Three Lengths Tell You Whether Points Make a Triangle or Just a Straight Line
Find the distance between any two points with the distance formula, test three points for collinearity, classify a triangle as isosceles, equilateral, scalene or right-angled, identify the quadrilateral four points make, and locate a point equidistant from two others.
How can you decide the shape of a figure without drawing it?
Give someone four pairs of numbers — , , and — and ask what shape they make. Drawing them on graph paper gives you a guess. Calculating six lengths gives you a proof.
Those six lengths are the four sides and the two diagonals, and in this case they come out as four equal sides and two equal diagonals. That is a square, and no amount of careful measurement with a ruler could establish it as certainly.
That is what coordinate geometry offers: geometry done with arithmetic instead of instruments. And the one tool that unlocks all of it in this part is a formula for the distance between two points, which is nothing more than the Pythagoras theorem written in coordinates.
From that single formula come four kinds of question, and each is worth marks in its own right:
- Collinearity — do three points lie on one straight line?
- Type of triangle — is it isosceles, equilateral, scalene or right-angled?
- Type of quadrilateral — square, rectangle, rhombus, parallelogram or none of these?
- Equidistant points — where is the point on an axis that is the same distance from two given points?
The method is always the same: compute lengths, then compare them. What changes from question to question is which comparison settles the matter, and knowing that in advance is what turns a long calculation into a short one.
This page covers the first part of the CBSE Class 10 Maths chapter on coordinate geometry: the distance formula, collinearity, classifying triangles and quadrilaterals, and equidistant points.
Those six lengths are the four sides and the two diagonals, and in this case they come out as four equal sides and two equal diagonals. That is a square, and no amount of careful measurement with a ruler could establish it as certainly.
That is what coordinate geometry offers: geometry done with arithmetic instead of instruments. And the one tool that unlocks all of it in this part is a formula for the distance between two points, which is nothing more than the Pythagoras theorem written in coordinates.
From that single formula come four kinds of question, and each is worth marks in its own right:
- Collinearity — do three points lie on one straight line?
- Type of triangle — is it isosceles, equilateral, scalene or right-angled?
- Type of quadrilateral — square, rectangle, rhombus, parallelogram or none of these?
- Equidistant points — where is the point on an axis that is the same distance from two given points?
The method is always the same: compute lengths, then compare them. What changes from question to question is which comparison settles the matter, and knowing that in advance is what turns a long calculation into a short one.
This page covers the first part of the CBSE Class 10 Maths chapter on coordinate geometry: the distance formula, collinearity, classifying triangles and quadrilaterals, and equidistant points.
Formula
What is the distance formula and where does it come from?
Subtract the coordinates, square both differences, add them, and take the square root.
For two points and ,
and the distance of a point from the origin is the special case with :
Where it comes from. Drop a horizontal line from and a vertical line from so that they meet at a point . Then is the horizontal gap , is the vertical gap , and the angle at is a right angle. **By the Pythagoras theorem, , which is the formula. So the distance formula is not a new idea at all — it is the Pythagoras theorem applied to a triangle you construct from the coordinates.
Why the order of subtraction does not matter. Both differences are squared, and a square is never negative. So , and you may subtract either way round. That is a genuine relief, and it is the one place in this chapter where sign errors cannot hurt you.
Worked example 1.** Find the distance between and .
**Leave it as **, not as and not as a decimal, unless the question asks for one.
Worked example 2. Find the distance of from the origin.
Notice that the minus sign vanished when squared, which is why the formula never needs you to worry about which quadrant a point is in.
Worked example 3 — a distance along an axis. Find the distance between and .
**For two points on the same horizontal line the formula reduces to the difference of the -coordinates, and similarly for a vertical line. Recognising that saves writing out the whole formula for something you can do in your head.
Worked example 4 — an unknown coordinate.** Find the values of for which the distance between and is units.
Square both sides of the formula to avoid working with a root:
Both values are valid, and it makes sense that there are two: a circle of radius centred at crosses the vertical line in two places.
**Check **: . **Check **: . Both correct.
The habit that saves time throughout this chapter. Work with the squares of the distances wherever you can, and only take square roots at the end if the question wants a length. Comparing with is easier and safer than comparing with , and every classification question in this chapter is a comparison.
For two points and ,
and the distance of a point from the origin is the special case with :
Where it comes from. Drop a horizontal line from and a vertical line from so that they meet at a point . Then is the horizontal gap , is the vertical gap , and the angle at is a right angle. **By the Pythagoras theorem, , which is the formula. So the distance formula is not a new idea at all — it is the Pythagoras theorem applied to a triangle you construct from the coordinates.
Why the order of subtraction does not matter. Both differences are squared, and a square is never negative. So , and you may subtract either way round. That is a genuine relief, and it is the one place in this chapter where sign errors cannot hurt you.
Worked example 1.** Find the distance between and .
**Leave it as **, not as and not as a decimal, unless the question asks for one.
Worked example 2. Find the distance of from the origin.
Notice that the minus sign vanished when squared, which is why the formula never needs you to worry about which quadrant a point is in.
Worked example 3 — a distance along an axis. Find the distance between and .
**For two points on the same horizontal line the formula reduces to the difference of the -coordinates, and similarly for a vertical line. Recognising that saves writing out the whole formula for something you can do in your head.
Worked example 4 — an unknown coordinate.** Find the values of for which the distance between and is units.
Square both sides of the formula to avoid working with a root:
Both values are valid, and it makes sense that there are two: a circle of radius centred at crosses the vertical line in two places.
**Check **: . **Check **: . Both correct.
The habit that saves time throughout this chapter. Work with the squares of the distances wherever you can, and only take square roots at the end if the question wants a length. Comparing with is easier and safer than comparing with , and every classification question in this chapter is a comparison.
How do you test three points for collinearity and classify a triangle?
Compute all three distances. If the two shorter ones add up to the longest, the points are collinear; if they do not, the points form a triangle, and comparing the three lengths tells you which kind.
The collinearity test, and why it works. Three points , , lie on one straight line exactly when one of them lies between the other two — and then the two short pieces make up the long one:
**If instead , the path through is a detour**, so is off the line and a genuine triangle exists.
Worked example 1 — collinear. Show that , and are collinear.
So the three points are collinear.
Notice how much the surd form helped. In surds the addition is immediate; in decimals you would be comparing against and trusting the rounding. Keep the surds and the test is exact.
Worked example 2 — not collinear. Are , and collinear?
Now , and . Then
The sum exceeds the longest side, so the points are not collinear. Here the surds could not be combined, so decimals were unavoidable — but the gap is large enough that rounding is not in question.
Classifying a triangle once you have the three lengths.
- All three equal — equilateral
- Exactly two equal — isosceles
- All three different — scalene
- The square of the longest equals the sum of the squares of the other two — right-angled, and this is a separate question from the first three
A triangle can be both, so check both properties. An isosceles right-angled triangle is entirely possible.
Worked example 3 — isosceles. Classify the triangle with vertices , and .
Two sides are equal, so the triangle is isosceles. Is it right-angled? The longest side is , and . No, so it is isosceles and not right-angled.
Worked example 4 — equilateral. Show that , and form an equilateral triangle.
**All three sides equal , so the triangle is equilateral.
Worked example 5 — right-angled.** Classify the triangle with vertices , and .
All three lengths differ, so the triangle is scalene. Testing for a right angle with the squares:
So the triangle is right-angled, and the right angle is at — the vertex opposite the longest side, since is the hypotenuse. The triangle is scalene and right-angled.
Naming the right-angle vertex is worth a mark, and the rule is simple: the right angle sits at the vertex not on the longest side.
The collinearity test, and why it works. Three points , , lie on one straight line exactly when one of them lies between the other two — and then the two short pieces make up the long one:
**If instead , the path through is a detour**, so is off the line and a genuine triangle exists.
Worked example 1 — collinear. Show that , and are collinear.
So the three points are collinear.
Notice how much the surd form helped. In surds the addition is immediate; in decimals you would be comparing against and trusting the rounding. Keep the surds and the test is exact.
Worked example 2 — not collinear. Are , and collinear?
Now , and . Then
The sum exceeds the longest side, so the points are not collinear. Here the surds could not be combined, so decimals were unavoidable — but the gap is large enough that rounding is not in question.
Classifying a triangle once you have the three lengths.
- All three equal — equilateral
- Exactly two equal — isosceles
- All three different — scalene
- The square of the longest equals the sum of the squares of the other two — right-angled, and this is a separate question from the first three
A triangle can be both, so check both properties. An isosceles right-angled triangle is entirely possible.
