Coordinate Geometry

A telephone company wants to place a relay tower between two towns so that it sits exactly twice as far from one town as from the other. A classroom teacher wants to know, without dragging out a ruler, whether four desks arranged on a graph-paper floor plan actually form a square. A surveyor wants to check whether three marked points on a map lie on a single straight road. Every one of these is a geometry question, yet none of them needs a single diagram drawn to scale — they can all be answered with nothing more than arithmetic on coordinates. That is the promise of coordinate geometry: translate a picture into a pair of numbers for each point, and geometric questions about distance, division and shape become algebraic questions about those numbers. This chapter builds exactly two tools for that translation, and then spends the rest of its time showing how far those two tools reach. The first is the distance formula, which measures the straight-line gap between any two points $(x_1, y_1)$ and $(x_2, y_2)$ by hiding a right triangle inside the picture and calling on the Pythagoras theorem you already trust. The second is the section formula, which locates the exact point that splits a segment in any chosen ratio — the relay-tower problem, solved in one line once the formula is in hand. A special case of the section formula, used constantly, is the midpoint formula, for the point exactly halfway between two others. With just these two results you will classify triangles as scalene, isosceles or right-angled without measuring a single angle; test whether four points form a square, rectangle, rhombus or parallelogram; decide whether three points are collinear; find a point equidistant from two others; locate the trisection points of a segment; and pin down an unknown coordinate from a geometric condition such as "these four points form a parallelogram." Every one of these lands you back at the same two formulas, applied with a little algebraic care — which is exactly why this short chapter carries so much weight in the exam and in every coordinate-based topic that follows it.

Key takeaways

Distance formula: for $P(x_1, y_1)$ and $Q(x_2, y_2)$ , $PQ = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$ — derived directly from the Pythagoras theorem.
Distance from the origin: the special case $P(x,y)$ from $O(0,0)$ gives $OP = \sqrt{x^2+y^2}$ .
The distance formula tests triangle type (scalene, isosceles, right-angled, using the converse of Pythagoras theorem) and collinearity (three points are collinear exactly when the sum of two of the distances equals the third).
Section formula: the point $P(x,y)$ dividing the segment joining $A(x_1,y_1)$ and $B(x_2,y_2)$ internally in the ratio $m_1:m_2$ is $\left(\dfrac{m_1x_2+m_2x_1}{m_1+m_2}, \dfrac{m_1y_2+m_2y_1}{m_1+m_2}\right)$ .
The same formula written with a single unknown ratio $k:1$ is $\left(\dfrac{kx_2+x_1}{k+1}, \dfrac{ky_2+y_1}{k+1}\right)$ — convenient whenever the ratio itself is what you need to find.
Midpoint formula: setting $k=1$ (ratio $1:1$ ) gives the midpoint $\left(\dfrac{x_1+x_2}{2}, \dfrac{y_1+y_2}{2}\right)$ , the special case used whenever a diagonal, median or perpendicular bisector is involved.
A quadrilateral's diagonals bisecting each other is the standard route to prove it is a parallelogram: equate the midpoint of one diagonal to the midpoint of the other.
To prove a quadrilateral is a square or rhombus, show all four sides equal using the distance formula, then use the diagonals (equal for a square, unequal for a rhombus) or the converse of Pythagoras theorem for the right angle.
Key idea: the distance formula (from Pythagoras theorem) measures how far apart two points are, and the section formula (from similar triangles) locates the point dividing a segment in any given ratio, and together these two coordinate tools classify triangles and quadrilaterals, test collinearity, and locate equidistant or dividing points without ever drawing an accurate diagram.

The core insight

Every distance question reduces to the distance formula $PQ = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$ built from the Pythagoras theorem, and every division-of-a-segment question reduces to the section formula $\left(\dfrac{m_1x_2+m_2x_1}{m_1+m_2}, \dfrac{m_1y_2+m_2y_1}{m_1+m_2}\right)$ built from similar triangles, with the midpoint formula as its $1:1$ special case; together they turn geometric shape and location questions into pure coordinate arithmetic.

