2. Find the area of the region bounded by the graphs of y = -x2 + 3x + 4 and y = 4. (5 points)
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The value of the variable x = 6 and y = 5. Then the position vector of P will be 6i + 5j.
<h3>What is the equation of a line passing through two points?</h3>The equation of line is given as
y = mx + c
Where m is the slope and c is the y-intercept.
PQRS is a parallelogram. the position vector of Q R and S are 5i +7j, -3i-8j and -4i-6j.
Then the point of Q, R, and S will be (5, 7), (-3, -8), and (-4, -6).
Let the point P be (x, y).
We know that the slope of the line QS and PQ will be same and line is passing through (-3, -8).
y = 1.444x + C
-8 = 1.444 (-3) + C
C = -3.667
Then the equation will be
y = 1.444x – 3.667 …1
The slope of the line PQ and RS will be same and line is passing through (5, 7).
y = -2x + D
7 = -2 (5) + D
D = 17
Then the equation will be
y = -2x + 17 …2
By solving equation 1 and 2, we have
x = 6 and y = 5
Then the position vector of P will be
⇒ 6i + 5j
The graph is given below.
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