Introduction to three-dimensional Geometry | ISC Class 11 Maths Notes

This note covers coordinate axes and planes in space, ordered coordinates, octants, points on axes and planes, perpendicular projections, distance between two points, internal and external section formulae, midpoints, and applications to collinearity, triangles and parallelograms.

How does coordinate geometry extend from a plane to space?

Coordinate geometry connects the position of a point with numbers measured relative to fixed reference lines. In a plane, two mutually perpendicular lines provide these references. Mutually perpendicular means meeting at right angles. A point in space requires a third coordinate to describe its position.

Consider the lowest tip of a bulb suspended from a room's ceiling. Its perpendicular distances from two adjacent walls locate it horizontally, but its height above the floor is also needed. The floor and those walls provide three mutually perpendicular reference planes.

What are the axes and the origin?

A plane is a flat surface extending indefinitely. Take three mutually perpendicular planes meeting at a point called O, the origin. Their lines of intersection form the three coordinate axes, named the x-axis, y-axis and z-axis. An axis is a reference line along which coordinates are measured.

The labels X, Y and Z indicate the positive directions of these axes from O. The labels X′, Y′ and Z′ indicate their opposite, negative directions. Thus X′OX, Y′OY and Z′OZ denote the complete axes, extending on both sides of the origin.

Definition: A rectangular Cartesian coordinate system in space consists of three mutually perpendicular coordinate axes with a common origin. Positions are specified relative to these fixed axes and the planes determined by their pairs.

Which planes do pairs of axes determine?

The names ZX-plane and XZ-plane refer to the same plane. Each name identifies its two coordinate axes. The third axis supplies the direction perpendicular to it. For example, motion parallel to the z-axis changes height relative to the XY-plane.

Three outlined planes meet at O. The diagram labels the positive directions X, Y and Z and their opposite directions X′, Y′ and Z′. The z-axis is drawn vertically through their intersection.

See Fig. 11.1 in your NCERT textbook

A drawing on paper represents a spatial arrangement. The coordinate axes are mutually perpendicular in space even though their drawn images do not all appear at right angles. Use the stated geometry of the system when interpreting the sketch.

How do three coordinates locate a point uniquely?

Write a point as P(x, y, z), where P names the point and x, y and z are its coordinates along the corresponding axes. The bracketed list is an ordered triplet: three numbers written in a fixed order. Its entries are real numbers, including positive numbers, negative numbers and zero.

The x-coordinate records signed displacement in the x-direction, the y-coordinate in the y-direction and the z-coordinate in the z-direction. A signed displacement includes both its size and its positive or negative direction. Interchanging coordinate positions generally changes the point.

How is the position constructed?

For a point P, let M be the foot of the perpendicular from P to the XY-plane. A foot of a perpendicular is the point where that perpendicular meets its target line or plane. Let L be the foot of the perpendicular from M to the x-axis.

1. Start at the origin O and locate L on the x-axis using the signed coordinate x.
2. From L, move parallel to the y-axis through signed displacement y to reach M in the XY-plane.
3. From M, move perpendicular to the XY-plane, parallel to the z-axis, through signed displacement z.
4. The endpoint is P(x, y, z); the intermediate point M has coordinates (x, y, 0).

The notation OL, LM and MP names the segments joining the indicated points. In the positive-octant construction, their lengths are x, y and z respectively. Negative coordinates require movement in the corresponding negative directions rather than negative physical lengths.

The figure marks L on the x-axis, M(x, y, 0) in the XY-plane, and P(x, y, z) above M. The horizontal construction from L to M is labelled y, and the vertical segment from M to P is labelled z.

See Fig. 11.2 in your NCERT textbook

What makes the description unique?

With the coordinate system fixed, each point has one ordered triplet, and each ordered triplet specifies one point. This is a one-to-one correspondence. The origin has coordinates (0, 0, 0), because no displacement along any axis is needed to reach it.

For example, to locate (2, 3, 4), move 2 units along the positive x-axis, then 3 units parallel to the positive y-axis, then 4 units parallel to the positive z-axis. The intermediate positions are (2, 0, 0) and (2, 3, 0).

The coordinate order also identifies reference planes: x corresponds to the YZ-plane, y to the ZX-plane, and z to the XY-plane. The magnitude of each coordinate gives the perpendicular distance from the corresponding plane; its sign identifies the side.