Worked example 3 — isosceles. Classify the triangle with vertices , and .
Two sides are equal, so the triangle is isosceles. Is it right-angled? The longest side is , and . No, so it is isosceles and not right-angled.
Worked example 4 — equilateral. Show that , and form an equilateral triangle.
**All three sides equal , so the triangle is equilateral.
Worked example 5 — right-angled.** Classify the triangle with vertices , and .
All three lengths differ, so the triangle is scalene. Testing for a right angle with the squares:
So the triangle is right-angled, and the right angle is at — the vertex opposite the longest side, since is the hypotenuse. The triangle is scalene and right-angled.
Naming the right-angle vertex is worth a mark, and the rule is simple: the right angle sits at the vertex not on the longest side.
How do you decide what kind of quadrilateral four points make?
Compute the four sides in order and then the two diagonals. The pattern of equalities identifies the shape.
The decision list, which is worth learning as a sequence of questions.
- All four sides equal and the diagonals equal — square
- All four sides equal and the diagonals unequal — rhombus
- Opposite sides equal and the diagonals equal — rectangle
- Opposite sides equal and the diagonals unequal — parallelogram
- None of these patterns — a general quadrilateral, or possibly a trapezium
The order of the four points matters absolutely. means the sides are , , , and the diagonals are and . Taking the points in a different order turns a square into a crossed figure, and a question that says "taken in order" is telling you not to rearrange them.
Worked example 1 — a square. Show that , , and are the vertices of a square.
The four sides:
The two diagonals:
All four sides are equal and both diagonals are equal, so it is a square.
One extra observation confirms it. For a square of side , the diagonal should be , so the squared diagonal should be twice the squared side: . It is — a check that takes one glance and rules out any arithmetic slip.
Worked example 2 — a rhombus. Identify the quadrilateral with vertices , , and .
The four sides:
The two diagonals:
All four sides are equal but the diagonals are not, so it is a rhombus and not a square.
That single comparison is the whole question. Had the diagonals matched, the answer would have been a square, and a student who computes only the sides cannot tell the two apart. Four equal sides never settle it on their own.
Worked example 3 — checking a parallelogram. If , , and are taken in order, what shape do they make?
Opposite sides are equal in pairs and the diagonals are unequal, so it is a parallelogram — neither a rectangle nor a rhombus.
Why opposite sides being equal is enough for a parallelogram. A quadrilateral whose opposite sides are equal in length must have those sides parallel as well, so the definition is met. But that is the weakest of the four conclusions, and you must go on to the diagonals and to the adjacent sides before claiming anything stronger.
The sequence to follow, every time.
- Are all four sides equal? If yes, it is a square or a rhombus — the diagonals decide
- If not, are opposite sides equal? If yes, it is a rectangle or a parallelogram — the diagonals decide
- If neither, it is none of the four named shapes, and you should say so rather than forcing a name
The decision list, which is worth learning as a sequence of questions.
- All four sides equal and the diagonals equal — square
- All four sides equal and the diagonals unequal — rhombus
- Opposite sides equal and the diagonals equal — rectangle
- Opposite sides equal and the diagonals unequal — parallelogram
- None of these patterns — a general quadrilateral, or possibly a trapezium
The order of the four points matters absolutely. means the sides are , , , and the diagonals are and . Taking the points in a different order turns a square into a crossed figure, and a question that says "taken in order" is telling you not to rearrange them.
Worked example 1 — a square. Show that , , and are the vertices of a square.
The four sides:
The two diagonals:
All four sides are equal and both diagonals are equal, so it is a square.
One extra observation confirms it. For a square of side , the diagonal should be , so the squared diagonal should be twice the squared side: . It is — a check that takes one glance and rules out any arithmetic slip.
Worked example 2 — a rhombus. Identify the quadrilateral with vertices , , and .
The four sides:
The two diagonals:
All four sides are equal but the diagonals are not, so it is a rhombus and not a square.
That single comparison is the whole question. Had the diagonals matched, the answer would have been a square, and a student who computes only the sides cannot tell the two apart. Four equal sides never settle it on their own.
Worked example 3 — checking a parallelogram. If , , and are taken in order, what shape do they make?
Opposite sides are equal in pairs and the diagonals are unequal, so it is a parallelogram — neither a rectangle nor a rhombus.
