Find the perimeter of triangleXYZ with vertices X(3,7), Y(-1,-5) & Z (6,-4)

Question

Need answer

Answer 1

Triangl XYZ with  X = (3,7)    Y = (-1,-5)    Z = (6,-4)

 

The three sides are created by the line segment XY, the line segment  XZ, and the line segment YZ.

We need the lengths of these three line segments which will be the distance from x to y, x to z, and y to z.

We need to find the 3 distances and then add them together.

 

The distance between two points is found from the distance formula

d = √((x₂-x₁)²+(y₂-y₁)²)

 

XY = √((-1-3)²+(-5-7)²) = √((-4)²+(-12)²) = √(16 + 144) = √(160) = 4√10

 

xz = √((6-3)²+(-4-7)²) = √(3² + (-11)²) = √(9+121) = √130

 

yz = √(((6-(-1))²+(-4-(-5))²) = √(7² +1²) = √50 = 5√2

 

We now have the lengths of each side of the triangle so add them to find perimeter

 

P = 4√10 + √130 + 5√2 ≅ 31.122

 

Answer 2

Given a point in the plane, (p,q) we can find the distance between it and the origin using the Pythagorean theorem:

 

Dist(p,q) = \sqrt(p^2+q^2)

 

If we shift our coordinates by moving h units right (left if h is negative) and k units up (down if k negative), then the distance between the translated points is the same...This shows us that the "distance between(h,k) and (h+p,k+q)" = \sqrt(p^2+q^2)

 

Introducing new variables a=h+p and b=k+q, we see that p=a-h and q=b-k,  This allows to say that the distance between points (h,k) and (a,b) in the plane is given by \sqrt{(a-h)^2+(b-k)^2}.

 

So we can find the 3 side lengths of triangle XYZ, using this formula.  We look at the difference between the coordinates and square them separately, then add up these square values and take a square root.

 

For example:

Side XY of the triangle has length = \sqrt{ (3- (-1))^2 + (7- (-5))^2} = \sqrt{4^2+12^2}  = \sqrt{160}

 

Do the same for YZ and ZX using the other pairs of points given, and add up the three sides to get the perimeter!