The correct option is A). 14.2
The correct option is B). (-7,-3)
Step-by-step explanation:
QA). The perimeter of triangle ABC with vertices A(3, 2), B(-2, 3), and C(2, 6)
Ans.
The distance between two points A(X1,Y1) and B(Y1,Y2) is
=[tex]\sqrt{(X1-X2)^{2}+(Y1-Y2)^{2}[/tex]
Now,
The distance between two points A(3, 2) and B(-2, 3) is
=[tex]\sqrt{(X1-X2)^{2}+(Y1-Y2)^{2}[/tex]
=[tex]\sqrt{(3-(-2))^{2}+(2-3)^{2}[/tex]
=[tex]\sqrt{25 +1}[/tex]
=[tex]\sqrt{26}[/tex]
The distance between two points B(-2, 3) and C(2, 6) is
=[tex]\sqrt{(X1-X2)^{2}+(Y1-Y2)^{2}[/tex]
=[tex]\sqrt{((-2)-(2))^{2}+(3-6)^{2}[/tex]
=[tex]\sqrt{16 +9}[/tex]
=[tex]\sqrt{25}[/tex]
=5
The distance between two points A(3, 2) and C(2, 6) is
=[tex]\sqrt{(X1-X2)^{2}+(Y1-Y2)^{2}[/tex]
=[tex]\sqrt{((3)-(2))^{2}+(2-6)^{2}[/tex]
=[tex]\sqrt{1 +16}[/tex]
=[tex]\sqrt{17}[/tex]
.
The perimeter of Triangle ABC is AB+BC+AC
=[tex]\sqrt{26}[/tex]+5+[tex]\sqrt{17}[/tex]
=5.099+5+4.1231
=14.2221
The correct option is A). 14.2
QB). Find the midpoint between (-4, -2) and (-10, -4)
Ans.
The midpoint of two points A(X1,Y1) and B(Y1,Y2) is
=[tex](\frac{(X1+X2)}{2},\frac{(Y1+Y2)}{2})[/tex]
The midpoint of two points (-4, -2) and (-10, -4) IS
=[tex](\frac{(X1+X2)}{2},\frac{(Y1+Y2)}{2})[/tex]
=[tex](\frac{((-4)+(-10))}{2},\frac{((-2)+(-4))}{2})[/tex]
=[tex](\frac{(-14)}{2},\frac{(-6)}{2})[/tex]
=(-7,-3)
The correct option is B). (-7,-3)