How do coordinate signs identify the eight octants?

The three coordinate planes divide space into eight regions called octants. Each coordinate plane separates positive and negative values of one coordinate. Combining the three signs identifies the region containing a point whose coordinates are all non-zero.

The numerals I to VIII below mean the first to eighth octants. In the table, + means a positive coordinate and − means a negative coordinate. Read the signs in the order x, then y, then z.

What is the octant sign table?

The first four octants have positive z-coordinates. The next four repeat the same x and y sign combinations with negative z-coordinates. Thus the fifth octant has positive x and y but negative z, while the seventh has all three coordinates negative.

Worked example 1. Identify the octants containing (−3, 1, 2) and (−3, 1, −2).

Answer: The point (−3, 1, 2) has signs (−, +, +), so it lies in octant II. The point (−3, 1, −2) has signs (−, +, −), so it lies in octant VI. Their x and y signs agree, but their z signs differ.

What happens when a coordinate is zero?

A zero coordinate places a point on a coordinate plane, which is a boundary between octants. Do not assign such a point to an open octant by treating zero as positive. Identify its plane or axis directly from the zero entries.

For (−3, 1, 2), the magnitude 3 does not determine the octant; the negative sign of its first coordinate does. Likewise, replacing the last coordinate by −2 changes the side of the XY-plane. Sign analysis and distance calculation answer different questions about a point.

To classify a point, first check for zeros, then write the three signs, then compare them with the complete table. This keeps the coordinate order visible and avoids confusing the second and sixth octants, which differ only in the sign of z.

How are points on axes, planes and their projections recognised?

A point on a coordinate plane has zero displacement perpendicular to that plane. A point on an axis has zero displacement in both other coordinate directions. These conditions describe all points on the named plane or axis, including the origin where appropriate.

Property: Zero-coordinate conditions

A perpendicular projection is the foot obtained by dropping a perpendicular to a line or plane. Projection onto a coordinate plane retains the two coordinates along that plane and sets the remaining coordinate to zero. Projection onto an axis retains just that axis's coordinate.

Worked example 2. For P(3, 4, 5), find the perpendicular projections onto the three coordinate axes and the three coordinate planes.

Answer: The projections onto the x-, y- and z-axes are (3, 0, 0), (0, 4, 0) and (0, 0, 5). The projections onto the XY-, YZ- and ZX-planes are (3, 4, 0), (0, 4, 5) and (3, 0, 5), respectively.

How are distances from planes interpreted?

For P(3, 4, 5), the distances from the XY-, YZ- and ZX-planes are 5, 3 and 4 units respectively. For general P(x, y, z), they are |z|, |x| and |y|. Here vertical bars denote absolute value, the non-negative magnitude of a real number.

Distinguish the projection, which is a point with coordinates, from the perpendicular distance, which is a length. The projection (3, 4, 0) and the distance 5 units describe different features of the same perpendicular from P to the XY-plane.

Note: The x-coordinate relates to perpendicular distance from the YZ-plane, not distance from the x-axis. An axis is a line, whereas a coordinate plane contains two axes.

In a projection onto the x-axis, both y and z disappear because the target line contains neither y-displacement nor z-displacement. In a projection onto the XY-plane, x and y remain because movement within that plane can occur in both directions.

How is the distance formula in space obtained and used?

Let P(x₁, y₁, z₁) and Q(x₂, y₂, z₂) be two points. The subscripts 1 and 2 distinguish the coordinates of the first and second points. The symbol PQ denotes their non-negative distance, and √ denotes the non-negative square root.

Theorem: Distance between two points

PQ = √[(x₂ − x₁)² + (y₂ − y₁)² + (z₂ − z₁)²]. A superscript ² means squaring, or multiplying a quantity by itself. Each difference compares matching coordinates: x with x, y with y, and z with z.

The result extends the plane distance formula by adding the squared difference of the third coordinates. If both points lie in the XY-plane, their z-coordinates are zero and this extra term vanishes. The spatial formula then gives the familiar distance within that plane.

Why does Pythagoras' theorem apply twice?

Pythagoras' theorem states that, in a right-angled triangle, the square of the hypotenuse equals the sum of the squares of the other two sides. The hypotenuse is the side opposite the right angle. A rectangular box provides two such triangles.

A rectangular box is drawn beside axes OX, OY and OZ. The points P and Q are joined across the box. Points A and N mark the auxiliary construction, with right-angle labels at A and N.