Why opposite sides being equal is enough for a parallelogram. A quadrilateral whose opposite sides are equal in length must have those sides parallel as well, so the definition is met. But that is the weakest of the four conclusions, and you must go on to the diagonals and to the adjacent sides before claiming anything stronger.
The sequence to follow, every time.
- Are all four sides equal? If yes, it is a square or a rhombus — the diagonals decide
- If not, are opposite sides equal? If yes, it is a rectangle or a parallelogram — the diagonals decide
- If neither, it is none of the four named shapes, and you should say so rather than forcing a name
How do you find a point equidistant from two given points?
Write the unknown point with the information the question gives you, set the two squared distances equal, and solve. Squaring first is what keeps the algebra clean.
**A point on the -axis has coordinates **, and a point on the -axis has coordinates . Recognising that is the whole of the setting-up step, and forgetting it is the most common reason these questions go wrong.
**Worked example 1 — on the -axis.** Find a point on the -axis that is equidistant from and .
Let the point be . Setting :
**The point is .
Check both distances.** From to :
and from to :
Equal, so the answer is right. Notice that the terms cancelled, leaving a linear equation — that always happens in this family, and if you find yourself with a quadratic then a squaring step went wrong.
**Worked example 2 — on the -axis.** Find a point on the -axis that is equidistant from and .
Let the point be . Then gives
**The point is .
Check**: from to is , and to is . Equal.
Worked example 3 — a general equidistant condition. Find the relation between and if the point is equidistant from and .
The answer is an equation, not a point — and that equation is the perpendicular bisector of the segment joining the two given points, since every point equidistant from two fixed points lies on it.
Check with the mid-point. The mid-point of and is , and substituting gives . The mid-point lies on the line, which it must, being equidistant from both ends.
Worked example 4 — an unknown coordinate from a condition. If the point is equidistant from and , find .
**So the point is , directly above the mid-point of the two given points — exactly where the perpendicular bisector runs.
Check**: from the origin, and from . Equal.
The single technique behind all four examples. Square the distances before equating them. The square roots disappear, the and terms cancel, and what remains is linear. Trying to work with the roots in place turns a three-line problem into an unmanageable one.
**A point on the -axis has coordinates **, and a point on the -axis has coordinates . Recognising that is the whole of the setting-up step, and forgetting it is the most common reason these questions go wrong.
**Worked example 1 — on the -axis.** Find a point on the -axis that is equidistant from and .
Let the point be . Setting :
**The point is .
Check both distances.** From to :
and from to :
Equal, so the answer is right. Notice that the terms cancelled, leaving a linear equation — that always happens in this family, and if you find yourself with a quadratic then a squaring step went wrong.
**Worked example 2 — on the -axis.** Find a point on the -axis that is equidistant from and .
Let the point be . Then gives
**The point is .
Check**: from to is , and to is . Equal.
Worked example 3 — a general equidistant condition. Find the relation between and if the point is equidistant from and .
The answer is an equation, not a point — and that equation is the perpendicular bisector of the segment joining the two given points, since every point equidistant from two fixed points lies on it.
Check with the mid-point. The mid-point of and is , and substituting gives . The mid-point lies on the line, which it must, being equidistant from both ends.
Worked example 4 — an unknown coordinate from a condition. If the point is equidistant from and , find .
**So the point is , directly above the mid-point of the two given points — exactly where the perpendicular bisector runs.
Check**: from the origin, and from . Equal.
The single technique behind all four examples. Square the distances before equating them. The square roots disappear, the and terms cancel, and what remains is linear. Trying to work with the roots in place turns a three-line problem into an unmanageable one.
Exam tip
Which habits keep a coordinate answer tidy?
Write the coordinates down with labels before calculating, and keep every length as a surd until the very end. Decimals are where certainty is lost in this chapter.
- **Label the points , , , in the order the question gives them, and keep that order
- Work with squared distances for every comparison, and take roots only if a length is asked for
- Simplify surds**: , ,
- For collinearity, add the two shorter lengths and compare with the longest — and keep them in surd form so the addition is exact
- Classify a triangle on two counts: equal sides first, then the Pythagoras test, since it can be both
- Name the right-angle vertex — it is the one not on the longest side
- For a quadrilateral, compute four sides and both diagonals. Sides alone cannot separate a square from a rhombus, or a rectangle from a parallelogram
- **A point on the -axis is ** and a point on the -axis is
- Square before equating in an equidistant problem; the terms will cancel
- Check your answer by computing both distances and confirming they match
The misconception to name. Four equal sides do not make a square. A rhombus has four equal sides too, and only the diagonals tell them apart: equal diagonals mean a square, unequal ones mean a rhombus. The two worked examples above differ in exactly that one comparison, and a student who stops after the sides will call both of them squares.