What this chapter covers

The chapter opens with a quick recap of how a point's position is fixed by its abscissa (distance from the $y$ -axis) and ordinate (distance from the $x$ -axis), then poses the town-distance problem to motivate the distance formula, first for points on an axis, then for points in a single quadrant using a hidden right triangle, and finally in full generality for any two points $P(x_1,y_1)$ and $Q(x_2,y_2)$ . Five worked examples show the formula testing triangle type, testing whether points form a square, testing collinearity, and finding a point equidistant from two given points. The chapter then turns to the section formula, built the same way — first through the relay-tower motivating example, then in full generality using similar triangles — and its special case, the midpoint formula. Five more worked examples apply the section formula to find a dividing point, recover an unknown ratio, locate trisection points, find where an axis cuts a segment, and pin down a missing vertex of a parallelogram using the fact that its diagonals bisect each other. The chapter's own summary lists exactly four results — the distance formula, the origin-distance formula, the section formula, and the midpoint formula — and a closing note previews external division, kept for a later class.

The core ideas

Fixing a point: abscissa, ordinate and the four quadrants

A point's position in the plane is completely fixed by an ordered pair $(x, y)$ : the abscissa $x$ is its distance from the $y$ -axis (with sign), and the ordinate $y$ is its distance from the $x$ -axis (with sign). A point on the $x$ -axis always has the form $(x, 0)$ , and a point on the $y$ -axis always has the form $(0, y)$ ; the origin is $(0,0)$ . Everything in this chapter is built by comparing the coordinates of two or more such points, so getting the sign and order of $x_1, y_1, x_2, y_2$ right at the start of a problem is worth the extra ten seconds it takes.

The distance formula

Consider two points on the $x$ -axis, say $A(4,0)$ and $B(6,0)$ : the distance between them is simply $AB = 6 - 4 = 2$ units, read straight off the axis. The same works on the $y$ -axis. The real question is what happens once the two points leave the axes and sit anywhere in the plane — and the answer is to build a right triangle out of the segment $PQ$ and apply the Pythagoras theorem. For $P(x_1,y_1)$ and $Q(x_2,y_2)$ ,

PQ = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}. \qquad (1)

where: $(x_1,y_1)$ and $(x_2,y_2)$ are the coordinates of the two points, and the positive square root is always taken, since a distance can never be negative. Because $(x_2-x_1)^2 = (x_1-x_2)^2$ , the formula gives the same value regardless of which point is called $P$ and which is called $Q$ — order does not matter. The special case of a point $P(x,y)$ measured from the origin $O(0,0)$ drops two of the four terms and leaves $OP = \sqrt{x^2+y^2}$ , a shortcut worth remembering on its own.

Figure

A right triangle formed by two points P and Q in the coordinate plane with a third point T making the right angle, showing the horizontal leg as the difference of the x coordinates and the vertical leg as the difference of the y coordinates with the hypotenuse PQ

The formula's most common use is not raw distance for its own sake but as a shape test. Compute all the side lengths (and, where useful, the diagonals) of a set of points, then read off the shape from the pattern of equal and unequal lengths: all four sides equal with equal diagonals signals a square; all four sides equal with unequal diagonals signals a rhombus; and the converse of the Pythagoras theorem — if the sum of the squares of two sides equals the square of the third, the angle between those two sides is $90^\circ$ — turns a plain triangle test into a right-triangle test. A third, quieter use is testing collinearity: three points $P, Q, R$ lie on a single straight line exactly when the sum of the two shorter distances among them equals the longest distance, for example $PQ + QR = PR$ .

