Let k > 4 / 5, then (5k-4) ^ 2 / k = 10, k = 8 / 5, k = 2 / 5 (rounding) Then y = 8 / 5x + 4 When 0

When crossing point a (- 5, - 4) on a straight line L, it intersects two coordinate axes and the area of the triangle enclosed by the two axes is 5, the equation for solving the straight line L is obtained 1. Let the slope of the line be K y+4=k(x+5) x=0,y=5k-4 y=0,x=4/k-5=(4-5k)/k So area = | 5k-4 | * | (4-5k) / K} / 2 = 5 |(5k-4)^2/k|=10 (5k-4)^2=±10k 25k^2-40k+16=±10k -No solution at 10K 25k^2-50k+16=0 (5k-8)(5k-2)=0 k=8/5,k=2/5 8x-5y+20=0 2x-5y-10=0 Or? 2. Let y = KX + 5k-4. (k is not equal to 0) Let y = 0, x = (4-5k) / K Let x = 0.y = 5k-4 S = 1 / 2 * {5k-4} * {(4-5k) / K} ({} is absolute value.) Let k > 4 / 5, then (5k-4) ^ 2 / k = 10, k = 8 / 5, k = 2 / 5 (rounding) Then y = 8 / 5x + 4 When 0

Let the slope of the line be KY + 4 = K (x + 5) x = 0, y = 5k-4y = 0, x = 4 / K-5 = (4-5k) / K, so the area = |5k-4 | * | (4-5k) / K} / 2 = 5 | (5k-4) ^ 2 / K | = 10 (5k-4) ^ 2 = ± 10k25k ^ 2-40k + 16 = ± 10k-10k, there is no solution 25K ^ 2-50k + 16 = 0 (5k-8) (5k-2) = 0k = 8 / 5, k = 2 / 58x-5y + 20 = 02x-5y-10 = 0