A second trap. Rearranging the four points to make the answer come out. **If a question names the vertices in order, the sides are , , , **, and computing , , , instead mixes sides with diagonals and produces a meaningless set of comparisons. The order is part of the data.
- **Label the points , , , in the order the question gives them, and keep that order
- Work with squared distances for every comparison, and take roots only if a length is asked for
- Simplify surds**: , ,
- For collinearity, add the two shorter lengths and compare with the longest — and keep them in surd form so the addition is exact
- Classify a triangle on two counts: equal sides first, then the Pythagoras test, since it can be both
- Name the right-angle vertex — it is the one not on the longest side
- For a quadrilateral, compute four sides and both diagonals. Sides alone cannot separate a square from a rhombus, or a rectangle from a parallelogram
- **A point on the -axis is ** and a point on the -axis is
- Square before equating in an equidistant problem; the terms will cancel
- Check your answer by computing both distances and confirming they match
The misconception to name. Four equal sides do not make a square. A rhombus has four equal sides too, and only the diagonals tell them apart: equal diagonals mean a square, unequal ones mean a rhombus. The two worked examples above differ in exactly that one comparison, and a student who stops after the sides will call both of them squares.
A second trap. Rearranging the four points to make the answer come out. **If a question names the vertices in order, the sides are , , , **, and computing , , , instead mixes sides with diagonals and produces a meaningless set of comparisons. The order is part of the data.
Did you know
Why is the distance formula just the Pythagoras theorem in disguise?
Take any two points and join them. Then draw a horizontal line from one and a vertical line from the other. They always meet at a right angle, and you have built a right triangle whose hypotenuse is the distance you want.
- The horizontal leg measures the difference of the -coordinates
- The vertical leg measures the difference of the -coordinates
- The hypotenuse is the distance
So , and that is the formula. There is genuinely nothing else in it — no new theorem, just the Pythagoras theorem with the legs read off the coordinate axes.
Which explains why the axes have to be perpendicular. The whole formula depends on that right angle existing, and if the two axes were drawn at to each other the formula would be wrong. The rectangular coordinate system is designed to make the Pythagoras theorem available, and that is the reason for its shape.
It also explains a pattern you may have noticed in the examples. The neat answers — from , in the earlier pole problem, appearing repeatedly — all come from Pythagorean triples, sets of three whole numbers satisfying :
-
-
-
-
Question setters choose coordinates whose differences form such a triple, so that the square root comes out whole. Recognising a triple mid-calculation is a real time-saver: seeing and writing without hesitating is faster than computing , and seeing and writing is faster still.
And it explains why the formula generalises so easily. In three dimensions the distance becomes
by applying the Pythagoras theorem twice — once in the horizontal plane and once vertically. The pattern of "square the differences, add, take the root" keeps working however many coordinates there are, which is why this one formula survives into every later branch of mathematics you will meet.
One boundary case worth stating. The formula gives a distance, which is never negative, and that is why the squares are there. It tells you how far apart two points are and nothing about direction — which is precisely the information the next part of the chapter adds, when the section formula asks not how far apart two points are but what lies between them.
- The horizontal leg measures the difference of the -coordinates
- The vertical leg measures the difference of the -coordinates
- The hypotenuse is the distance
So , and that is the formula. There is genuinely nothing else in it — no new theorem, just the Pythagoras theorem with the legs read off the coordinate axes.
Which explains why the axes have to be perpendicular. The whole formula depends on that right angle existing, and if the two axes were drawn at to each other the formula would be wrong. The rectangular coordinate system is designed to make the Pythagoras theorem available, and that is the reason for its shape.
It also explains a pattern you may have noticed in the examples. The neat answers — from , in the earlier pole problem, appearing repeatedly — all come from Pythagorean triples, sets of three whole numbers satisfying :
-
-
-
-
Question setters choose coordinates whose differences form such a triple, so that the square root comes out whole. Recognising a triple mid-calculation is a real time-saver: seeing and writing without hesitating is faster than computing , and seeing and writing is faster still.
