I only need number 4 and number 8

  4.  (2√10 +2√17) units

  8.  25/2 square units

Step-by-step explanation:

4.  The length of JK is given by the distance formula ...

  d = √((x2 -x1)^2 +(y2 -y1)^2)

  d = √((3-2)^2 +(6-3)^2) = √(1+9) = √10

The length of KL is given by the same formula:

  d = √(4^2 +1^2) = √17

The perimeter is twice the sum of these lengths:

  P = 2(JK +KL)

  P = 2√10 +2√17

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8. The area can be computed from coordinates* using the formula ...

  A = (1/2)|x1(y2 -y3) +x2(y3 -y1) +x3(y1 -y2)|

  A = (1/2)|3(4 -0) +7(0 -7) +4(7 -4)| = (1/2)|12 -49 +12|

  A = 25/2 = 12.5 . . . . square units

* One way to look at the problem is as a set of triangles and a trapezoid. The x-axis is the base of the right triangles with line PQ and PR as hypotenuses. The trapezoid is a right trapezoid with parallel bases extending from the x-axis to points Q and R.

The left triangle area is 1/2(1)(7) = 7/2; the right triangle area is 1/2(3)(4) = 6. The area of the trapezoid is 1/2(4+7)(4) = 22. The area of the triangle of interest is the difference ...

  area PQR = trapezoid area - (sum of triangle areas) = 22 -19/2 = 25/2

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If point P were not on the x-axis, then the figure could be resolved into 3 right trapezoids in similar fashion. If you were to use points (x1, y1), (x2, y2), and (x3, y3) instead of numerical values, you could obtain the formula shown above.

step-by-step explanation: