Imagine you're telling a friend where you left your book in a library. You don't say "near the window" — you say "third shelf, second row, fourth book from the left." You're using numbers to pin down an exact location.
Coordinate geometry does the same thing, but for points on a flat surface. It gives every point a precise address — a pair of numbers — so we can describe shapes, distances, and positions using algebra.
The Big Idea
Before coordinate geometry, geometry was about drawing shapes and proving things with logic alone. Algebra was about numbers and equations. These two worlds seemed separate.
Then René Descartes (a French mathematician) had a simple but revolutionary idea: draw two perpendicular number lines that cross at zero. Now every point on the plane has a unique pair of numbers — its coordinates.
That's it. That's the entire foundation.
The Coordinate System
Take a horizontal line — call it the x-axis. Take a vertical line — call it the y-axis. They cross at a point called the origin, labelled O.
Any point P is located by two numbers:
Its x-coordinate: how far right (positive) or left (negative) from the origin
Its y-coordinate: how far up (positive) or down (negative) from the origin
We write this as an ordered pair: (x,y).
Note
The order matters. (3,5) is not the same point as (5,3). The first number is always the horizontal position; the second is always the vertical.
A Concrete Example
Plot the point A(2,3):
Start at the origin (0,0).
Move 2 units to the right along the x-axis.
From there, move 3 units up (parallel to the y-axis).
Mark the point.
Now plot B(−1,4):
Start at the origin.
Move 1 unit left (negative x-direction).
Move 4 units up.
Mark the point.
Every point on the plane has exactly one such address. And every pair of numbers corresponds to exactly one point. This one-to-one matching is what makes coordinate geometry powerful.
The Four Quadrants
The axes divide the plane into four regions, called quadrants:
Quadrant
x-sign
y-sign
Example
I
+
+
(2,3)
II
−
+
(−1,4)
III
−
−
(−3,−2)
IV
+
−
(5,−1)
Points on the axes themselves (where either coordinate is zero) don't belong to any quadrant.
Why This Matters
Once every point has a number address, we can:
Calculate distances between points using the Pythagorean theorem
Find midpoints by averaging coordinates
Describe lines with equations like y=mx+c
Solve geometric problems using algebra instead of drawing
Important
The distance between two points (x1,y1) and (x2,y2) is:
d=(x2−x1)2+(y2−y1)2
This is just the Pythagorean theorem in disguise.
The Precise Statement
Coordinate geometry (also called analytic geometry) is the study of geometry using a coordinate system. It establishes a correspondence between:
Points on a plane and ordered pairs of real numbers
Geometric figures (lines, circles, curves) and algebraic equations …
We check whether the three points form a right triangle by computing the squared distances between each pair and testing if the Pythagorean relation holds. The distances are AB=714, BC=6686, and CA=3101; none of the squared sums match, so the triangle is not right-angled.
The question is: do three given points in 3D space form a right-angled triangle? In coordinate geometry, the cleanest way to test this is to compute the side lengths (using the distance formula) and then check if the square of the longest side equals the sum of the squares of the other two. That is the Pythagorean theorem in reverse — if it holds, the triangle is right-angled; if not, it isn’t.
We don’t need to find angles or slopes. Just distances. And because we’re in three dimensions, the distance between two points (x1,y1,z1) and (x2,y2,z2) is:
d=(x2−x1)2+(y2−y1)2+(z2−z1)2
We’ll compute AB, BC, and CA, then compare squares.
Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.
Showing the 12 most recent of 22 on this concept.
KEAM 2026Set eng-2026-04174 marksMCQ
Q.Let O be the origin and P be a point on a line such that OP is perpendicular to that line. If OP makes an obtuse angle α with the x-axis, OP=5 and sinα=53 , then the equation of the line is
(A) 3x−4y−25=0
(B) 4x+3y+25=0
(C) 3x−4y+25=0
(D) 4x−3y−25=0
(E) 4x−3y+25=0
›Reveal solutionSolution
The line is the perpendicular to OP at its foot; use the normal form xcosα+ysinα=p with p=OP=5.
Since OP is perpendicular to the line, OP is the normal from the origin and p=OP=5. The angle is obtuse with sinα=53, so cosα=−1−259=−54. …
Q.If the foci of the hyperbola a2x2−b2y2=1 coincide with the foci of the ellipse 49x2+36y2=1 , then the value of a2+b2 is equal to
(A) 25
(B) 36
(C) 85
(D) 169
(E) 13
›Reveal solutionSolution
The shared focal distance c satisfies c2=13; for the hyperbola a2+b2=c2.
Q.If the distance between the foci of an ellipse is 4 and its eccentricity is 21 , then the length of its latus rectum is
(A) 4
(B) 8
(C) 6
(D) 12
(E) 10
›Reveal solutionSolution
From 2c=4 and e=21 get a,b2, then a2b2.
Distance between foci 2c=4⇒c=2. Eccentricity e=ac=21⇒a=4.
Q.A point C lies on the perpendicular bisector of the straight-line segment joining the points A(−3,−6) and B(13,−6). If the point C lies in the first quadrant and the distance between the point C and the midpoint of AB is 8 units, then the coordinates of C are
(A) (5,1)
(B) (5,2)
(C) (5,3)
(D) (5,4)
(E) (5,5)
›Reveal solutionSolution
The perpendicular bisector is x=5; the point 8 units from (5,−6) in the first quadrant is (5,2).
A(−3,−6) and B(13,−6) share y=−6, so their midpoint is M=(5,−6) and the perpendicular bisector is the vertical line x=5. …
Q.If the eccentricity and the length of latus rectum of an ellipse are, respectively, 51 and 548, then the length of the major axis of the ellipse is
(A) 5
(B) 6
(C) 8
(D) 10
(E) 12
›Reveal solutionSolution
Combine LR =a2b2 with b2=a2(1−e2) to get a=5, so major axis 2a=10.
Q.If the length of the major axis of an ellipse is thrice the length of the minor axis, then its eccentricity is equal to
(A) 322
(B) 32
(C) 21
(D) 21
(E) 221
›Reveal solutionSolution
Major axis =3×minor axis means a=3b, giving e=322.
Length of major axis =2a, minor axis =2b. Given 2a=3(2b), so a=3b. Then …
Q.The x-intercept and y-intercept of a line are three times and four times of the x-intercept and y-intercept of the line 3x+2y=6, respectively. Then the equation of the line is
(A) 2x−y=12
(B) 2x+y=12
(C) 2x−y=−12
(D) x+2y=12
(E) 2x+2y=12
›Reveal solutionSolution
Get the base line's intercepts, scale them, then form the intercept-form equation.
For 3x+2y=6: x-intercept =2 (set y=0), y-intercept =3 (set x=0).
New x-intercept =3×2=6; new y-intercept =4×3=12. …
Q.The equation of a hyperbola is 9x2−16y2=144. If A and S are, respectively, the focus and the vertex of one section of the hyperbola, then the length of AS is
(A) 25
(B) 23
(C) 21
(D) 2
(E) 1
›Reveal solutionSolution
Put the hyperbola in standard form, find a and c, then AS=c−a.
Q.The length of major axis and minor axis of an ellipse are, respectively, m and n. If m2−n2=45 and the eccentricity of the ellipse is 35, then the length of the major axis is
(A) 13
(B) 6
(C) 12
(D) 18
(E) 9
›Reveal solutionSolution
The major axis length is 9.
Concept and Intuition
With major axis m=2a and minor axis n=2b, m2−n2=4(a2−b2)=4a2e2. Use the given eccentricity to solve for a.