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SSC CGL Quantitative Aptitude · Formula sheet

Coordinate Geometry formulas

8 formulas and 19 common traps for SSC CGL Quantitative Aptitude Coordinate Geometry, grouped by subtopic.

Full notes with worked examples

Distance, Section Formula and Area

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Distance between two points

Distance formula

d=(x2−x1)2+(y2−y1)2dorigin=x2+y2d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \qquad d_{\text{origin}} = \sqrt{x^2 + y^2}

Midpoint, section formula and centroid

Midpoint, section point and centroid

M=(x1+x22,y1+y22)P=(mx2+nx1m+n,my2+ny1m+n)G=(x1+x2+x33,y1+y2+y33)M = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right) \qquad P = \left(\frac{m x_2 + n x_1}{m + n}, \frac{m y_2 + n y_1}{m + n}\right) \qquad G = \left(\frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3}\right)
  • m:nm : nthe ratio in which P divides the segment, measured from the first point
  • GGthe centroid, where the three medians meet

Area of a triangle from its corners

Area of a triangle from coordinates

Area=12∣x1(y2−y3)+x2(y3−y1)+x3(y1−y2)∣\text{Area} = \frac{1}{2}\left|x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)\right|

Common traps

Losing the minus sign when a coordinate is negative

The gap between x=−2x = -2 and x=4x = 4 is 4−(−2)=64 - (-2) = 6, not 2. Write the subtraction with brackets whenever a coordinate is negative.

Adding the gaps instead of squaring them

Gaps of 3 and 4 give a distance of 5, not 7. The distance is the square root of the sum of the squares of the gaps, never the plain sum.

Dividing by 2 for a centroid

The centroid averages three corners, so the sum is divided by 3. Dividing by 2 gives a point that is not on any median.

Taking the centroid one third of the way from the corner

The centroid is two thirds of the way along a median from the corner, and one third of the way from the midpoint of the opposite side. The ratio from the corner is 2:12 : 1.

Using the midpoint of the hypotenuse for any triangle

The point equidistant from three corners is the midpoint of the longest side only in a right triangle. Check first that two sides lie along perpendicular lines, such as the two axes.

Forgetting the half

The bracket gives twice the area. An option equal to the full bracket is usually printed to catch the student who stops before halving.

Reporting a negative area

Going round the corners in the other order flips the sign of the bracket. Area is never negative: take the absolute value before halving.

Slope and Equation of a Line

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Slope of a line

Slope from two points and from an equation

m=y2−y1x2−x1y=mx+cax+by+c=0⇒m=−abm = \frac{y_2 - y_1}{x_2 - x_1} \qquad y = mx + c \qquad ax + by + c = 0 \Rightarrow m = -\frac{a}{b}
  • mmthe slope
  • ccin y = mx + c, the y-intercept

Where a line cuts the axes

Intercept form of a line

xp+yq=1\frac{x}{p} + \frac{y}{q} = 1
  • ppthe x-intercept
  • qqthe y-intercept

Equation of a line from a point and its slope

Point-slope form

y−y1=m(x−x1)y - y_1 = m(x - x_1)

Common traps

Dropping the minus sign in minus a over b

For ax+by+c=0ax + by + c = 0 the slope is −ab-\frac{a}{b}. The positive fraction ab\frac{a}{b} and the upside-down ba\frac{b}{a} are always among the options.

Subtracting in a different order on top and bottom

Rise over run needs the same point first in both differences: y2−y1x2−x1\frac{y_2 - y_1}{x_2 - x_1}. Mixing the order flips the sign of the slope.

Putting the wrong coordinate to zero

The x-intercept comes from y=0y = 0, not x=0x = 0. Putting x=0x = 0 gives the y-intercept, which the options usually print beside the right answer.

Stopping at the first option that fits

When 'all of the above' is an option, a point that fits does not end the question. Test every listed point; if they all fit, the answer is the 'all' option.

Swapping the coordinates when substituting

In (3,17)(3, 17) the 3 is x and the 17 is y. Putting 17 in for x gives a different, wrong value of the unknown.

Keeping the wrong sign inside the bracket

For the point (−4,−6)(-4, -6) the form is y+6=m(x+4)y + 6 = m(x + 4): subtracting a negative coordinate turns into adding. Writing y−6y - 6 or x−4x - 4 gives a line through a different point.

Parallel and Perpendicular Lines

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Slopes of parallel and perpendicular lines

Parallel and perpendicular conditions

m1=m2m1m2=−1a1a2=b1b2a1a2+b1b2=0m_1 = m_2 \qquad m_1 m_2 = -1 \qquad \frac{a_1}{a_2} = \frac{b_1}{b_2} \qquad a_1a_2 + b_1b_2 = 0
  • m1,m2m_1, m_2the slopes of the two lines
  • a,ba, bthe coefficients of x and y in ax + by + c = 0

Shortest distance from a point to a line

Distance from a point to a line

d=∣ax0+by0+c∣a2+b2dparallel=∣c1−c2∣a2+b2d = \frac{|ax_0 + by_0 + c|}{\sqrt{a^2 + b^2}} \qquad d_{\text{parallel}} = \frac{|c_1 - c_2|}{\sqrt{a^2 + b^2}}
  • (x0,y0)(x_0, y_0)the point

Common traps

Taking only the reciprocal, or only the negative

The perpendicular slope to 6 is −16-\frac{1}{6}. The options print 16\frac{1}{6}, −6-6 and 6 as well; check that the two slopes multiply to exactly −1-1.

Keeping both roots when the unknown is squared

A parallel condition such as 2k+1=k+18\frac{2}{k + 1} = \frac{k + 1}{8} gives k+1=±4k + 1 = \pm 4, two values of kk. Keep only the one the question allows (for example k>0k > 0), and check it does not make the two equations the same line.

Drawing the bisector through an end instead of the midpoint

The perpendicular bisector passes through the midpoint of the segment. A line through one end with the perpendicular slope is perpendicular, but it is not the bisector.

Giving the bisector the segment's own slope

The bisector crosses the segment at a right angle, so its slope is the negative reciprocal of the segment's slope. A line with the segment's own slope through the midpoint is the segment's line itself.

Leaving out the square root on the bottom

The bottom is a2+b2\sqrt{a^2 + b^2}, not a2+b2a^2 + b^2 and not a+ba + b. For 12x+5y=3912x + 5y = 39 it is 13, and the distance from the origin is 3913=3\frac{39}{13} = 3.

Using unequal coefficients for parallel lines

2x+3y=12x + 3y = 1 and 4x+6y=94x + 6y = 9 are parallel, but subtract the constants only after rewriting the first as 4x+6y=24x + 6y = 2. The distance is then 752\frac{7}{\sqrt{52}}.

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