Distance, Section Formula and Area of a Triangle | NIMCET

Quick answer: For points $P(x_1, y_1)$ and $Q(x_2, y_2)$, the distance is $PQ = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$. The point dividing PQ internally in the ratio $m : n$ is $\left(\frac{mx_2 + nx_1}{m + n}, \frac{my_2 + ny_1}{m + n}\right)$. The area of a triangle with vertices $(x_1, y_1)$, $(x_2, y_2)$, $(x_3, y_3)$ is $\frac{1}{2}\lvert x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)\rvert$, and the points are collinear when this area is 0.

The coordinate plane

A point is written $(x, y)$: x is the abscissa and y is the ordinate.

Distance formula

$PQ = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

The distance of $(x, y)$ from the origin is $\sqrt{x^2 + y^2}$.

Using distances to classify a triangle or quadrilateral:

Section formula

Ratio in which an axis divides PQ:

• The x-axis divides PQ in the ratio $-y_1 : y_2$.
• The y-axis divides PQ in the ratio $-x_1 : x_2$.

A positive ratio means internal division; a negative ratio means external.

Area of a triangle

$\Delta = \frac{1}{2}\left\lvert x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)\right\rvert$

For a quadrilateral, split it into two triangles along a diagonal, or use the "shoelace" pattern with the vertices taken in order.

Collinearity: three points are collinear if and only if the area expression equals 0.

Worked examples

Example 1: distance

Find the distance between $(2, -3)$ and $(-4, 5)$.

$\sqrt{(-6)^2 + 8^2} = \sqrt{100} = 10$

Example 2: internal division

Find the point dividing the join of $A(1, 2)$ and $B(7, 8)$ internally in the ratio $1 : 2$.

$\left(\frac{1 \cdot 7 + 2 \cdot 1}{3}, \frac{1 \cdot 8 + 2 \cdot 2}{3}\right) = (3, 4)$

Example 3: ratio from the axis

In what ratio does the x-axis divide the join of $(2, -3)$ and $(5, 6)$?

The ratio is $-y_1 : y_2 = 3 : 6 = 1 : 2$, internally. The point is $\left(\frac{5 + 4}{3}, 0\right) = (3, 0)$.

Example 4: area and collinearity

Find k if $(1, 2)$, $(3, k)$ and $(5, 6)$ are collinear.

Step 1: Set the area expression to 0: $1(k - 6) + 3(6 - 2) + 5(2 - k) = 0$.

Step 2: $k - 6 + 12 + 10 - 5k = 0$, so $-4k + 16 = 0$ and $k = 4$.

Check: $(3, 4)$ is the midpoint of $(1, 2)$ and $(5, 6)$ ✓.

Summary table

Exam tips

• In the section formula, m multiplies the second point's coordinates.
• For "ratio in which a line divides a segment", assume $k : 1$, substitute, and solve for k.
• Area questions often hide a collinearity condition; set the area to 0.
• Always take the absolute value for area.

Next, study centroid, incentre, orthocentre and circumcentre, or see the NIMCET Coordinate Geometry and Vectors chapter.

Common mistakes

• Swapping m and n in the section formula.
• Forgetting the absolute value in the area formula and reporting a negative area.
• Using the internal formula when the point lies outside the segment.
• Concluding a quadrilateral is a square from equal sides alone; check the diagonals too.

Frequently Asked Questions

What is the distance formula?

The distance between (x₁, y₁) and (x₂, y₂) is √((x₂ − x₁)² + (y₂ − y₁)²).

What is the section formula?

The point dividing the join of (x₁, y₁) and (x₂, y₂) internally in ratio m : n is ((mx₂ + nx₁)/(m + n), (my₂ + ny₁)/(m + n)).

How do you check whether three points are collinear?

Find the area of the triangle they form. If the area is zero, the points are collinear.

In what ratio does the x-axis divide a line segment?

It divides the join of (x₁, y₁) and (x₂, y₂) in the ratio −y₁ : y₂.

How do you find the area of a triangle from its vertices?

Use ½ |x₁(y₂ − y₃) + x₂(y₃ − y₁) + x₃(y₁ − y₂)|.