See Fig. 11.4 in your NCERT textbook

A rectangular parallelepiped is a box-shaped solid with rectangular faces. In this construction, A and N are auxiliary points chosen so that PA, AN and NQ are parallel to coordinate axes. The segment PQ is a diagonal joining opposite vertices.

1. In right-angled triangle PAQ, apply Pythagoras' theorem to obtain PQ² = PA² + AQ².
2. In right-angled triangle ANQ, apply it again to obtain AQ² = AN² + NQ².
3. Substitution gives PQ² = PA² + AN² + NQ², the sum of the three squared edge lengths.
4. These squared lengths equal (y₂ − y₁)², (x₂ − x₁)² and (z₂ − z₁)². Taking the non-negative square root gives the distance formula.

Worked example 3. Find the distance between P(1, −3, 4) and Q(−4, 1, 2).

Answer: The coordinate differences are −4 − 1 = −5, 1 − (−3) = 4, and 2 − 4 = −2. Therefore PQ = √[25 + 16 + 4] = √45 = 3√5 units.

Result: Distance from the origin

Since O has coordinates (0, 0, 0), the distance from O to P(x, y, z) is OP = √(x² + y² + z²). Coordinate signs disappear on squaring, but this does not erase their importance when locating the point.

Keep subtraction inside brackets when a coordinate is negative. Complete all three squares before adding, and take the square root last. Reversing the order of the two points changes the signs of the differences but leaves their squares and the distance unchanged.

How can distances establish collinearity and triangle properties?

Collinear points lie on the same straight line. For three distinct points, equality between the largest pairwise distance and the sum of the other two establishes collinearity. It also identifies which point lies between the other two.

How is collinearity tested?

Worked example 4. Show that P(−2, 3, 5), Q(1, 2, 3) and R(7, 0, −1) are collinear.

Answer: PQ² = 9 + 1 + 4 = 14, QR² = 36 + 4 + 16 = 56, and PR² = 81 + 9 + 36 = 126. Thus PQ = √14, QR = 2√14 and PR = 3√14. Since PQ + QR = PR, the points are collinear, with Q between P and R.

The symbols PQ, QR and PR denote the three pairwise distances. In this test, add the lengths, not their squares. The equality of lengths describes a straight journey from P through Q to R, with no change of direction.

Do not stop after calculating three distances. The final comparison is the reason the calculation establishes the geometric claim. Identify the longest distance and compare it with the sum of the other two, retaining exact square roots where possible.

How is a right-angled isosceles triangle recognised?

An isosceles triangle has two equal sides. A right-angled triangle has one right angle. To test the latter using distances, compare the largest squared side with the sum of the other two squared sides, using the converse of Pythagoras' theorem.

Worked example 5. Show that P(0, 7, 10), Q(−1, 6, 6) and R(−4, 9, 6) form a right-angled isosceles triangle.

Answer: PQ² = 1 + 1 + 16 = 18; QR² = 9 + 9 + 0 = 18; PR² = 16 + 4 + 16 = 36. Hence PQ = QR = 3√2 units and PR = 6 units. Since PQ² + QR² = PR², the right angle is at Q, where the equal sides meet.

This comparison uses squares because the required relation is Pythagorean. By contrast, the collinearity test uses the sum of actual lengths. Writing the intended relation before calculating helps distinguish the two methods and makes the conclusion precise.

For a right angle, name its vertex. The side opposite it is the hypotenuse: here PR is opposite Q. The equalities show both requested properties separately, so neither the right-angle conclusion nor the isosceles conclusion is left implicit.

How does the internal section formula divide a segment?

A line segment is the finite part of a line between two endpoints. Let P(x₁, y₁, z₁) and Q(x₂, y₂, z₂) be distinct endpoints. Suppose R is between them and PR : RQ = m : n, where m and n are positive numbers expressing the ratio of those lengths.

This is internal division. The colon means “in the ratio”; the first number belongs to PR and the second to RQ. The section formula finds the coordinates of R from the endpoints and the specified ratio.

Result: Internal section formula

R = ((mx₂ + nx₁)/(m + n), (my₂ + ny₁)/(m + n), (mz₂ + nz₁)/(m + n)). Here R on the left abbreviates the coordinate triplet of the dividing point. The slash means division, and each denominator divides its entire bracketed numerator.

