Please help me this.Thank you
1 answer:
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Given equation of the line ,
Simplify ,
Transpose terms to RHS ,
Divide both sides by 5 ,
Now we know the slope intercept form of the line as we have ,
As we know that the slope of all parallel lines are same . Henceforth ,
Answer:
The maximum area of the triangular region is 2 square units
Step-by-step explanation:
we know that
The area of the triangular region is equal to
-----> equation A
where
x is the x-coordinate of point P
y is the y-coordinate of point P
we have
The equation of the graph of the line is
-----> equation B
substitute equation B in equation A
This is a vertical parabola open downward (the leading coefficient is negative)
The vertex represent a maximum
The y-coordinate of the vertex represent the maximum area of the triangular region
Find the vertex
Convert the quadratic equation in vertex form
Factor -1/2
Complete the square
Rewrite as perfect squares
The vertex is the point (2,2)
so
The maximum area is the y-coordinate of the vertex
The maximum area of the triangular region is 2 square units
Find the coordinates of point P for the maximum area
x=2 (x-coordinate of the vertex)
Find the y-coordinate of point P
substitute in equation B
y=-2+4=2
The coordinates of point P for the maximum area is P(2,2)