The section formula and its midpoint special case

Now flip the question: instead of measuring the gap between two fixed points, find the exact point that splits that gap in a chosen ratio. If $P(x,y)$ divides the segment joining $A(x_1,y_1)$ and $B(x_2,y_2)$ internally so that $PA:PB = m_1:m_2$ , then

P(x,y) = \left(\dfrac{m_1x_2+m_2x_1}{m_1+m_2},\ \dfrac{m_1y_2+m_2y_1}{m_1+m_2}\right). \qquad (2)

where: $m_1$ is the part of the ratio nearer $B$ 's coordinates in the numerator pattern and $m_2$ the part nearer $A$ 's — notice each coordinate of $A$ is paired with $m_2$ and each coordinate of $B$ with $m_1$ , the opposite way round from what many students first guess. Written with a single unknown $k$ in place of the ratio $m_1:m_2 = k:1$ , the same point is $\left(\dfrac{kx_2+x_1}{k+1}, \dfrac{ky_2+y_1}{k+1}\right)$ , which is the form to reach for whenever the ratio itself, not the point, is what a problem asks you to find. Setting the ratio to $1:1$ — the point exactly halfway along — collapses the formula to the midpoint formula,

\text{midpoint of } AB = \left(\dfrac{x_1+x_2}{2},\ \dfrac{y_1+y_2}{2}\right). \qquad (3)

where: $(x_1,y_1)$ and $(x_2,y_2)$ are again the two endpoints. The midpoint formula is used constantly on its own, most often through the fact that the diagonals of a parallelogram bisect each other: if $ABCD$ is a parallelogram, the midpoint of diagonal $AC$ equals the midpoint of diagonal $BD$ , an equation that pins down any single unknown coordinate among the four vertices.

Figure

A line segment AB with a point P dividing it internally in the ratio m one to m two, with perpendiculars dropped to the x axis creating two similar right triangles that give the section formula

Two worked derivations

Derivation ADeriving the distance formula from the Pythagoras theorem

The goal is to build the distance formula $PQ = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$ from scratch for two completely general points $P(x_1,y_1)$ and $Q(x_2,y_2)$ , using nothing beyond the Pythagoras theorem, by manufacturing a right triangle whose hypotenuse is exactly the segment $PQ$ we want to measure, and whose two legs are horizontal and vertical differences that can be read straight off the coordinates.

Step 1. Drop a perpendicular from $P$ to the $x$ -axis, meeting it at $R(x_1, 0)$ , and a perpendicular from $Q$ to the $x$ -axis, meeting it at $S(x_2, 0)$ . Then draw a line from $P$ parallel to the $x$ -axis, meeting the vertical line through $Q$ at a point $T$ . Because $PT$ is horizontal and $QT$ is vertical, angle $PTQ$ is exactly $90^\circ$ , so $PTQ$ is a right triangle with hypotenuse $PQ$ — the segment we are trying to find.

Step 2. Read off the two legs from the coordinates. Since $T$ lies directly below $Q$ and on the same horizontal level as $P$ , its coordinates are $T(x_2, y_1)$ . The horizontal leg is $PT = RS = x_2 - x_1$ (the gap between the two points' abscissas), and the vertical leg is $QT = y_2 - y_1$ (the gap between their ordinates). No measurement is needed for either leg — both come directly from subtracting coordinates.

Step 3. Apply the Pythagoras theorem to the right triangle $PTQ$ : $PQ^2 = PT^2 + QT^2 = (x_2-x_1)^2 + (y_2-y_1)^2$ . Squaring the differences automatically removes any concern about sign, since a negative difference squared is still positive — the formula works identically whichever point is labelled $P$ and whichever is labelled $Q$ .

Step 4. Take the positive square root, since distance is never negative: $PQ = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$ , which is equation $(1)$ . Setting $P$ to the origin $(0,0)$ and $Q$ to a general point $(x,y)$ collapses this to $OP = \sqrt{x^2+y^2}$ , the distance-from-origin shortcut used whenever one of the two points is the origin itself.

