The line $y = b-x$ with $0 < b < 4$ intersects the $y$-axis at $p$ and the line $x=4$ at $s$. if the ratio of the area of triangle $qrs$ to the area of triangle $qop$ is 9: 25, what is the value of $b$? express the answer as a decimal to the nearest tenth.

b=2.5

Step-by-step explanation:

The given line has equation:

The x-intercept is (b,0)

The y-intercept is (0,b)

The area of triangle QOP

The area of triangle QRS

The area of triangle QRS and QOP are in the ratio 9:25

We must negate the area under the x-axis.

But 0<b<4

Therefore b=2.5

VERIFY

The area of triangle QOP

The area of triangle QRS

We must take absolute value because this area is below the x-axis.

there are lots of ways.   pick's theorem tells us if the vertices are lattice points the area a is

a = b/2 + i - 1

where b is the number of boundary lattice points and i is the number of interior lattice points.

in the figure i count b=16, i=50 so

a = 8 + 50 - 1 = 57

let's try the shoelace formula.   the area is half the sum of the cross products of the sides.  

a(-2,6)   -2(2)-6(-5) = 26

b(-5,2)   -5(-3) - 2(-2) = 19

c(-2,-3) -2(-3) - (-3)(4) = 18

d(4,-3)   4(3) - (-3)(5) = 27

e(5,3)   5(3) - 3(1) = 12

f(1,3)     1(6) - 3(-2) = 12

a(-2,6)

each of the cross products is twice the area of the triangle with the associated side and one vertex the origin. so the area of abo=26/2=13.

area = (1/2) (26 + 19 +   18 + 27 + 12 + 12) = 57

that checks!

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step-by-step explanation:

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