Find area of the triangle ABC, given the coordinates of vertices in plane $A, B$ and $C$ are the points $(7,3), (-4,1)$ and $(-3,-2)$ respectively. Find the area of the triangle $ABC$.


I've worked out the lengths of each side of the triangle which are $AB=5\sqrt5$, $BC=\sqrt10$ and $AC=5\sqrt5$.

I know that the formula for the area of a triangle is $\frac12hb$ but when I checked the solutions the answer to the area of this triangle is $17\frac12$.

I do not understand how this answer is achieved.
Since you have obtained the length of each side, using Heron's Formula is a natural way to find the area.  Let's consider the approach suomynonA suggested in the comments. Consider the figure below.



We can find the area of $\triangle ABC$ by subtracting the sum of the areas of the three right triangles $ABD$, $ACF$, and $BCE$ from the area of rectangle $ADEF$.  I will leave the details of the calculations to you.
I need anwer .A(6,2),B(2,4) and C(8,-4) are three in a plane in the line BC and Q is a point in the plane, such that PAQB is a rectangle. Find the (I) equation of the sides AP and BP. (ii) the coordinate of the points P (iii) the area of the rectangle