One Right Triangle Hides Between Any Two Points
Derive the distance formula from the Pythagoras theorem, measure from the origin and the axes, prove points collinear or a triangle isosceles or right-angled, and find a missing coordinate or an equidistant point.
How do you measure the distance between two points that are not in a line?
If two points share a height, the distance between them is easy: and are units apart, because only the x-coordinate changed. If they share an x-coordinate, the same applies vertically.
But and share neither. The gap between them is slanting, and the grid cannot count it directly.
So build a right-angled triangle. Go across from to , a horizontal step of units, then up from to , a vertical step of units. The slanting gap is the hypotenuse of that triangle, so by the Pythagoras theorem
Every distance question in coordinate geometry is that one triangle. The horizontal step is the difference of the x-coordinates, the vertical step is the difference of the y-coordinates, and the distance is the hypotenuse. Nothing new is being introduced — the Pythagoras theorem is simply being written in coordinates.
This page covers the ICSE Class 9 Mathematics chapter on the distance formula: the formula and its derivation, distances from the origin and the axes, proving figures from side lengths, and finding a missing coordinate or an equidistant point.
But and share neither. The gap between them is slanting, and the grid cannot count it directly.
So build a right-angled triangle. Go across from to , a horizontal step of units, then up from to , a vertical step of units. The slanting gap is the hypotenuse of that triangle, so by the Pythagoras theorem
Every distance question in coordinate geometry is that one triangle. The horizontal step is the difference of the x-coordinates, the vertical step is the difference of the y-coordinates, and the distance is the hypotenuse. Nothing new is being introduced — the Pythagoras theorem is simply being written in coordinates.
This page covers the ICSE Class 9 Mathematics chapter on the distance formula: the formula and its derivation, distances from the origin and the axes, proving figures from side lengths, and finding a missing coordinate or an equidistant point.
Formula
What is the distance formula and how is it derived?
The distance between two points is the square root of the sum of the squares of the differences of their coordinates.
The derivation is the triangle from the previous section, written with letters. Take and , and let be the point — directly across from and directly below . Then is horizontal and is vertical, so , and
Applying the Pythagoras theorem to gives , which is the formula.
Worked example 1. Find the distance between and .
Worked example 2 — with negatives. Find the distance between and .
The squaring destroys the sign, which is why the order of subtraction does not matter. Doing it the other way round gives , the same answer. But be consistent within one calculation — subtracting and then is where errors creep in.
Worked example 3 — an answer in surd form. Find the distance between and ... and between and .
**Leave the answer as unless a decimal is asked for, and simplify the surd as you learned in the surds chapter.
One check that catches sign errors.** The distance must be at least as large as either individual step. In example 2 the steps were and and the distance was , which is bigger than both. And it can never be more than their sum — — which is the triangle inequality with the right angle as the equality-free case. If your answer falls outside that range, a sign has gone wrong.
The derivation is the triangle from the previous section, written with letters. Take and , and let be the point — directly across from and directly below . Then is horizontal and is vertical, so , and
Applying the Pythagoras theorem to gives , which is the formula.
Worked example 1. Find the distance between and .
Worked example 2 — with negatives. Find the distance between and .
The squaring destroys the sign, which is why the order of subtraction does not matter. Doing it the other way round gives , the same answer. But be consistent within one calculation — subtracting and then is where errors creep in.
Worked example 3 — an answer in surd form. Find the distance between and ... and between and .
**Leave the answer as unless a decimal is asked for, and simplify the surd as you learned in the surds chapter.
One check that catches sign errors.** The distance must be at least as large as either individual step. In example 2 the steps were and and the distance was , which is bigger than both. And it can never be more than their sum — — which is the triangle inequality with the right angle as the equality-free case. If your answer falls outside that range, a sign has gone wrong.
How do you find the distance from the origin or from a point on an axis?
**From the origin, the formula collapses to **, because and are both zero.
Worked example 1. Find the distance of from the origin.
Worked example 2. Find the distance of from the origin.
The signs vanished, so , , and are all units from the origin — four points, one distance. **They lie on a circle of radius centred at the origin, which is the first hint of the equation of a circle you meet in Class 11.
