Question

PLS ANSWER RN

What is the area of AJKL?

Answer 1

8.5

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 =
`√((7-3)^2 + (0-1)^2) = √(16 + 1) = √(17)`

KJ =
`√((4-3)^2 + (5-1)^2)= √(1 + 16) = √(17)`

area = (√17)^2/2 = 17/2 = 8.5

Answer 2

8.5



….hope this helps!