Apply the same weights separately to x, y and z. Notice the opposite-endpoint weighting: m multiplies the coordinates of Q, while n multiplies those of P. Writing the ratio with named segments before substitution prevents accidentally reversing these weights.

How does a fraction of the journey give a ratio?

If a point is five-sixths of the way from the first endpoint to the second, five of the six equal parts have been traversed and one part remains. The internal ratio is therefore 5 : 1, rather than 5 : 6. The symbol × below means multiplication.

Worked example 6. Find P, five-sixths of the way from A(−2, 0, 6) to B(10, −6, −12).

Answer: AP : PB = 5 : 1. Thus the x-coordinate is [5 × 10 + 1 × (−2)]/6 = 8; the y-coordinate is [5 × (−6) + 1 × 0]/6 = −5; the z-coordinate is [5 × (−12) + 1 × 6]/6 = −9. Therefore P = (8, −5, −9).

Here A and B name the endpoints, while P names the dividing point. The change of point labels does not change the formula: the first endpoint receives weight 1 and the second receives weight 5.

For a check, each resulting coordinate lies between the corresponding endpoint coordinates. Also, a point five-sixths of the way from A is closer to B. These observations support the substitution but do not replace writing all three calculated coordinates.

How does external division differ from internal division?

In external division, the dividing point lies on the line through the endpoints but outside the segment between them. Let distinct endpoints be P(x₁, y₁, z₁) and Q(x₂, y₂, z₂), with external dividing point R satisfying PR : RQ = m : n.

The numbers m and n represent positive lengths in ratio form. They must be unequal for a finite external dividing point. The coordinate calculation uses differences of weighted coordinates and a difference in the denominator.

Result: External section formula

R = ((mx₂ − nx₁)/(m − n), (my₂ − ny₁)/(m − n), (mz₂ − nz₁)/(m − n)), with m ≠ n. The symbol ≠ means “is not equal to”. As with internal division, m weights the second endpoint and n weights the first.

Do not change the denominator alone. External division changes the corresponding addition in every numerator to subtraction as well. If m = n, the displayed denominator is zero, so this formula does not provide a finite point.

How can a coordinate plane determine the dividing ratio?

Worked example 7. Find the ratio in which the XZ-plane divides the line through P(2, 4, 5) and Q(3, 5, −4), and state whether the division is internal or external.

Answer: On the XZ-plane, y = 0. Using the external formula, 0 = (5m − 4n)/(m − n), so 5m = 4n and m : n = 4 : 5. Both endpoint y-coordinates are positive, so their intersection with y = 0 is outside the segment. The division is external in ratio 4 : 5.

Another method uses a signed ratio, allowing a negative value to signal external division. Write the internal-form expression with ratio k : 1, where k is the unknown signed ratio parameter. Its y-coordinate is (5k + 4)/(k + 1).

Setting that coordinate to zero gives k = −4/5. The signed result −4 : 5 represents external division in the positive length ratio 4 : 5. It does not mean that a geometric length is negative.

In a plane-intersection problem, begin with the coordinate forced to zero: y for the XZ-plane, x for the YZ-plane, or z for the XY-plane. Solve for the ratio and identify the type of division before calculating any further coordinates requested.

How do midpoints help solve problems about parallelograms?

The midpoint of a segment is the point halfway between its endpoints, dividing it internally in the ratio 1 : 1. Substituting equal weights into the internal section formula gives the arithmetic mean of each pair of corresponding coordinates.

Result: Midpoint formula

For endpoints P(x₁, y₁, z₁) and Q(x₂, y₂, z₂), let M denote their midpoint. Then M = ((x₁ + x₂)/2, (y₁ + y₂)/2, (z₁ + z₂)/2). An arithmetic mean of two numbers is their sum divided by two.

The midpoint formula uses sums of coordinates, while the distance formula uses squared differences. Both concern the same endpoints but answer different questions. A midpoint answer is an ordered triplet; a distance answer is a non-negative length.

How are diagonal midpoints compared?

A parallelogram is a quadrilateral with both pairs of opposite sides parallel. A quadrilateral has four sides, and a diagonal joins opposite vertices. The diagonals of a parallelogram bisect each other: each passes through the midpoint of the other.

Worked example 8. Show that A(3, −1, −1), B(5, −4, 0), C(2, 3, −2) and D(0, 6, −3), taken in that order, are vertices of a parallelogram.