Worked example 3 — from a point on an axis.** Find the distance between and .
Worked example 4 — a point on an axis with an unknown. Find a point on the x-axis that is units from .
A point on the x-axis has the form , so
So , giving or . There are two such points, and , one on each side.
Check both: from the steps are and , distance ; from they are and again, distance . Both answers are required — reporting only one is a half answer, and the square root is what creates the pair.
The geometric reason for two answers. A circle of radius centred at cuts the x-axis at two places, because the centre is only units above the axis and . **If the centre had been units up there would be no solution at all**, and the equation would give a negative value for — which is the arithmetic telling you the circle misses the axis entirely.
Worked example 1. Find the distance of from the origin.
Worked example 2. Find the distance of from the origin.
The signs vanished, so , , and are all units from the origin — four points, one distance. **They lie on a circle of radius centred at the origin, which is the first hint of the equation of a circle you meet in Class 11.
Worked example 3 — from a point on an axis.** Find the distance between and .
Worked example 4 — a point on an axis with an unknown. Find a point on the x-axis that is units from .
A point on the x-axis has the form , so
So , giving or . There are two such points, and , one on each side.
Check both: from the steps are and , distance ; from they are and again, distance . Both answers are required — reporting only one is a half answer, and the square root is what creates the pair.
The geometric reason for two answers. A circle of radius centred at cuts the x-axis at two places, because the centre is only units above the axis and . **If the centre had been units up there would be no solution at all**, and the equation would give a negative value for — which is the arithmetic telling you the circle misses the axis entirely.
How do you prove points collinear, or a triangle isosceles or right-angled?
Compute all three distances, then compare them. Which comparison you make depends on what you are proving.
To prove three points collinear: show that the sum of two of the distances equals the third. Three points in a straight line cannot enclose a triangle, so the triangle inequality becomes an equality.
Worked example 1. Show that , and are collinear.
So the three points lie in a straight line, with between and .
Which point is in the middle matters. It is the one not in the longest distance — here is the longest, so lies between them. If you add the wrong two distances you get no equality and may wrongly conclude that the points are not collinear. Order the three lengths before adding.
To prove a triangle isosceles: show two sides equal. Equilateral: all three equal. Right-angled: show the Pythagoras relation holds for the three lengths.
Worked example 2 — isosceles. Show that , and form an isosceles triangle.
, so the triangle is isosceles with as its base.
Worked example 3 — right-angled. Show that , and form a right-angled triangle and find its area.
By the converse of the Pythagoras theorem the triangle is right-angled at , and
Worked example 4 — equilateral. Show that , and form an equilateral triangle.
All three sides are units, so the triangle is equilateral, with area square units.
The order to work in. Compute all three distances first, as exact surds, and write them down; then decide what the comparison shows. Students who compute one distance and then guess the answer lose the marks for the other two — and the proof is the comparison, not the arithmetic.
To prove three points collinear: show that the sum of two of the distances equals the third. Three points in a straight line cannot enclose a triangle, so the triangle inequality becomes an equality.
Worked example 1. Show that , and are collinear.
So the three points lie in a straight line, with between and .
Which point is in the middle matters. It is the one not in the longest distance — here is the longest, so lies between them. If you add the wrong two distances you get no equality and may wrongly conclude that the points are not collinear. Order the three lengths before adding.
To prove a triangle isosceles: show two sides equal. Equilateral: all three equal. Right-angled: show the Pythagoras relation holds for the three lengths.
Worked example 2 — isosceles. Show that , and form an isosceles triangle.
, so the triangle is isosceles with as its base.
Worked example 3 — right-angled. Show that , and form a right-angled triangle and find its area.
By the converse of the Pythagoras theorem the triangle is right-angled at , and
Worked example 4 — equilateral. Show that , and form an equilateral triangle.
All three sides are units, so the triangle is equilateral, with area square units.
The order to work in. Compute all three distances first, as exact surds, and write them down; then decide what the comparison shows. Students who compute one distance and then guess the answer lose the marks for the other two — and the proof is the comparison, not the arithmetic.
How do you find a missing coordinate or a point equidistant from two others?
Write the distance condition as an equation, square both sides to clear the roots, and solve. Squaring is always safe here because a distance is never negative.