Go deeper · Derivation B · 1 minDeriving the section formula from similar triangles

Derivation BDeriving the section formula from similar triangles

The goal is to find the coordinates of the point $P(x,y)$ that divides the segment joining $A(x_1,y_1)$ and $B(x_2,y_2)$ internally in a given ratio $m_1:m_2$ — meaning $PA:PB = m_1:m_2$ — using the AA similarity criterion on two right triangles constructed from perpendiculars to the $x$ -axis, the same construction NCERT uses to justify equation $(2)$ .

Step 1. Drop perpendiculars from $A$ , $P$ and $B$ to the $x$ -axis, meeting it at $R$ , $S$ and $T$ respectively. Draw $AQ$ and $PC$ parallel to the $x$ -axis, meeting the perpendiculars through $P$ and $B$ at $Q$ and $C$ . This produces two right triangles, $PAQ$ and $BPC$ , stacked one after the other along the line $AB$ .

Step 2. Show the two triangles are similar. Angle $AQP$ and angle $PCB$ are both $90^\circ$ by construction (they are the angles between a horizontal and a vertical segment). Angle $PAQ$ and angle $BPC$ are equal because $AB$ is a single straight line cutting the two parallel horizontal segments $AQ$ and $PC$ , making these corresponding angles. Two equal angles are enough: by the AA similarity criterion, $\triangle PAQ \sim \triangle BPC$ .

Step 3. Similar triangles have proportional corresponding sides, so $\dfrac{PA}{BP} = \dfrac{AQ}{PC} = \dfrac{PQ}{BC}$ . Since $PA:PB = m_1:m_2$ by assumption, this whole chain of ratios equals $m_1/m_2$ . Now express each length in coordinates: $AQ = x - x_1$ (horizontal gap from $A$ to $P$ ), $PC = x_2 - x$ (horizontal gap from $P$ to $B$ ), $PQ = y - y_1$ , and $BC = y_2 - y$ .

Step 4. Solve the two proportions separately. From $\dfrac{m_1}{m_2} = \dfrac{x-x_1}{x_2-x}$ , cross-multiplying gives $m_1(x_2-x) = m_2(x-x_1)$ , so $m_1x_2 - m_1x = m_2x - m_2x_1$ , and collecting the $x$ terms gives $x = \dfrac{m_1x_2+m_2x_1}{m_1+m_2}$ . The identical algebra on $\dfrac{m_1}{m_2} = \dfrac{y-y_1}{y_2-y}$ gives $y = \dfrac{m_1y_2+m_2y_1}{m_1+m_2}$ . Together these are exactly equation $(2)$ , the section formula, derived without assuming any particular numbers for $m_1$ , $m_2$ or the coordinates.

Go deeper · Derivation C · 1 minFrom the ratio form to the k to 1 form and the midpoint formula

Derivation CFrom the ratio form to the k to 1 form and the midpoint formula

The goal is to show that the two other forms used constantly in this chapter — the single-unknown $k:1$ form of the section formula, and the midpoint formula — are not separate results to memorise but direct algebraic consequences of the general section formula in equation $(2)$ , so that only one formula ever needs to be trusted.

Step 1. Start from the general section formula with ratio $m_1:m_2$ . Since only the ratio between $m_1$ and $m_2$ matters (doubling both leaves $P$ unchanged), divide both terms of the ratio by $m_2$ and rename $k = m_1/m_2$ , so the ratio becomes $k:1$ .

Step 2. Substitute $m_1 = k$ and $m_2 = 1$ directly into equation $(2)$ : $x = \dfrac{kx_2 + 1\cdot x_1}{k+1}$ and $y = \dfrac{ky_2+1\cdot y_1}{k+1}$ , which is exactly the $k:1$ form quoted earlier. This form is preferred whenever a problem gives you the point and asks for the ratio (or asks which axis divides a segment and in what ratio), because there is only one unknown, $k$ , to solve for instead of two.

