Answer:
B = (-b, a)
C = (a, 0)
D = (-b, 0)
Step-by-step explanation:
Line 1(L₁) passes through the point O(0, 0) and A(a, b). Hence the equation of L₁ is:
[tex]y-y_1=\frac{y_2-y_}{x_2-x_1}(x-x_1) \\\\y-0=\frac{b-0}{a-0} (x-0)\\\\ay -bx=0[/tex]
Point C has the same x coordinate as point A. Since point C is on the origin line (y = 0 line), its y coordinate is 0.
The coordinate of C = (a, 0)
L₁ has a slope of b/a. Since L₂ is perpendicular to L₁, the product of their slope is -1. Hence slope of L₂ = -a/b
L₂ has a slope of -a/b and pass through the origin. The equation of L₂ is:
[tex]y-y_1=m(x-x_1)\\\\y-0=\frac{-a}{b}(x-0)\\\\y= \frac{-a}{b}x[/tex]
But OD = AC
AC = b
Hence OD = AC = b
The y coordinate of D is 0. Let its x coordinate be -x. Hence D = (-x, 0)
[tex]OD=\sqrt{(-x-0)^2+(0-0)^2} \\\\OD=x\\\\OD=x=b\\\\x=b;-x=-b[/tex]
The coordinate of D is (-b, 0)
B has the same x coordinate as D. The y coordinate of B is gotten by using L₂ equation and substituting x = -b
y = (-a/b)(-b)
y = a
Coordinate of B = (-b, a)
