If the slope of a straight line (m<0) and the line passes through point A(1;8). How do I represent the line's equation using m?

Using the point-slope form of linear equation, you can represent the equation of the line as :
y - 8 = m(x - 1)
y - 8 = mx - m
mx - y - (m - 8) = 0


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When x = 0 : y = -(m - 8)
When y = 0 : x = (m - 8)/m

Area of the triangle that the line creates with the x and y axis :
(1/2) × [-(m - 8)] × [(m - 8)/m] = 16
- (m - 8)² = 32m
- m² + 16m - 64 = 32m
m² + 16m + 64 = 0
(m + 8)² = 0
m = -8

　
Since slope is negative, and line passes through quadrant 1, then
x-intercept = a > 0
y-intercept = b > 0

Area of triangle = 1/2 * a * b = 16 ----> ab = 32

Equation of line in intercept-intercept form:
x/a + y/b = 1

Since line passes through point (1,8) we get:
1/a + 8/b = 1
b + 8a = ab
8a + b = 32
b = 32−8a
ab = 32
a(32−8a) = 32
32a − 8a² = 32
8a² − 32a + 32 = 0
8 (a² − 4a + 4) = 0
8 (a − 2) = 0
a = 2
b = 16

m = −b/a = −8

Equation of line:
x/2 + y/16 = 1
8x + y = 16
y = −8x + 16

You can check that this line does indeed pass through point (1,8)

you are STUPID