Worked example 1 — a missing coordinate. The distance between and is units. Find .
So , giving or . Both are valid, and a complete answer gives both.
**Check **: the steps are and , so the distance is , as required.
Worked example 2 — equidistant on the x-axis. Find the point on the x-axis equidistant from and .
Let the point be and set the two squared distances equal:
So the point is .
Check: the distance to is , and to it is . Equal, as required.
**The always cancels, which is why an equidistance condition gives a linear equation with exactly one answer — unlike a fixed-distance condition, which is quadratic and gives two.
Worked example 3 — equidistant on the y-axis.** Find the point on the y-axis equidistant from and .
Let it be :
So the point is .
Check: to the distance is ; to it is . Equal, as required.
Worked example 4 — a point equidistant from three points. Find the centre of the circle through , and .
A centre is equidistant from all three, so take and equate squared distances to the first two:
Now to the first and third:
So the centre is , at a distance from each of the three points.
And that agrees with the circle chapter. Those three points form a right-angled triangle with hypotenuse , and the circumcentre of a right-angled triangle is the mid-point of the hypotenuse — which is indeed , at a distance of . Two different chapters, one answer, which is the best possible check.
Worked example 1 — a missing coordinate. The distance between and is units. Find .
So , giving or . Both are valid, and a complete answer gives both.
**Check **: the steps are and , so the distance is , as required.
Worked example 2 — equidistant on the x-axis. Find the point on the x-axis equidistant from and .
Let the point be and set the two squared distances equal:
So the point is .
Check: the distance to is , and to it is . Equal, as required.
**The always cancels, which is why an equidistance condition gives a linear equation with exactly one answer — unlike a fixed-distance condition, which is quadratic and gives two.
Worked example 3 — equidistant on the y-axis.** Find the point on the y-axis equidistant from and .
Let it be :
So the point is .
Check: to the distance is ; to it is . Equal, as required.
Worked example 4 — a point equidistant from three points. Find the centre of the circle through , and .
A centre is equidistant from all three, so take and equate squared distances to the first two:
Now to the first and third:
So the centre is , at a distance from each of the three points.
And that agrees with the circle chapter. Those three points form a right-angled triangle with hypotenuse , and the circumcentre of a right-angled triangle is the mid-point of the hypotenuse — which is indeed , at a distance of . Two different chapters, one answer, which is the best possible check.
Exam tip
What layout keeps distance-formula work accurate?
Write the formula, then the substitution with brackets around every difference, then the value. The brackets are what protect you from sign errors with negative coordinates.
- Name the points with letters and coordinates: *let and *. Then every difference can be checked against the naming
- Keep brackets on every subtraction: , not . The second is a different number entirely
- Subtract in the same order for both coordinates within one calculation
- Leave surds exact and simplify them:
- For a proof, compute all three distances first and write them down before comparing
- Order the three lengths before testing collinearity; the sum of the two smaller ones must equal the largest
- Square both sides early in an equation, since a distance cannot be negative
- Expect two answers from a fixed-distance condition and one from an equidistance condition
- Always check by substituting back and confirming the two distances are equal
The distinction worth naming. A fixed distance condition, such as * units from , is quadratic and usually has two solutions. An equidistance condition, such as the same distance from as from *, is linear because the squared terms cancel — and it has exactly one solution on a given axis. Count the answers you expect before you start solving, and a missing root becomes obvious.
- Name the points with letters and coordinates: *let and *. Then every difference can be checked against the naming
- Keep brackets on every subtraction: , not . The second is a different number entirely
- Subtract in the same order for both coordinates within one calculation
- Leave surds exact and simplify them:
- For a proof, compute all three distances first and write them down before comparing
- Order the three lengths before testing collinearity; the sum of the two smaller ones must equal the largest
- Square both sides early in an equation, since a distance cannot be negative
- Expect two answers from a fixed-distance condition and one from an equidistance condition
- Always check by substituting back and confirming the two distances are equal
The distinction worth naming. A fixed distance condition, such as * units from , is quadratic and usually has two solutions. An equidistance condition, such as the same distance from as from *, is linear because the squared terms cancel — and it has exactly one solution on a given axis. Count the answers you expect before you start solving, and a missing root becomes obvious.
