Find equation of straight line ,coordinates and distance. 30 pointss please help

We'll do it their way, but first let's find OC another way.

AB is the hypotenuse, so AB² = 5² + (5×2)² = 5²(1+2²) = 5³

AB = 5√5

OC is an altitude so the area of the triangle is

(1/2) OA OB = (1/2) OC AB

5(10) = OC (5√5)

OC = 10/√5 = 2√5

OK,  we'll check that later.

a)

Equation of AB.  Point point form for the join of (a,b) and (c,d) is

(c-a)(y-b)=(d-b)(x-a)

A(0,5), B(10,0)

(10 - 0)(y - 5) = (0 - 5)(x - 0)

10y - 50 = -5x

 AB is x + 2y = 10

OC is perpendicular so we swap the coefficients on x and y, negating one.  Through the origin means a zero constant.

OC is 2x - y = 0

b)

We find the meet, second equation times 2:

4x - 2y = 0

Add first equation,

5x = 10

x = 2

y = 2x = 4

Check 2(2) -4 = 0, good

C is (2,4)

distance d satisfies

d² = 2² + 4² = 4 + 16 = 20

d = 2√5

Distance 2√5

That's what we got at the outset.  Math works!

answer:   28.56 feet (approx)

step-by-step explanation:

since, triangle abc has measure of angle c equal to 55 degrees, measure of angle abc equal to 90 degrees, and length of bc equal to 20 feet.

we have to find out the measurement of segment ab.

therefore, tan 55° =

but bc = 20 feet.

⇒ tan 55 ° =  

⇒ ab = 20 × tan 55 °

⇒ ab = 20 × 1.42814800674

⇒ ab = 28.5629601348   ≈ 28.56 feet