Answer: The midpoint of AC is ((3 + 2)/2, (−1 + 3)/2, (−1 − 2)/2) = (5/2, 1, −3/2). The midpoint of BD is ((5 + 0)/2, (−4 + 6)/2, (0 − 3)/2) = (5/2, 1, −3/2). The diagonals bisect each other, so ABCD is a parallelogram.

Pair opposite vertices according to their stated order. For quadrilateral ABCD, the diagonals are AC and BD; AB and CD are opposite sides. Comparing the wrong pair of midpoints would answer a different geometric question.

If a fourth vertex is unknown, give its coordinates separate unknown symbols, equate the two diagonal midpoint triplets, and solve the three coordinate equations. Equality of ordered triplets means equality of the first entries, the second entries and the third entries separately.

A common midpoint establishes the parallelogram property. It does not by itself establish that the shape is a rectangle, a parallelogram with right angles. Additional distance comparisons, such as checking its diagonals, are needed for that further conclusion.

Glossary

• Coordinate axis — A reference line used to measure one coordinate of a point in space.
• Origin — The common intersection of the three coordinate axes, with coordinates (0, 0, 0).
• Coordinate plane — A plane determined by two coordinate axes and perpendicular to the third axis.
• Ordered triplet — Three numbers listed in a fixed order to specify the coordinates of a point.
• Octant — One of the eight regions into which the coordinate planes divide three-dimensional space.
• Perpendicular projection — The foot obtained by dropping a perpendicular from a point to a line or plane.
• Absolute value — The non-negative magnitude of a real number, written using vertical bars around it.
• Collinear points — Points that lie on the same straight line in a plane or in space.
• Internal division — Division of a segment by a point lying between its two distinct endpoints.
• External division — Division on the line through two endpoints by a point outside their segment.
• Midpoint — The point dividing a line segment internally into two parts of equal length.
• Diagonal — A segment joining opposite vertices of a quadrilateral, or opposite vertices of a rectangular box.

Common errors and misconceptions

• Misconception: Two coordinates locate every point in space. Correct: Three coordinates are needed relative to three fixed axes; two suffice within a specified coordinate plane.
• Misconception: The x-coordinate gives distance from the x-axis. Correct: Its absolute value gives perpendicular distance from the YZ-plane.
• Misconception: A point on the XZ-plane has z = 0. Correct: It has y = 0; its x- and z-coordinates can vary within that plane.
• Misconception: A zero coordinate can be treated as positive when choosing an octant. Correct: The point lies on a coordinate-plane boundary and is not inside an open octant.
• Misconception: PQ² + QR² = PR² proves collinearity. Correct: Collinearity uses PQ + QR = PR when Q lies between P and R; the squared relation tests a right angle.
• Misconception: In internal division PR : RQ = m : n, m multiplies P's coordinates. Correct: It multiplies Q's coordinates, while n multiplies P's coordinates.
• Misconception: External division changes only the denominator of the internal formula. Correct: Use subtraction in each weighted numerator and m − n in the denominator, with m ≠ n.
• Misconception: A common diagonal midpoint proves that a quadrilateral is a rectangle. Correct: It establishes the parallelogram property; further conditions are needed to establish a rectangle.

Exam-style questions with model answers

Q1. For P(3, 4, 5), find its perpendicular projection onto the XY-plane and its perpendicular distance from that plane. [2 marks]

1. The projection is (3, 4, 0), because projection onto the XY-plane keeps x and y unchanged and sets z to zero.
2. The perpendicular distance is the magnitude of the z-coordinate, so the distance is 5 units.

Q2. Find the distance between P(1, −3, 4) and Q(−4, 1, 2). [3 marks]

1. Use the spatial distance formula: square each difference of matching coordinates, add the three squares, and take the non-negative square root.
2. The differences are −4 − 1 = −5, 1 − (−3) = 4, and 2 − 4 = −2. Their squares are 25, 16 and 4.
3. Hence PQ = √(25 + 16 + 4) = √45 = 3√5 units, taking the positive root because PQ is a length.