Did you know
Why does every equidistant point lie on one straight line?
In the worked examples you found single points equidistant from two others — one on the x-axis, one on the y-axis. But those are not the only such points, and the full set has a shape you already know.
Take the pair and from worked example 2. The point was equidistant from both. So is their mid-point, — obviously, since it is the same distance along the segment from each end.
Both of those points, and every other equidistant point, lie on the perpendicular bisector of the segment joining the two. That is the same statement you used to find a circumcentre in the circle chapter: the set of points equidistant from two given points is the perpendicular bisector of the segment joining them.
Check it. The segment from to has mid-point and its own steepness is
A line perpendicular to it must have steepness , since . Starting from with that steepness, moving units left drops the height by , landing at — exactly the point the algebra produced.
So the two methods agree, and that is why the equidistance equation came out linear: its solutions form a line, and restricting to the x-axis or the y-axis picks out the single place where that line crosses.
The same idea answers a practical question. If two villages need a shared water point at equal distance from both, every acceptable site lies on one straight line between them — and choosing among those sites is then about the road, the ground and the depth of water, not about geometry. Coordinate geometry does not just answer a question; it tells you how many answers there are, and here the answer is a whole line rather than a point.
Take the pair and from worked example 2. The point was equidistant from both. So is their mid-point, — obviously, since it is the same distance along the segment from each end.
Both of those points, and every other equidistant point, lie on the perpendicular bisector of the segment joining the two. That is the same statement you used to find a circumcentre in the circle chapter: the set of points equidistant from two given points is the perpendicular bisector of the segment joining them.
Check it. The segment from to has mid-point and its own steepness is
A line perpendicular to it must have steepness , since . Starting from with that steepness, moving units left drops the height by , landing at — exactly the point the algebra produced.
So the two methods agree, and that is why the equidistance equation came out linear: its solutions form a line, and restricting to the x-axis or the y-axis picks out the single place where that line crosses.
The same idea answers a practical question. If two villages need a shared water point at equal distance from both, every acceptable site lies on one straight line between them — and choosing among those sites is then about the road, the ground and the depth of water, not about geometry. Coordinate geometry does not just answer a question; it tells you how many answers there are, and here the answer is a whole line rather than a point.
Exam relevance
How is the distance formula used in JEE-level questions?
This is foundation work that becomes one of the most frequently used single lines in the JEE Mathematics syllabus.
Where it leads. Class 10 adds the section formula and the area of a triangle from coordinates, which together with the distance formula complete the toolkit. Class 11 Straight Lines uses distance for the perpendicular distance from a point to a line; Circles are defined entirely by it, since is the statement that a point is a fixed distance from the origin; and Conic Sections are defined by distance conditions — an ellipse is the set of points whose distances to two fixed points add to a constant. Every one of those is a distance-formula condition with a different constraint.
Where the classification technique reappears. Deciding whether a triangle from three coordinates is isosceles, equilateral or right-angled is a standard JEE Main item, and so is showing that four points form a parallelogram, rhombus or square — which combines this chapter with the diagonal tests from the quadrilaterals chapter. Equal sides plus equal diagonals means a square; equal sides with unequal diagonals means a rhombus.
Where it appears in Physics. Displacement is a distance between two positions, and in three dimensions the formula simply gains a term: . The magnitude of a vector is the same calculation, and the work done by a force over a displacement needs it. JEE Main and NEET Physics use it in kinematics and in electrostatics, where the inverse-square law needs the distance between two charges.
Question types to expect. At this level: find a distance, prove a figure, find a missing coordinate or an equidistant point. In competitive papers: the locus of a point satisfying a distance condition, the type of quadrilateral formed by four points, and the radius and centre of a circle from three points.
That last one you have already done. Worked example 4 of the previous section found the centre and radius of the circle through three points using nothing but equal distances — which is precisely how the Class 11 method works before it is compressed into a formula.
The single trap that costs marks. Losing a sign inside a bracket, and dropping the second root of a quadratic condition. In JEE, a locus question whose answer is a pair of lines or two possible points is deliberately set so that candidates who find one answer choose a wrong option. Expect two answers whenever you have squared something.
