Coordinate Geometry of Lines and Triangles

01 / UNDERSTAND & APPLY

1. Choose coordinates that preserve the geometry

A point P=(x,y) records signed displacements along two perpendicular axes. In ordinary Euclidean coordinate geometry, the axes use the same length unit. A plotted graph with unequal display scales can still encode correct coordinates, but its visible angles and lengths cannot be measured directly from the picture.

For A=(x₁,y₁) and B=(x₂,y₂), the displacement B−A is (x₂−x₁,y₂−y₁). Its length is √[(x₂−x₁)²+(y₂−y₁)²], from the Pythagorean theorem. The midpoint averages coordinates because it lies halfway along the displacement in each component.

A translation of axes can simplify a problem by placing a convenient vertex at the origin. A rotation can align a side with an axis. Such rigid changes preserve distances, angles and areas, but the transformed coordinates must be used consistently throughout the calculation.

Before solving, check whether the geometric object is nondegenerate. Two identical points do not determine a unique line, and three collinear points do not form a nondegenerate triangle. A formula whose denominator vanishes often signals exactly such a missing condition.

02 / UNDERSTAND & APPLY

2. A normal vector gives a line without a slope exception

The equation ax+by+c=0 represents a line when (a,b)≠(0,0). The vector (a,b) is normal to the line, while (−b,a) is a direction vector. Multiplying all coefficients by the same nonzero constant gives the same line, so coefficients are not unique.

If b≠0, the slope is −a/b and the line can be written y=mx+k. If b=0, the line is vertical and its slope is undefined. A horizontal line has slope zero. Treating an undefined slope as an ordinary large number leads to invalid algebra.

Through a point (x₀,y₀), a line with normal (a,b) has equation a(x−x₀)+b(y−y₀)=0. A line with direction (u,v) can instead be parametrized as (x,y)=(x₀,y₀)+t(u,v), with (u,v)≠(0,0). These forms work for every orientation.

Through distinct points A,B, use the determinant or cross-multiplied form (y−y₁)(x₂−x₁)=(x−x₁)(y₂−y₁). Unlike the two-point slope fraction, it remains valid when x₂=x₁. Intercept form x/p+y/q=1 is useful only when the two nonzero finite axis intercepts p,q exist.

03 / UNDERSTAND & APPLY

3. Direction and normal tests avoid dividing by slopes

Two line equations a₁x+b₁y+c₁=0 and a₂x+b₂y+c₂=0 have a unique intersection when a₁b₂−a₂b₁≠0. If that determinant is zero, their normals are parallel; the lines are either distinct parallel lines or coincident. Compare constants with the same scaling to distinguish those cases.

The lines are perpendicular exactly when a₁a₂+b₁b₂=0, since rotating both normal vectors by a right angle preserves their dot product. This test covers vertical and horizontal lines. The slope product m₁m₂=−1 is a convenient equivalent only when both finite slopes exist.

The acute or right angle θ between two unoriented lines satisfies cos θ=|a₁a₂+b₁b₂|/[√(a₁²+b₁²)√(a₂²+b₂²)]. The absolute value selects the smaller angle between the lines. For oriented directions or a specified interior triangle angle, use the actual direction vectors and the appropriate sign instead.

When neither line is vertical and 1+m₁m₂≠0, tan θ=|(m₂−m₁)/(1+m₁m₂)| for the acute angle. A zero denominator indicates perpendicular lines, not a failure of the geometry. Normal-vector formulas are usually safer when parameters might create these exceptional cases.

04 / UNDERSTAND & APPLY

4. Perpendicular distance is a normalized signed residual

For P=(x₀,y₀) and line ax+by+c=0, the distance is |ax₀+by₀+c|/√(a²+b²). The numerator is the equation residual at P; dividing by the normal length removes the arbitrary scaling of the line equation. Omitting that normalization makes the answer depend on how the same line was written.

Let s=(ax₀+by₀+c)/(a²+b²). The perpendicular foot is H=P−s(a,b). This moves P along the normal until the line residual becomes zero. The reflection P′=P−2s(a,b) moves twice as far, placing H at the midpoint of P and P′.

The signed quantity (ax₀+by₀+c)/√(a²+b²) distinguishes the two sides relative to a chosen normal orientation. Distance uses its absolute value. Multiplying a line equation by a negative constant reverses that signed orientation but does not change the geometric distance.

For two parallel lines written with identical a,b coefficients, their separation is |c₁−c₂|/√(a²+b²). First rescale to a common normal. Subtracting constants from differently scaled equations is not meaningful.

05 / UNDERSTAND & APPLY

5. Families encode a shared geometric condition

If two nonparallel lines L₁=0 and L₂=0 meet at O, every equation λL₁+μL₂=0 with (λ,μ)≠(0,0) describes a line through O. Conversely every line through O can be represented this way. The one-parameter form L₁+kL₂=0 omits L₂=0, so remember that member when completeness matters.

The loci of points equidistant from two intersecting lines are their two angle bisectors. Their equations are L₁/√(a₁²+b₁²)=±L₂/√(a₂²+b₂²). Normalization is required before comparing distances. Both signs are needed to obtain both bisectors.

To select an internal triangle angle bisector rather than the external one, identify which branch passes through the triangle interior or use a suitable test point. The signs of raw line equations depend on how the equations were written, so a memorized plus-versus-minus rule without orientation information is unsafe.

