Coordinate Geometry — SSC CGL Quantitative Aptitude

Coordinate geometry is the most formula-direct topic in the paper: no construction, no cases — just substitute the given points. The entire syllabus is six formulas (distance, midpoint, section, centroid, area, slope) plus the parallel/perpendicular slope rules. Learn which phrase triggers which formula and these become ten-second marks.

1. What SSC actually asks

Tier 1: 1–2 Q · Tier 2: 1–2 Q. Types: distance between two points, midpoint/section point, centroid of a triangle, area of a triangle (and the collinearity test area = 0), slope, and equation-of-line questions with parallel/perpendicular conditions or distance from a line.

2. The point formulas

For points $A(x_1,y_1)$ and $B(x_2,y_2)$:

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

$\text{Midpoint}=\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)$

$\text{Section (ratio }m:n\text{ internally)}=\left(\frac{m{x}_2+n{x}_1}{m+n},\frac{m{y}_2+n{y}_1}{m+n}\right)$

• Distance from the origin to $(x,y)$ is $\sqrt{x^2+y^2}$.
• Midpoint is the section formula with $m:n=1:1$.
• Careful with the cross-pairing in the section formula: $m$ multiplies the second point's coordinate.

3. Triangle formulas

$\text{Centroid}=\left(\frac{x_1+x_2+x_3}{3},\frac{y_1+y_2+y_3}{3}\right)$

$\text{Area}=\frac{1}{2}|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)|$

• The centroid is just the average of the three vertices.
• Area = 0 ⇔ the three points are collinear — the standard collinearity test.
• A triangle with a vertex at the origin and legs on the axes has area $\frac{1}{2}\times \text{base}\times \text{height}$ directly.

4. Slope and lines

$\text{Slope }m=\frac{y_2-y_1}{x_2-x_1},\text{line: }y=mx+c\text{ or }ax+by+c=0$

• Parallel lines: equal slopes. Perpendicular lines: $m_1{m}_2=-1$ (so slope $2\perp$ slope $-\frac{1}{2}$).
• For $ax+by+c=0$, slope $=-\frac{a}{b}$.
• Distance of point $(x_0,y_0)$ from $ax+by+c=0$: $\frac{|a{x}_0+b{y}_0+c|}{\sqrt{a^2+b^2}}$.

5. Quadrants and axes (quick facts)

• Quadrant signs: I $(+,+)$, II $(-,+)$, III $(-,-)$, IV $(+,-)$.
• On the x-axis $y=0$; on the y-axis $x=0$.
• Collinearity, quadrant identification and "which axis" questions are free marks — read the signs.

6. Solved PYQ-style examples

Q1. Distance between $(2,3)$ and $(5,7)$? Solution. $\sqrt{3^2+4^2}=\sqrt{25}=$ 5.

Q2. The point dividing the join of $(1,2)$ and $(4,5)$ internally in the ratio $2:1$? Solution. $\left(\frac{2\cdot 4+1\cdot 1}{3},\frac{2\cdot 5+1\cdot 2}{3}\right)=\left(\frac{9}{3},\frac{12}{3}\right)=$ $(3,4)$.

Q3. Area of the triangle with vertices $(0,0),(4,0),(0,3)$? Solution. Right triangle on the axes: $\frac{1}{2}\times 4\times 3=$ 6.

Q4. For what value of $k$ are $(2,3),(4,k),(6,-3)$ collinear? Solution. Slopes equal: $\frac{k-3}{2}=\frac{-3-3}{4}=-\frac{3}{2}\Rightarrow k-3=-3\Rightarrow k=$ 0.

Q5. Distance of the point $(3,4)$ from the line $3x+4y-5=0$? Solution. $\frac{|3(3)+4(4)-5|}{\sqrt{9+16}}=\frac{20}{5}=$ 4.

7. Exam protocol

1. Match the phrase to the formula: "distance" → distance, "divides in ratio" → section, "middle" → midpoint.
2. In the section formula, $m$ pairs with the far point's coordinate — write it carefully.
3. Centroid = average of vertices; area uses the shoelace formula with the absolute value.
4. Collinear? Set the triangle area to 0 (or equate two slopes).
5. Perpendicular means slope product $-1$; parallel means equal slopes. Read $ax+by+c=0$ as slope $-a/b$.