Board versus competitive emphasis. ICSE marks the formula, the bracketed substitution, the exact surd and the comparison that completes a proof; a competitive paper marks the classification or the coordinate. The transferable habit is computing every distance before drawing any conclusion.
Where it leads. Class 10 adds the section formula and the area of a triangle from coordinates, which together with the distance formula complete the toolkit. Class 11 Straight Lines uses distance for the perpendicular distance from a point to a line; Circles are defined entirely by it, since is the statement that a point is a fixed distance from the origin; and Conic Sections are defined by distance conditions — an ellipse is the set of points whose distances to two fixed points add to a constant. Every one of those is a distance-formula condition with a different constraint.
Where the classification technique reappears. Deciding whether a triangle from three coordinates is isosceles, equilateral or right-angled is a standard JEE Main item, and so is showing that four points form a parallelogram, rhombus or square — which combines this chapter with the diagonal tests from the quadrilaterals chapter. Equal sides plus equal diagonals means a square; equal sides with unequal diagonals means a rhombus.
Where it appears in Physics. Displacement is a distance between two positions, and in three dimensions the formula simply gains a term: . The magnitude of a vector is the same calculation, and the work done by a force over a displacement needs it. JEE Main and NEET Physics use it in kinematics and in electrostatics, where the inverse-square law needs the distance between two charges.
Question types to expect. At this level: find a distance, prove a figure, find a missing coordinate or an equidistant point. In competitive papers: the locus of a point satisfying a distance condition, the type of quadrilateral formed by four points, and the radius and centre of a circle from three points.
That last one you have already done. Worked example 4 of the previous section found the centre and radius of the circle through three points using nothing but equal distances — which is precisely how the Class 11 method works before it is compressed into a formula.
The single trap that costs marks. Losing a sign inside a bracket, and dropping the second root of a quadratic condition. In JEE, a locus question whose answer is a pair of lines or two possible points is deliberately set so that candidates who find one answer choose a wrong option. Expect two answers whenever you have squared something.
Board versus competitive emphasis. ICSE marks the formula, the bracketed substitution, the exact surd and the comparison that completes a proof; a competitive paper marks the classification or the coordinate. The transferable habit is computing every distance before drawing any conclusion.
Key takeaways
What should you be able to do with the distance formula?
One formula, one theorem behind it, and four question types built on it.
- **, which is the Pythagoras theorem applied to the horizontal and vertical steps
- From the origin** it becomes , and the four sign combinations of a point all give the same distance
- Keep brackets on every difference and subtract in a consistent order
- Leave answers as simplified surds unless a decimal is asked for
- Collinear: the two smaller distances add to the largest — order the three before testing
- Isosceles means two equal sides, equilateral three, and right-angled is tested by the Pythagoras relation on the three lengths
- A fixed-distance condition is quadratic and usually gives two answers; an equidistance condition is linear and gives one, because the squared terms cancel
- A point equidistant from two others lies on the perpendicular bisector of the segment joining them
- A point equidistant from three points is a circumcentre, found by equating squared distances twice
The sharpest self-test is the circumcentre: find the point equidistant from , and , then check your answer against the right-angled-triangle shortcut from the circle chapter — the mid-point of the hypotenuse.
- **, which is the Pythagoras theorem applied to the horizontal and vertical steps
- From the origin** it becomes , and the four sign combinations of a point all give the same distance
- Keep brackets on every difference and subtract in a consistent order
- Leave answers as simplified surds unless a decimal is asked for
- Collinear: the two smaller distances add to the largest — order the three before testing
- Isosceles means two equal sides, equilateral three, and right-angled is tested by the Pythagoras relation on the three lengths
- A fixed-distance condition is quadratic and usually gives two answers; an equidistance condition is linear and gives one, because the squared terms cancel
- A point equidistant from two others lies on the perpendicular bisector of the segment joining them
- A point equidistant from three points is a circumcentre, found by equating squared distances twice
The sharpest self-test is the circumcentre: find the point equidistant from , and , then check your answer against the right-angled-triangle shortcut from the circle chapter — the mid-point of the hypotenuse.