In plane 2

A triangle ABC is located in the plane with a right angle at vertex C, for which the following holds: A(1, 2), B(5, 2), C(x, x+1), where x > -1.
a) determine the value of x
b) determine the coordinates of point M, which is the midpoint of line segment AB
c) prove that vectors AB and CM are perpendicular
d) determine the size of the angle CAB
e) calculate the perimeter of triangle ABC

Final Answer:

x =  3
Mx =  3
My =  2
o =  9.6569
CAB =  45 °

Step-by-step explanation:

$$\begin{gathered} A=(1,2) \\ B=(5,2) \\ C=(x,x+1) \\ {c}^2=a^2+b^2 \\ |AB{|}^2=|BC{|}^2+|AC{|}^2 \\ (Bx-Ax{)}^2+(By-Ay{)}^2=(x-Bx{)}^2+(x+1-By{)}^2+(x-Ax{)}^2+(x+1-Ay{)}^2 \\ (5-1{)}^2+(2-2{)}^2=(x-5{)}^2+(x+1-2{)}^2+(x-1{)}^2+(x+1-2{)}^2 \\ -4x^2+16x-12=0 \\ 4x^2-16x+12=0 \\ 4=2^2 \\ 16=2^4 \\ 12=2^2\cdot 3 \\ \text{GCD}(4,16,12)=2^2=4 \\ {x}^2-4x+3=0 \\ a=1;b=-4;c=3 \\ D=b^2-4ac=4^2-4\cdot 1\cdot 3=4 \\ D>0 \\ {x}_{1,2}=\frac{-b\pm \sqrt{D}}{2a}=\frac{4\pm \sqrt{4}}{2} \\ {x}_{1,2}=\frac{4\pm 2}{2} \\ {x}_{1,2}=2\pm 1 \\ {x}_1=3 \\ {x}_2=1 \\ x>0;x>-1 \\ x=x_1=3 \end{gathered}$$

Our quadratic equation calculator calculates it.

$Mx=M_x=\frac{A_x+B_x}{2}=\frac{1+5}{2}=3$

$My=M_y=\frac{A_y+B_y}{2}=\frac{2+2}{2}=2$

$$\begin{gathered} C=(3,4) \\ a=dist(B,C)=|BC|=\sqrt{(B_x-C_x{)}^2+(B_y-C_y{)}^2}=\sqrt{(5-3{)}^2+(2-4{)}^2}=2\sqrt{2}\doteq 2.8284 \\ b=dist(A,C)=|AC|=\sqrt{(A_x-C_x{)}^2+(A_y-C_y{)}^2}=\sqrt{(1-3{)}^2+(2-4{)}^2}=2\sqrt{2}\doteq 2.8284 \\ c=dist(A,B)=|AB|=\sqrt{(A_x-B_x{)}^2+(A_y-B_y{)}^2}=\sqrt{(1-5{)}^2+(2-2{)}^2}=4 \\ o=a+b+c=2.8284+2.8284+4=9.6569 \end{gathered}$$

$$\begin{gathered} \sin CAB=a:c \\ \alpha =\arcsin (a/c)=\arcsin (2.8284/4)\doteq 0.7854\text{rad} \\ CAB=\angle CAB=\alpha \to \text{°}=\alpha \cdot \frac{180}{\pi }\text{°}=0.7854\cdot \frac{180}{\pi }\text{°}=45\text{°}=45^{\circ } \end{gathered}$$

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