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!

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