Step 3. Now specialise further to the case $PA = PB$ , the midpoint. This means $m_1 = m_2$ , so $k = m_1/m_2 = 1$ . Substituting $k=1$ into the $k:1$ form gives $x = \dfrac{1\cdot x_2+x_1}{1+1} = \dfrac{x_1+x_2}{2}$ and $y = \dfrac{y_1+y_2}{2}$ , exactly the midpoint formula of equation $(3)$ .

Step 4. Read the logical chain both ways. Going down, the midpoint formula is simply "ratio $1:1$ " plugged into the section formula — nothing new is being assumed. Going up, if a problem states two coordinates are midpoints of each other's segment, that single fact is enough to recover the full section-formula machinery if a different, non-equal ratio is later needed for the same pair of points. This is exactly why the diagonals-bisect-each-other property of a parallelogram (an equal-ratio, midpoint statement) and the trisection-point problems (unequal-ratio, general section-formula statements) are treated as two faces of the very same tool.

Worked examples

Board level - classifying a triangle by its side lengths (Example 1 pattern)

Do the points $P(3,2)$ , $Q(-2,-3)$ and $R(2,3)$ form a triangle? If so, name the type of triangle formed.

Step 1. Apply the distance formula to all three pairs. $PQ = \sqrt{(3-(-2))^2+(2-(-3))^2} = \sqrt{5^2+5^2} = \sqrt{50} \approx 7.07$ .

Step 2. $QR = \sqrt{(-2-2)^2+(-3-3)^2} = \sqrt{(-4)^2+(-6)^2} = \sqrt{16+36} = \sqrt{52} \approx 7.21$ , and $PR = \sqrt{(3-2)^2+(2-3)^2} = \sqrt{1+1} = \sqrt{2} \approx 1.41$ .

Step 3. Check the triangle inequality: the sum of any two of these distances must exceed the third. Since $PR + PQ = 1.41 + 7.07 = 8.48 > QR$ , and similarly for the other two pairs, the three points genuinely form a triangle rather than lying on one line.

Step 4. Test for a right angle using the converse of the Pythagoras theorem: $PQ^2 + PR^2 = 50 + 2 = 52 = QR^2$ . Since the square of one side equals the sum of the squares of the other two, the angle opposite the longest side $QR$ — which is angle $P$ — is $90^\circ$ . Answer: PQR is a right-angled triangle, right-angled at P.

Board level - the ratio in which the y-axis divides a segment (Example 9 pattern)

Find the ratio in which the $y$ -axis divides the line segment joining $A(5,-6)$ and $B(-1,-4)$ . Also find the point of intersection.

Step 1. Since the ratio itself is unknown, use the $k:1$ form of the section formula. Let the $y$ -axis divide $AB$ in the ratio $k:1$ at the point $\left(\dfrac{k(-1)+5}{k+1}, \dfrac{k(-4)+(-6)}{k+1}\right)$ .

Step 2. Any point on the $y$ -axis has abscissa $0$ , so set the $x$ -coordinate to zero: $\dfrac{-k+5}{k+1} = 0$ , which forces $-k + 5 = 0$ , so $k = 5$ . The ratio is therefore $5:1$ .

Step 3. Substitute $k=5$ into the $y$ -coordinate: $y = \dfrac{5(-4)+(-6)}{5+1} = \dfrac{-20-6}{6} = \dfrac{-26}{6} = -\dfrac{13}{3}$ .

Step 4. Since $k = 5 > 0$ , the point genuinely lies between $A$ and $B$ (internal division), confirming the ratio is valid. Answer: the y-axis divides AB in the ratio 5 : 1, at the point (0, −13/3).

Board HOTS / foundation JEE-NEET - locating a fractional point and verifying it by distance (Exercise 7.2 pattern, extending Example 6)

If $A(-2,-2)$ and $B(2,-4)$ , find the coordinates of $P$ on segment $AB$ such that $AP = \dfrac{3}{7}AB$ . Verify your answer using the distance formula.