And it explains why the formula generalises so easily. In three dimensions the distance becomes
by applying the Pythagoras theorem twice — once in the horizontal plane and once vertically. The pattern of "square the differences, add, take the root" keeps working however many coordinates there are, which is why this one formula survives into every later branch of mathematics you will meet.
One boundary case worth stating. The formula gives a distance, which is never negative, and that is why the squares are there. It tells you how far apart two points are and nothing about direction — which is precisely the information the next part of the chapter adds, when the section formula asks not how far apart two points are but what lies between them.
Exam relevance
How is the distance formula used in JEE?
This is foundation work for Class 11 Straight Lines and Conic Sections, and it is used in almost every coordinate-geometry question in JEE Main and JEE Advanced.
Where the formula leads. Class 11 defines a circle as the set of points at a fixed distance from a centre, which turns the distance formula directly into . The parabola, ellipse and hyperbola are all defined by conditions on distances — equal distances to a point and a line, a constant sum of distances, a constant difference — so every conic-section derivation begins with the formula you learn here.
Where the collinearity test leads. Class 11 replaces the length comparison with a much faster condition: three points are collinear when the area of the triangle they form is zero, which is a determinant. JEE Main uses the determinant form, but the geometric meaning is the one you meet here, and the determinant is preferred precisely because it avoids the surds that made the second worked example messy.
Where the equidistant condition leads. "Equidistant from two points" becomes the perpendicular bisector, and Class 11 finds it directly from the mid-point and the slope. JEE Advanced uses the same reasoning for the locus questions — describe a condition on distances, then find the equation it forces — and worked example 3 above is a locus problem in Class 10 clothing.
Where the classification questions lead. Class 11 uses slopes instead of lengths to detect right angles and parallel sides, which is usually quicker. But the length method never fails and needs no new formula, so it remains the reliable fallback when slopes are undefined — a vertical side, for instance, has no finite slope, and that is exactly where a slope-based solution breaks and a distance-based one does not.
Where the three-dimensional extension leads. Class 11 and Class 12 use it for Three-Dimensional Geometry and Vectors, where the magnitude of a vector is the same square-root-of-a-sum-of-squares.
Question types to expect. At this level: find a distance, prove collinearity, classify a triangle or quadrilateral, and find an equidistant point. In competitive papers: loci defined by distance conditions, equations of circles and conics, and collinearity by determinant.
The single trap that costs marks. Comparing decimals instead of surds in a collinearity test. ** is exact**, while relies on rounding that could hide a genuine gap — and at JEE level the same carelessness appears as accepting a numerically close value as an exact one.
A second trap. Taking the square root too early. Every comparison in this chapter works just as well on squared distances, and in Class 11 the squared form is the one that appears in the equation of a circle. Rooting early creates surds you then have to carry through the algebra for no reason.
Board versus competitive emphasis. The CBSE paper marks the formula, each length, the comparison and the named conclusion; a competitive paper marks the equation or the single value. The transferable habit is squaring the distance condition and simplifying before interpreting it — because that is how every locus, circle and conic equation in the next two years is derived.
Where the formula leads. Class 11 defines a circle as the set of points at a fixed distance from a centre, which turns the distance formula directly into . The parabola, ellipse and hyperbola are all defined by conditions on distances — equal distances to a point and a line, a constant sum of distances, a constant difference — so every conic-section derivation begins with the formula you learn here.
Where the collinearity test leads. Class 11 replaces the length comparison with a much faster condition: three points are collinear when the area of the triangle they form is zero, which is a determinant. JEE Main uses the determinant form, but the geometric meaning is the one you meet here, and the determinant is preferred precisely because it avoids the surds that made the second worked example messy.
Where the equidistant condition leads. "Equidistant from two points" becomes the perpendicular bisector, and Class 11 finds it directly from the mid-point and the slope. JEE Advanced uses the same reasoning for the locus questions — describe a condition on distances, then find the equation it forces — and worked example 3 above is a locus problem in Class 10 clothing.
Where the classification questions lead. Class 11 uses slopes instead of lengths to detect right angles and parallel sides, which is usually quicker. But the length method never fails and needs no new formula, so it remains the reliable fallback when slopes are undefined — a vertical side, for instance, has no finite slope, and that is exactly where a slope-based solution breaks and a distance-based one does not.
