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Coordinate Geometry

Learning Goals

By the end of this chapter, you will be able to:

• Find distances, midpoints, and section points between any two coordinates
• Calculate the slope, equation, and angle of any straight line
• Work confidently with circles, parabolas, ellipses, and hyperbolas
• Compute areas of triangles and polygons using coordinates
• Solve locus problems — the favourite hunting ground of CAT and SSC CGL Tier-II

The Hook: A Lost Ship and a Genius in Bed

In 1637, a French mathematician named René Descartes was lying sick in bed, watching a fly crawl across his ceiling. He wondered: *"How can I describe the fly's exact position using only numbers?"* His answer — two perpendicular number lines — became the Cartesian plane, and it changed mathematics forever.

Why should you care? Because every GPS satellite, every Google Maps route, every missile trajectory, and every CAT question on "the locus of a point" runs on Descartes' idea. Coordinate Geometry is not a chapter — it is the bridge between algebra and geometry. Master it, and geometry problems become pure calculation. No diagrams to guess, no constructions to fumble. Just numbers.

Why Students Fear This Chapter (And Why You Won't)

Most aspirants struggle here for one reason: they memorise 40 formulas without seeing the one idea behind all of them — *every geometric object is just a set of points satisfying an equation*. A line is not a drawing; it is the set of all (x, y) where ax + by + c = 0. Once this clicks, formulas stop being memory burdens and become obvious consequences.

We will build the chapter in small chunks. Read each chunk, pause, and test yourself at every 🧠 prompt.

Chunk 1: The Distance Formula — Pythagoras in Disguise

The distance between two points A(x₁, y₁) and B(x₂, y₂) is:

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

This is nothing but the Pythagoras theorem. The horizontal gap (x₂ − x₁) and vertical gap (y₂ − y₁) are the two legs of a right triangle; the distance is the hypotenuse.

Example: Distance between (2, 3) and (6, 6) = √[(6−2)² + (6−3)²] = √(16+9) = √25 = 5.

🧠 Active Recall: Without looking back — what is the distance between (0, 0) and (3, 4)? Why does this formula remind you of a right triangle?

Chunk 2: Section Formula — Dividing a Journey

If point P divides the line joining A(x₁, y₁) and B(x₂, y₂) in the ratio m : n (internally):

$$P = \left(\frac{mx_2 + nx_1}{m+n}, \frac{my_2 + ny_1}{m+n}\right)$$

Notice the cross pattern: m multiplies x₂ (the *far* end), n multiplies x₁. The midpoint is the special case m = n = 1:

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

For external division, replace n with −n (or m with −m). Examiners love asking: "In what ratio does the x-axis divide the segment joining (2, −3) and (5, 6)?" Trick: on the x-axis, y = 0, so set the y-part of the section formula to zero and solve for the ratio.

🧠 Active Recall: In the section formula, which coordinate does 'm' multiply — the near point or the far point? What is the midpoint of (4, 8) and (10, 2)?

Chunk 3: Area of a Triangle — The Shoelace

For vertices (x₁, y₁), (x₂, y₂), (x₃, y₃):

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

Mnemonic — "The Shoelace": Write the coordinates in a column, repeat the first point at the bottom, then cross-multiply diagonally down (add) and diagonally up (subtract), and halve the result. It looks like lacing a shoe!

Killer application: If the area comes out zero, the three points are collinear. This is the fastest collinearity test in the entire syllabus — far quicker than checking slopes.

🧠 Active Recall: What does a zero area tell you about three points? Compute the area of the triangle with vertices (0,0), (4,0), (0,3).

Chunk 4: Slope — The Personality of a Line

The slope (gradient) of a line through (x₁, y₁) and (x₂, y₂):

$$m = \frac{y_2 - y_1}{x_2 - x_1} = \tan\theta$$

where θ is the angle the line makes with the positive x-axis.

• m > 0 → line rises left to right
• m < 0 → line falls left to right
• m = 0 → horizontal line
• m undefined → vertical line

Parallel lines: m₁ = m₂. Perpendicular lines: m₁ × m₂ = −1.

Mnemonic — "Perpendicular = Negative Reciprocal = Flip and Switch": If one slope is 2/3, the perpendicular slope is −3/2. Flip the fraction, switch the sign.

🧠 Active Recall: A line has slope 4. What is the slope of (a) a line parallel to it, (b) a line perpendicular to it?