Step 1. Convert the fractional condition into a ratio. Since $AP = \dfrac{3}{7}AB$ , the remaining part is $PB = AB - AP = \dfrac{4}{7}AB$ , so $AP:PB = 3:4$ — that is, $m_1:m_2 = 3:4$ in equation $(2)$ , with $A$ playing the role of $(x_1,y_1)$ and $B$ the role of $(x_2,y_2)$ .

Step 2. Apply the section formula: $x = \dfrac{3(2)+4(-2)}{3+4} = \dfrac{6-8}{7} = -\dfrac{2}{7}$ , and $y = \dfrac{3(-4)+4(-2)}{3+4} = \dfrac{-12-8}{7} = -\dfrac{20}{7}$ . So $P = \left(-\dfrac{2}{7}, -\dfrac{20}{7}\right)$ .

Step 3. Verify by computing $AB$ and $AP$ directly with the distance formula and checking their ratio. $AB = \sqrt{(2-(-2))^2+(-4-(-2))^2} = \sqrt{4^2+(-2)^2} = \sqrt{16+4} = \sqrt{20} = 2\sqrt5$ .

Step 4. $AP = \sqrt{\left(-\dfrac{2}{7}-(-2)\right)^2 + \left(-\dfrac{20}{7}-(-2)\right)^2} = \sqrt{\left(\dfrac{12}{7}\right)^2+\left(-\dfrac{6}{7}\right)^2} = \sqrt{\dfrac{144+36}{49}} = \sqrt{\dfrac{180}{49}} = \dfrac{6\sqrt5}{7}$ . Compare with $\dfrac{3}{7}AB = \dfrac{3}{7}\times 2\sqrt5 = \dfrac{6\sqrt5}{7}$ — the two values match exactly. Answer: P = (−2/7, −20/7), and AP = (6√5)/7 correctly equals (3/7)AB. Timing note: budget about three minutes. The trap is stopping after Step 2 without noticing that the verification step is where marks are actually won in a HOTS-style question — examiners reward the explicit closing comparison $AP = \frac{3}{7}AB$ over simply quoting coordinates. The efficient route is to convert every fractional or "in what ratio" condition into a clean $m_1:m_2$ pair before touching the section formula, and to keep fractions unsimplified until the very last line so that the verification arithmetic stays exact rather than approximate.

Common traps

The single most frequent slip is mixing up which point's coordinates pair with which part of the ratio in the section formula: $A(x_1,y_1)$ 's coordinates are multiplied by $m_2$ , not $m_1$ , and $B(x_2,y_2)$ 's coordinates are multiplied by $m_1$ — reversing this silently swaps the answer. A second common error is testing only that four side lengths are equal and concluding "square," when equal sides alone only guarantee a rhombus — the diagonals must also be checked equal (or one angle shown to be $90^\circ$ via the converse of Pythagoras theorem) before "square" is justified. A third slip appears in collinearity questions: students compute all three pairwise distances correctly but forget to check which sum should equal the longest distance, sometimes testing $PQ + PR = QR$ when $QR$ is not actually the largest — always identify the longest distance first. Finally, in "in what ratio" problems, forgetting that $k$ coming out negative means the division is external, not internal, leads to accepting a meaningless answer without a second glance — a positive $k$ confirms the point lies between $A$ and $B$ .

Common trapEqual sides alone never prove a square is a square — always check the diagonals or an angle too

The biggest misconception in this chapter is stopping the moment all four sides of a quadrilateral come out equal using the distance formula and declaring victory for "square." Four equal sides by themselves only establish that the quadrilateral could be a rhombus — a shape that is deliberately not a rectangle and has no right angles. To upgrade a rhombus to a square, additionally show the two diagonals are equal in length (a square's diagonals are equal; a general rhombus's are not) or use the converse of the Pythagoras theorem to confirm one interior angle is exactly $90^\circ$ . Whenever a question says "show ABCD is a square/rhombus/rectangle," plan on computing sides and diagonals before writing a final answer.

