A line contains points (4,a), has direction (a,-3) and has x-intercept 5.?

Question

A line contains points (4,a), has direction (a,-3) and has x-intercept 5. One possible value for "a" rounded to 2 decimals is ...

Answer 1

Since the x-intercept is 5 the line must pass through(5,0).
Slope of line = (a-0)/(4-5) = -a
Slope of line also = (0-(-3))/(5-a) = 3/(5-a)
So -a =3/(5-a)
-5a+a^2 =3
a^2-5a-3=0
a =[5 +/- sqrt(25+12)]/2
a = 2.5 +.5 sqrt(37) = 5.54

Answer 2

The slope of the line is given by the two points

(4,a) and (5,0)

and can be calculated as

m = ∆y / ∆x = (a - 0) / (4 - 5) = a/-1 = -a

But the slope of the line is also represented by the direction
(a,-3) which implies the slope is

m = -3/a

Setting the two equations for m equal we have:

-a = -3/a
a² = 3
a = √3 ≈ 1.73