Coordinate Geometry - Triangle

Area of a Triangle

We can find the area of a triangle if we know the coordinates of its vertices. Let’s see how.

If the vertices of a ∆ABC are A ($x_1$, $y_1$), B ($x_2$, $y_2$) and C ($x_3$, $y_3$), then:

Area of the triangle, ∆ = $\frac{1}{2}\left|\begin{array}{ccc}{x}_1& y_1& 1\\ x_2& y_2& 1\\ x_3& y_3& 1\end{array}\right|$ = $\frac{1}{2}|x_1(y_2–y_3)+x_2(y_3–y_1)+x_3(y_1–y_2)|$ = $\frac{1}{2}|(x_1{y}_2–x_2{y}_1)+(x_2{y}_3–x_3{y}_2)+(x_3{y}_1–x_1{y}_3)|$

Coordinates of Important Points in a Triangle

Coordinates of Centroid

If the vertices of a ∆ABC are A ($x_1$, $y_1$), B ($x_2$, $y_2$) and C ($x_3$, $y_3$), then:
Coordinates of its Centroid = $(\frac{x_1+x_2+x_3}{3},\frac{y_1+y_2+y_3}{3})$

Coordinates of In-Centre

If the vertices of a ∆ABC are A ($x_1$, $y_1$), B ($x_2$, $y_2$) and C ($x_3$, $y_3$), and the length of their opposite sides are a, b and c, then:
Coordinates of its In-Centre = $(\frac{a{x}_1+b{x}_2+c{x}_3}{a+b+c},\frac{a{y}_1+b{y}_2+c{y}_3}{a+b+c})$

In-Centre is the point of intersection of the internal bisectors of the angles of a triangle.

Coordinates of Circumcenter

If the vertices of a ∆ABC are A ($x_1$, $y_1$), B ($x_2$, $y_2$) and C ($x_3$, $y_3$), then:
Coordinates of its Circumcenter = $(\frac{x_{1}sin2A+x_{2}sin2B+x_{3}sin2C}{sin2A+sin2B+sin2C},\frac{y_{1}sin2A+y_{2}sin2B+y_{3}sin2C}{sin2A+sin2B+sin2C})$

Circumcenter is the point of intersection of the perpendicular bisectors of the sides of a triangle.

Coordinates of Orthocenter

If the vertices of a ∆ABC are A ($x_1$, $y_1$), B ($x_2$, $y_2$) and C ($x_3$, $y_3$), then:
Coordinates of its Orthocenter = $(\frac{x_{1}tanA+x_{2}tanB+x_{3}tanC}{tanA+tanB+tanC},\frac{y_{1}tanA+y_{2}tanB+y_{3}tanC}{tanA+tanB+tanC})$

Orthocenter is the point of intersection of the altitudes of a triangle.