Where the three-dimensional extension leads. Class 11 and Class 12 use it for Three-Dimensional Geometry and Vectors, where the magnitude of a vector is the same square-root-of-a-sum-of-squares.
Question types to expect. At this level: find a distance, prove collinearity, classify a triangle or quadrilateral, and find an equidistant point. In competitive papers: loci defined by distance conditions, equations of circles and conics, and collinearity by determinant.
The single trap that costs marks. Comparing decimals instead of surds in a collinearity test. ** is exact**, while relies on rounding that could hide a genuine gap — and at JEE level the same carelessness appears as accepting a numerically close value as an exact one.
A second trap. Taking the square root too early. Every comparison in this chapter works just as well on squared distances, and in Class 11 the squared form is the one that appears in the equation of a circle. Rooting early creates surds you then have to carry through the algebra for no reason.
Board versus competitive emphasis. The CBSE paper marks the formula, each length, the comparison and the named conclusion; a competitive paper marks the equation or the single value. The transferable habit is squaring the distance condition and simplifying before interpreting it — because that is how every locus, circle and conic equation in the next two years is derived.
Key takeaways
What must you be able to do from this part?
One formula and four comparisons.
- Distance formula: , and from the origin
- It is the Pythagoras theorem with the horizontal and vertical gaps as the legs
- The order of subtraction does not matter, since both differences are squared
- ** to is ; is from the origin
- Work with squared distances and take roots only at the end
- Collinear** when the two shorter lengths add to the longest: proves , , collinear
- **, , are not collinear**, since exceeds
- Triangle types: three equal sides is equilateral, two equal is isosceles, all different is scalene — and check the Pythagoras condition separately, since a triangle can be both
- **, , is isosceles** with two sides ; **, , is equilateral** with all sides
- **, , is scalene and right-angled**, since , with the right angle at
- The right angle sits at the vertex not on the longest side
- Quadrilaterals need four sides and both diagonals: equal sides with equal diagonals is a square, equal sides with unequal diagonals is a rhombus, opposite sides equal with equal diagonals is a rectangle, opposite sides equal with unequal diagonals is a parallelogram
- **, , , is a square** with sides and diagonals ; **, , , is a rhombus** with sides and diagonals and
- Take the vertices in the order given — , , , are the sides and , the diagonals
- **A point on the -axis is **, so the point equidistant from and is ; on the -axis, the point equidistant from and is
- An equidistant condition on a general point gives the perpendicular bisector, so and give
- An unknown coordinate at a given distance has two answers — at distance from gives or
The best self-test is a shape you invent. Write down four points you believe make a rhombus, compute all four sides and both diagonals, and see whether the diagonals come out unequal — if they match, you accidentally drew a square.
- Distance formula: , and from the origin
- It is the Pythagoras theorem with the horizontal and vertical gaps as the legs
- The order of subtraction does not matter, since both differences are squared
- ** to is ; is from the origin
- Work with squared distances and take roots only at the end
- Collinear** when the two shorter lengths add to the longest: proves , , collinear
- **, , are not collinear**, since exceeds
- Triangle types: three equal sides is equilateral, two equal is isosceles, all different is scalene — and check the Pythagoras condition separately, since a triangle can be both
- **, , is isosceles** with two sides ; **, , is equilateral** with all sides
- **, , is scalene and right-angled**, since , with the right angle at
- The right angle sits at the vertex not on the longest side
- Quadrilaterals need four sides and both diagonals: equal sides with equal diagonals is a square, equal sides with unequal diagonals is a rhombus, opposite sides equal with equal diagonals is a rectangle, opposite sides equal with unequal diagonals is a parallelogram
- **, , , is a square** with sides and diagonals ; **, , , is a rhombus** with sides and diagonals and
- Take the vertices in the order given — , , , are the sides and , the diagonals
- **A point on the -axis is **, so the point equidistant from and is ; on the -axis, the point equidistant from and is
- An equidistant condition on a general point gives the perpendicular bisector, so and give
- An unknown coordinate at a given distance has two answers — at distance from gives or
The best self-test is a shape you invent. Write down four points you believe make a rhombus, compute all four sides and both diagonals, and see whether the diagonals come out unequal — if they match, you accidentally drew a square.