You should be able to…

State the distance formula and the section formula precisely, including which coordinates pair with which part of the ratio.
Derive the distance formula from the Pythagoras theorem and the section formula from similar triangles, without needing to memorise either as a bare formula.
Use the distance formula to classify a triangle as scalene, isosceles or right-angled, applying the converse of the Pythagoras theorem where needed.
Use the distance formula to test whether three points are collinear, and to test whether four points form a square, rectangle, rhombus or parallelogram.
Apply the section formula, in both the $m_1:m_2$ form and the single-unknown $k:1$ form, to find a dividing point or an unknown ratio.
Recognise the midpoint formula as the special case of the section formula at ratio $1:1$ , and use it via "diagonals of a parallelogram bisect each other" to find an unknown vertex.
Find the trisection points of a segment by applying the section formula twice, at ratios $1:2$ and $2:1$ .
Find a point equidistant from two given points, either on a stated axis or through a general relation between $x$ and $y$ .
Recognise when a computed ratio is negative and understand this signals external division, outside this chapter's scope.

Ask the tutor: Want to work through finding the fourth vertex of a parallelogram, or proving a trickier quadrilateral is a rhombus, step by step together?

Why do we need both the distance formula and the section formula, since they both use the same two coordinates?

They answer two genuinely different questions. The distance formula tells you how far apart two fixed points are — a single number, useful for measuring sides, testing shapes and checking collinearity. The section formula tells you where a new point sits between two given points for a chosen ratio — it produces a brand-new pair of coordinates, not a number. You reach for the distance formula whenever a question is about length, equality of sides, or "are these points on one line," and for the section formula whenever a question is about locating a dividing point, a midpoint, a trisection point, or recovering an unknown ratio. Many harder problems, like the third worked example above, use both formulas one after another: the section formula to find a new point, then the distance formula to check or use that point.

In the section formula, why does A's coordinates get multiplied by m2 and not m1 — it feels backwards?

It only feels backwards because of which label sits closer to which point, but the derivation settles it: in the similar triangles used to prove the formula, the gap nearer $A$ (namely $AQ = x - x_1$ ) is proportional to $m_1$ , while the gap nearer $B$ (namely $PC = x_2-x$ ) is proportional to $m_2$ — cross-multiplying and solving for $x$ then places $m_1$ next to $x_2$ and $m_2$ next to $x_1$ . A quick memory check: whichever number is written second in the ratio ends up multiplying the coordinates written first. If it still feels shaky, test with ratio $1:1$ — both forms must collapse to the same symmetric midpoint formula, which they do.

When should I use the m1 to m2 form of the section formula and when should I use the k to 1 form?

Use whichever form has fewer unknowns for what the question is actually asking. If the ratio is already given as a pair of numbers, like $3:4$ or $2:1$ , plug straight into the $m_1:m_2$ form — there is nothing to solve for beyond $x$ and $y$ . If instead the ratio itself is unknown — for instance, "in what ratio does the $y$ -axis divide this segment" or "in what ratio does point $(-4,6)$ divide $AB$ " — switch to the $k:1$ form, because it has only one unknown, $k$ , to solve for from a single equation (usually by setting a known coordinate, like $x=0$ on the $y$ -axis, equal to the formula). Trying to solve for both $m_1$ and $m_2$ separately when only their ratio matters is unnecessary extra work.

Does the section formula still work if the ratio comes out as a negative number or a fraction like 3 is to 4?

A fractional ratio like $3:4$ is completely normal — it simply means the dividing point is closer to $A$ than to $B$ , since $3 < 4$ . A negative value of $k$ , however, is a signal, not an error: it means the point does not lie between $A$ and $B$ at all, but on the extension of the line beyond one endpoint — what this chapter calls external division, deliberately left for a later class. If your Class 10 calculation produces a negative $k$ , re-read the question rather than reporting it as a final internal-division answer.