Area of Triangle, Midpoint, and Reflection of Points

Area of a triangle from coordinates. For a triangle with vertices A($x_1$, $y_1$), B($x_2$, $y_2$) and C($x_3$, $y_3$), the area is

$\text{Area}=\dfrac12\left|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)\right|$.

The vertical bars mean absolute value: an area can never be negative, so if the bracket comes out negative we drop the sign. The expression has a neat cyclic pattern — each $x$ is multiplied by the difference of the other two $y$-values, and the subscripts cycle $1\to2\to3\to1$. Writing the numbers in that cycle helps avoid mistakes.

Condition for collinearity. Three points lie on one straight line precisely when the "triangle" they form has zero area. So setting

$x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)=0$

is the standard test for collinearity, and the same equation lets you find an unknown coordinate or an unknown $k$ that makes three points collinear.

Area of a quadrilateral. A quadrilateral ABCD has no single area formula at this level. Split it along a diagonal — say AC — into two triangles ABC and ACD, find each area with the triangle formula, and add. Take the vertices in order (going round the boundary) so the two triangles do not overlap.

Midpoint as an average. The midpoint of A($x_1$, $y_1$) and B($x_2$, $y_2$) is $\left(\dfrac{x_1+x_2}{2},\dfrac{y_1+y_2}{2}\right)$ — literally the average of the endpoints. It recovers a missing endpoint when the midpoint is known, locates the centre of a circle from a diameter, and tests for a parallelogram (diagonals share a midpoint).

Reflection of a point. Reflecting a point flips it across a line, like a mirror. The four reflections you must know:

Reflecting in the x-axis keeps the horizontal position and flips the vertical one (sign of $y$ changes); reflecting in the y-axis does the opposite. Reflecting in the origin flips both, and reflecting in $y=x$ simply swaps the two coordinates.

Why the area formula works. One clean derivation drops perpendiculars from A, B and C to the x-axis and adds or subtracts the areas of the three trapezia so formed. Each trapezium has area $\dfrac12(\text{sum of parallel sides})\times(\text{horizontal width})$, and when you combine them the $x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)$ pattern appears, with the $\dfrac12$ in front. The sign of the bracket tells you the orientation (clockwise versus anticlockwise) in which the vertices were listed; since orientation does not change the size of the region, we take the absolute value. This is also why a zero result means the three points never "enclose" anything — they are collinear.

Using area to find an unknown. Two question types reuse the area formula. (1) "Find $k$ so that three points are collinear" — set the area to 0 and solve the linear equation in $k$. (2) "Find a point/coordinate given a fixed area" — keep the $\dfrac12|\dots|$, set it equal to the given area, and solve; the absolute value usually yields two answers. Always substitute the vertices in a consistent cyclic order to keep the bookkeeping clean.

Common mistakes to avoid.

• Forgetting the factor $\dfrac12$, or forgetting the absolute-value bars (giving a negative area).
• Pairing each $x$ with its own $y$-difference instead of the difference of the other two $y$'s.
• For a quadrilateral, taking the vertices out of order — the two triangles then overlap and the area is wrong.
• Mixing up the reflection rules: x-axis changes the sign of $y$ (not $x$).