In plane 2

Triangle ABC lies in the plane with a right angle at vertex C, where A(1, 2), B(5, 2), C(x, x+1), and x > −1.
a) Determine the value of x.
b) Determine the coordinates of point M, the midpoint of segment AB.
c) Prove that vectors AB and CM are perpendicular.
d) Determine the size of 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.

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

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

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

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

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