PLS ANSWER RN
What is the area of AJKL?

Answer:
8.5
Step-by-step explanation:
slope of the line that contains KL
(y2 - y1)/(x2-x1)
(0-1)/(7-3)
m = -1/4
slope of the line that contains JK
(5-1)/(4-3)
m = 4
the linea are perpendicular so triangle is right
area = (leg1 x leg2)/2
KL = [tex]\sqrt{(7-3)^2 + (0-1)^2} = \sqrt{16 + 1} = \sqrt{17}[/tex]
KJ = [tex]\sqrt{(4-3)^2 + (5-1)^2}= \sqrt{1 + 16} = \sqrt{17}[/tex]
area = (√17)^2/2 = 17/2 = 8.5