Question

The point B is the image of A in the
line x + y + 4 = 0 and C is the image
of B in the line
2x – y + 2 = 0. If A = (1, 2) then
circum diameter of triangle ABC is​

Answer 1

Answer:

Solve the both equations

Answer 2

Given: B is the image of A on line x + y + 4 = 0.

C is the image of b on line 2x – y + 2 = 0.

A = (1, 2)

Find: Calculate the circum diameter of triangle ABC.

Solution:

B is the image of A(1,2) with respect to x+y+4=0

$\frac{h-x_{1} }{a} =\frac{k-y_{b} }{b} =\frac{-2(ax_{1}+by_{1}+c) }{a^{2}+ b^{2} }$

$\frac{h-1}{1}=\frac{k-2}{1} =\frac{-2(1+2+4)}{2}$

h-1=k-2= -7

h-1=-7

h= -7+1

h= -6

k-2= -7

k= -7+2

k= -5

B(-6,-5)

C is the image of B(-6,-5) with respect to 2x-y+2=0

$\frac{h+6}{2}=\frac{k+5}{-1}=\frac{-2(-12+5+2)}{5}$

$\frac{h+6}{2}= \frac{k+5}{-1}=2$

h+6=4

h=4-6

h=-2

k+5=-2

k= -2-5

k= -7

A(1,2) B(-6,-5) C(-2,-7)

In triangle ABC

AB=$\sqrt{49+49}$

    = $7\sqrt{2}$

BC=$\sqrt{16+4}$

    =$2\sqrt{5}$

CA=$\sqrt{9+81}$

     =$3\sqrt{10}$

Slope of AB=$\frac{y2-y1}{x2-x1}$

                   = -7/-7

                   = 1

Slope of BC= 2/-4

                   = -1/2

Slope of AC= 9/3

                    = 3

Circum diameter (R)= abc/4Δ

                                = $7\sqrt{2}$X$2\sqrt{5}$X$3\sqrt{10}$/4Δ

                                = 420/4Δ

                                = 105/Δ

#SPJ3