Chunk 5: Equations of a Line — Five Costumes, One Actor

The same straight line can be written in five standard forms:

1. Slope-intercept: y = mx + c (c = y-intercept)
2. Point-slope: y − y₁ = m(x − x₁)
3. Two-point: y − y₁ = [(y₂ − y₁)/(x₂ − x₁)](x − x₁)
4. Intercept form: x/a + y/b = 1 (a, b = x and y intercepts)
5. General form: ax + by + c = 0, where slope = −a/b

Exam gold: From ax + by + c = 0, read the slope instantly as −a/b. No rearranging needed. For 3x + 4y − 12 = 0, slope = −3/4.

Distance of a point (x₁, y₁) from line ax + by + c = 0:

$$d = \frac{|ax_1 + by_1 + c|}{\sqrt{a^2 + b^2}}$$

Distance between parallel lines ax + by + c₁ = 0 and ax + by + c₂ = 0:

$$d = \frac{|c_1 - c_2|}{\sqrt{a^2 + b^2}}$$

🧠 Active Recall: What is the slope of 5x − 2y + 7 = 0? What goes in the numerator of the point-to-line distance formula?

Chunk 6: Angle Between Two Lines

$$\tan\theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right|$$

If m₁m₂ = −1, the denominator vanishes → θ = 90°. This single formula quietly contains the perpendicularity condition — a classic examiner trap is to ask for the angle when the lines are actually perpendicular.

Chunk 7: The Circle — One Fixed Point, One Fixed Distance

A circle is the locus of all points at a fixed distance (radius r) from a fixed point (centre).

• Standard form: (x − h)² + (y − k)² = r², centre (h, k)
• General form: x² + y² + 2gx + 2fy + c = 0, centre (−g, −f), radius = √(g² + f² − c)

Mnemonic — "Centre flips, radius squares": From the general form, the centre coordinates are the *negatives* of half the coefficients of x and y, and the radius needs g² + f² − c under the root.

Condition check: If g² + f² − c < 0, no real circle exists. CAT and CGL Tier-II both test this boundary.

Tangent length from external point (x₁, y₁): √(x₁² + y₁² + 2gx₁ + 2fy₁ + c) — just substitute the point into the left side of the general equation and take the root.

🧠 Active Recall: For x² + y² − 6x + 4y − 12 = 0, what are the centre and radius? (Hint: 2g = −6, 2f = 4.)

Chunk 8: Conic Sections — The Eccentricity Family

Cut a cone with a plane at different angles and you get the conic family. The unifying parameter is eccentricity (e) — how "stretched" the curve is:

Mnemonic — "0-1-less-more: Circle, Parabola, Ellipse, Hyperbola" — say it as a rhythm: *zero, one, less, more*.

Parabola essentials (y² = 4ax): focus (a, 0), directrix x = −a, latus rectum = 4a.

Ellipse essentials: for a > b, foci at (±ae, 0), e = √(1 − b²/a²), and the defining property — the sum of distances from any point on the ellipse to the two foci is constant = 2a.

Hyperbola essentials: foci (±ae, 0), e = √(1 + b²/a²), and the difference of distances to the foci is constant = 2a.

Mnemonic — "Ellipse Adds, Hyperbola Subtracts" — sum for ellipse, difference for hyperbola.

🧠 Active Recall: Which conic has e = 1? For an ellipse, is the sum of focal distances constant, or the difference?

Chunk 9: Locus — The Detective Work

A locus is the path traced by a point moving under a given condition. The method never changes:

1. Let the moving point be (h, k).
2. Translate the given condition into an equation in h and k.
3. Replace (h, k) with (x, y) — that equation is the locus.

Example: A point moves so that its distance from (2, 0) always equals its distance from the y-axis. Then √[(h−2)² + k²] = |h| → h² − 4h + 4 + k² = h² → k² = 4h − 4 → y² = 4(x − 1), a parabola. Notice how the locus method *gave birth* to the conic — this is why examiners adore it.

🧠 Active Recall: What are the three steps of every locus problem?

Summary Table

🔁 Review This After 2 Days

Close the book and write from memory: the distance formula, the section formula, the shoelace area, the point-to-line distance, the circle's centre-radius from the general form, and the eccentricity values of all four conics. Then re-attempt every 🧠 prompt in this chapter. What you cannot recall after 2 days is exactly what the exam will punish — revise those chunks first.