Plugging All the Intercepts for Each Line:
y = 3/4x + 6
(0 , 6) and (-8 , 0)
y = 3/4x - 6
(0 , -6) and (8 , 0)
y = (-)3/4x + 6
(0 , 6) and (8 , 0)
y = (-)3/4x - 6
(0 , -6) and (-8 , 0)
4 Vertices bound the Quadrilateral:
(-8 , 0)
(0 , 6)
(8 , 0)
(0 , -6)
Method 1: You can Add up the AREAS of the 4 Right Triangles that are formed by Origin (0 ,0) as the Right Angle
4 * [ (1/2) * (8) * (6) ] = 96
Method 2:
Rule: In a Square, the 2 Diagonal are CONGRUENT and are Perpendicular, Bisectors of each Other
While the 2 Diagonals are Perpendicular to Each Other and DO Bisect Each Other:
Diagonal 1 runs from (-8 , 0) -------> to (8 , 0) = 16 Units ----- 8 on Each Side of the Origin running on the X-Axis
Diagonal 2 runs from (0 , -6) -------> to (0 , 6) = 12 Units ---- 6 on Each Side of the Origin running on the Y-Axis
Rule: in a Rhombus, the 2 Diagonals are NON-Congruent ---- but the 2 Diagonals are Perpendicular, Bisectors of Each Other
Based on these Fact, the Quadrilateral is a Rhombus
Area of a Rhombus = (Diagonal 1) * (Diagonal 2) * (1/2)
Area of this Rhombus = (16) * (12) * (1/2) = 16 * 6 =
96
-C-