Find the equation of this line

Find the equation of the line that has slope -5 and forms the hypotenuse of a triangle lying in Quadrant I with area is 10 with legs that lie on the x and y axes.

It should be written as y=

Please tell me how you did it, thank you

let's assume the coordinates of the hypotenuse: a(x1 , 0) , b(0 , y1)
m = (y1 - 0)/(0 - x1)
= -y1/x1 = -5
y1/x1 = 5 => y1 = 5x1
A = 1/2 * x1 * y1
1/2 * x1 * y1 = 10
x1 * y1 = 20 => y1 = 20/x1
5x1 = 20/x1
x1 = 2
y1 = 10
y = -5(x - 2)
y = -5x + 10

Edit: Werther (above) got the first correct answer let's check:
y = -5x + 10 => slope = -5 , y-intercept (0 , 10) , x-intercept(2 , 0)
Area = 1/2 * 2 * 10 = 10 units^2 => checks good.

y = -5x + 2 => slope = -5 ,y-intercept (0 , 2) , x-intercept(2/5 , 0)
Area = 1/2 * 2 * 2/5 = 2/5 => fails to verify.

You are welcome, and thanks for the BA.

Regards.

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Line equations have the form: y = mx + n
Here m is slope, so m = -5
Now we have y = -5x + n

It forms a triangle in Q1, so it passes from the points (a, 0) and (0, b). Replace them to the equation:

0 = -5a + n
b = -5*0 + n => b = n , replace b
-5a + b = 0 => b = 5a

a and b are side lengths of the triangle, therefore Area = 10 = (a*b) / 2 => a*b = 20
a*(5a) = 20 => a = 2 => b = 10 (We know a and b positive, since they lay on Q1)

n = b = 10

We get the equation: y = -5x + 10

a = 1/2bh
10 = 1/2bh
20= bh

h = 20/b

slope = change y/ change x = - 5
slope = change h/ change b = - 5

use = 5 to find b then determine the +- to fit the correct quad which is 1.

(20/b) / b = 5
20 /b^2 = -5
20 = 5b^2
4 = b^2
2 = b

It has to be + 2 to because the triangle lies in quad 1 and the negative slope goes from upper left to lower
right side of graph.
y = - 5x + 2