A locus problem asks for all points satisfying a geometric condition. Derive a necessary equation, then check its converse: every point on the resulting equation should satisfy the original condition, except any explicitly excluded points. Squaring a distance equality is safe when both sides are nonnegative, but later divisions can still lose cases.

06 / UNDERSTAND & APPLY

6. Affine combinations locate points; determinants measure area

For A,B and a point P dividing AB internally in the ratio AP:PB=m:n with m,n>0, P=(nA+mB)/(m+n). The weight of B is m because a larger AP moves P closer to B. The midpoint is the special case m=n.

An external division with ratio m:n, m≠n, is represented by P=(mB−nA)/(m−n). The denominator can vanish when the ratio is 1:1; there is then no finite external division point for distinct endpoints. A signed affine parameter P=A+t(B−A) handles internal points 0<t<1 and external points t<0 or t>1 uniformly.

For triangle vertices A,B,C, the signed doubled area is det(B−A,C−A). The geometric area is half its absolute value. Swapping two vertices reverses the sign but does not change the area. A zero determinant means collinearity and a degenerate triangle.

The determinant formula is translation invariant because it uses displacement vectors. It is also equivalent to the coordinate expression (1/2)|x₁(y₂−y₃)+x₂(y₃−y₁)+x₃(y₁−y₂)|. Use the form that keeps arithmetic simplest and retain the absolute value for ordinary area.

07 / UNDERSTAND & APPLY

7. The centroid is an equal-weight average

For a nondegenerate triangle with position vectors A,B,C, the centroid is G=(A+B+C)/3. Let M=(B+C)/2 be the midpoint of BC. Then G=A+(2/3)(M−A), so G lies on the median from A and divides it in ratio AG:GM=2:1.

The same calculation works for medians from B and C, proving their concurrence at G rather than merely observing it in a diagram. The centroid lies inside every nondegenerate triangle because its three vertex weights are positive and sum to 1.

A median joins a vertex to the midpoint of the opposite side. It need not be perpendicular to that side and need not bisect the vertex angle. Those coincidences occur in special symmetric triangles, not as part of the median definition.

The section formula also shows that the three medians divide the triangle into six equal-area small triangles. A median first halves the area; the 2:1 division at the centroid then supplies the remaining equal-area relations. Distinguish equal areas from equal shapes or equal side lengths.

08 / UNDERSTAND & APPLY

8. Construct centers from the conditions they satisfy

The circumcenter O is equidistant from the three vertices. Its coordinates can be found by solving |O−A|²=|O−B|² and |O−A|²=|O−C|². The quadratic terms cancel, leaving two linear equations: these are perpendicular bisectors of two sides. Noncollinearity ensures a unique solution.

The orthocenter H is the common point of the three altitudes. An altitude passes through a vertex and is perpendicular to the opposite side. To compute H, write two such lines using dot products or normal vectors and intersect them; verify the third altitude if desired.

For a right triangle, the orthocenter is the right-angle vertex and the circumcenter is the midpoint of the hypotenuse. In an obtuse triangle both can lie outside the triangle. The centroid remains inside, so a picture placing every center inside every triangle is misleading.

The Euler-line identity H=A+B+C−2O follows from a dot-product calculation: (H−A)·(C−B)=(B+C−2O)·(C−B)=|C−O|²−|B−O|²=0. Thus the proposed H lies on the altitude from A, and similarly on the others. Since 3G=A+B+C, H=3G−2O, giving OG:GH=1:2 when the centers are distinct.

09 / UNDERSTAND & APPLY

9. The incenter uses side-length weights

The incenter I is the intersection of the internal angle bisectors and is equidistant from all three sides. If a=|BC|, b=|CA| and c=|AB|, then I=(aA+bB+cC)/(a+b+c). Each vertex is weighted by the length of its opposite side, not by an adjacent side chosen arbitrarily.

One way to justify the weights is through areas. If the common perpendicular distance from I to each side is r, the three subtriangle areas are ar/2,br/2,cr/2. Their ratios are a:b:c, which are the corresponding barycentric weights. All are positive, so I lies inside the triangle.

The total area Δ equals r(a+b+c)/2=rs, where s is the semiperimeter. Thus r=Δ/s. This gives a useful independent check after finding I: its distances to the three side lines should all equal that value.

The incenter usually does not lie on the Euler line. Do not interchange it with the circumcenter: one is equidistant from side lines, the other from vertices. Both definitions involve equal distances, but to different geometric objects.

10 / UNDERSTAND & APPLY

10. Original practice with geometric checks

Use normal vectors for exceptional slopes and verify a computed center against its defining distances or perpendicularities.

11 / UNDERSTAND & APPLY

11. Let the geometry choose the algebra

Use a point and a direction for motion along a line, a normal for perpendicularity and distance, and determinants for orientation and area. These representations describe the same objects but make different constraints easier to enforce.

For a triangle center, start from its definition rather than a remembered picture. Medians identify G, perpendicular bisectors identify O, altitudes identify H and internal angle bisectors identify I. Their locations can differ dramatically, especially for right or obtuse triangles.

The frequent errors are dividing by a zero slope denominator, using unnormalized line residuals as distances, forgetting the absolute value in area, reversing section weights, treating a median as an altitude, and failing to check collinearity before solving for a circumcenter.

Conclude with direct checks: a point on a line must satisfy its equation, a foot must lie on the line and have a normal displacement, a reflection must have that foot as midpoint, and a center must meet its defining equal-distance or perpendicular conditions. Those checks turn coordinate arithmetic back into geometry.