Question

The quadratic function f and △MPQ are graphed in the standard (x,y) coordinate plane below. Points M(2a,5b) , N(4a,9b) , and P(6a,5b) are on f . Point Q(4a,0) is NOT on f . In terms of a and b , what is the area, in square coordinate units, of △MPQ ?

Answer 1

Answer:

  10|ab|

Step-by-step explanation:

You want the area of ∆MPQ, given coordinates M(2a, 5b), P(6a, 5b), Q(4a, 0) and N(4a, 9b) where M, N, P are on the graph of quadratic function f and Q is not.

Area

The area can be found as ...

  A = 1/2(MP×MQ)

where ...

  MP = P -M = (6a, 5b) -(2a, 5b) = (4a, 0)

  MQ = Q -M = (4a, 0) -(2a, 5b) = (2a, -5b)

The cross product is the determinant of the matrix with rows MP and MQ:

  $A=\dfrac{1}{2}\left|\left|\begin{array}{cc}4a&0\\2a&-5b\end{array}\right|\right|=\left|\dfrac{(4a)(-5b)-(2a)(0)}{2}\right|=\boxed{10|ab|}$

The area of triangle MPQ is 10|ab|.

Restriction

We don't have the graph to help out here, but we can solve for the quadratic f using points M, N, P. We get ...

  f(x) = (-b/a²)x² +(8b/a)x -7b

Then the restriction that Q(4a, 0) is not a point on the curve resolves to the requirement that b≠0. We already know that a≠0, or the quadratic would be undefined.