Q3. Show that P(−2, 3, 5), Q(1, 2, 3) and R(7, 0, −1) are collinear, and identify the point between the other two. [4 marks]

1. The squared distance PQ² is (1 + 2)² + (2 − 3)² + (3 − 5)² = 14, so PQ = √14.
2. Similarly QR² = (7 − 1)² + (0 − 2)² + (−1 − 3)² = 56, giving QR = 2√14.
3. The remaining distance satisfies PR² = (7 + 2)² + (0 − 3)² + (−1 − 5)² = 126, so PR = 3√14.
4. Therefore PQ + QR = PR. The points lie on one straight line, with Q between P and R.

Q4. A point P is five-sixths of the way from A(−2, 0, 6) to B(10, −6, −12). Find its coordinates using internal division. [5 marks]

1. Five of the six equal parts from A to B are before P and one remains after P. Therefore the internal ratio AP : PB is 5 : 1.
2. The section formula weights B's coordinates by 5 and A's coordinates by 1, dividing each weighted sum by the total weight 6.
3. The x-coordinate of P is [5 × 10 + 1 × (−2)]/6 = 48/6 = 8.
4. The y-coordinate is [5 × (−6) + 1 × 0]/6 = −5, and the z-coordinate is [5 × (−12) + 1 × 6]/6 = −9.
5. Combining the coordinates in x, y, z order gives P(8, −5, −9). Each coordinate lies between the corresponding endpoint coordinates, consistent with internal division.

Q5. The XZ-plane meets the line through P(2, 4, 5) and Q(3, 5, −4) at R. Find PR : RQ and state the type of division. [3 marks]

1. Every point on the XZ-plane has y = 0. The endpoint y-coordinates, 4 and 5, are both positive, so R is outside the segment and the division is external.
2. Let PR : RQ = m : n, where m and n are positive. The external section formula gives 0 = (5m − 4n)/(m − n).
3. Thus 5m = 4n, giving m : n = 4 : 5. The required external ratio PR : RQ is 4 : 5.

Q6. Show that A(3, −1, −1), B(5, −4, 0), C(2, 3, −2) and D(0, 6, −3), taken in order, form a parallelogram by comparing diagonal midpoints. [4 marks]

1. The diagonals are AC and BD. Calculate each midpoint by averaging the corresponding coordinates of its two endpoints.
2. The midpoint of AC is ((3 + 2)/2, (−1 + 3)/2, (−1 − 2)/2) = (5/2, 1, −3/2).
3. The midpoint of BD is ((5 + 0)/2, (−4 + 6)/2, (0 − 3)/2) = (5/2, 1, −3/2).
4. All three coordinates agree, so the diagonals have a common midpoint and bisect each other. Therefore ABCD is a parallelogram.

Key takeaways

• Three mutually perpendicular axes and their three coordinate planes provide the fixed reference system for locating points in space.
• An ordered triplet records coordinates in x, y, z order; each entry's sign identifies a direction relative to the origin.
• The eight octants correspond to the eight sign combinations of three non-zero coordinates; coordinate planes form their boundaries.
• Projection onto a coordinate plane sets one coordinate to zero; projection onto an axis sets the other two to zero.
• Distance between two points is the square root of the sum of the three squared differences of corresponding coordinates.
• Internal division uses weighted sums divided by m + n; external division uses weighted differences divided by m − n.
• The midpoint formula averages corresponding coordinates, and equal diagonal midpoints establish that a quadrilateral is a parallelogram.
• Use sums of lengths for collinearity and sums of squared lengths for a right-angle test.

Test yourself

Which coordinate is zero for every point in the YZ-plane?

The x-coordinate is zero; the general coordinate form is (0, y, z).

What are the coordinates of a general point on the z-axis?

They are (0, 0, z), because its x- and y-coordinates are both zero.

In which octant does (−3, 1, −2) lie?

It lies in octant VI, whose coordinate signs are negative, positive and negative.

What is the perpendicular projection of (3, 4, 5) onto the x-axis?

It is (3, 0, 0): retain the x-coordinate and set the other two to zero.

What is the distance of P(x, y, z) from the origin O?

OP = √(x² + y² + z²), obtained by using (0, 0, 0) as one endpoint.

A point is five-sixths of the way from A to B. In what ratio does it divide AB?

The internal ratio is 5 : 1, because one-sixth of the segment remains after the point.

Why does the external section formula require unequal ratio terms?

Its denominator is their difference, which would be zero if the ratio terms were equal.

What does equality of the midpoints of a quadrilateral's diagonals establish?

The diagonals bisect each other, establishing that the quadrilateral is a parallelogram.