Respuesta :

caylus
Hello,
Let's be F the intersection of BO and its perpendicular passing through A

BF=k
Thalès tr BAF and BDO ==>AB/4AB=k/(k+6)==>k+6=4k==>k=2
AF/DO=BF/BO==>DO=AF*BO/BF=2*8/2=8

Coordinate of D=(-8,0)
BD=√8²+8²=8√2

Line BC
Coordinate of C:
DA/AB=DO/OC
==>OC=DO*AB/DA=8/4=2

C=(2,0)
B=(0,8)
slope=(8-0)/(0-2)=-4
y-0=-4(x-2)==>y=-4x+8




dalendrk
Look at the picture.

[tex]a)\\\Delta DOB\ and\ \Delta AEB\ are\ similar\ therefore\\\\\dfrac{x}{2}=\dfrac{4a}{a}\to\dfrac{x}{2}=4\ \ \ |multiply\ both\ sides\ by\ 2\\\\\boxed{x=8}\\\\therefore\ \boxed{D(-8;\ 0)}\\\\\boxed{Answer:(-8;\ 0)}[/tex]


[tex]b)\\(i)\\\dfrac{y}{2}=\dfrac{y+6}{8}\ \ \ \ |cross\ multiply\\\\8y=2(y+6)\\8y=2y+12\ \ \ |subtract\ 2y\ from\ both\ sides\\6y=12\ \ \ \ |divide\ both\ sides\ by\ 6\\y=2\\\\therefore\ B(0;\ 6+2)\to B(0;\ 8)\\---------------\\The\ distance\ between\ A(x_A;\ y_A)\ and\ B(x_B;\ y_B)\\\\d=\sqrt{(x_B-x_A)^2+(y_B-y_A)^2}\\\\subtitute\ B(0;\ 8);\ D(-8;\ 0)\\\\|BD|=\sqrt{(0-8)^2+(-8-0)^2}=\sqrt{(-8)^2+(-8)^2}\\\\=\sqrt{64+64}=\sqrt{64\cdot2}=\sqrt{64}\cdot\sqrt2=8\sqrt2\\\\\boxed{Answer:|BD|=8\sqrt2\ units.}[/tex]

[tex](ii)\\[/tex]
OA is parallel to CB therefore line OA and line CB have the same slope.

[tex]A(x_1;\ y_1);\ B(x_2;\ y_2)\\\\the\ slope-intercept\ form:y=mx+b\\\\the\ slope:m=\dfrac{y_2-y_1}{x_2-x_1}\\b\ is\ y-intercept\\---------------------\\The\ slope\ of\ the\ line\ OA\\\\O(0;\ 0);\ A(-2;\ 6)\\\\m=\dfrac{6-0}{-2-0}=\dfrac{6}{-2}=-3\\------------------\\The\ line\ BC:y=-3x+b\\\\B(0;\ 8)-y-intercept\\\\therefore\\\\\boxed{Answer:y=-3x+8}[/tex]

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