Chapter 7 · Coordinate Geometry

7.1 The Distance Formula

Describe every point by two numbers and the straight-line distance between any two points follows from Pythagoras: $d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$. With one formula you can measure a map, test whether three points lie on a line, and recognise a square from its corners.

At the start of this chapter a flag stood at $(24, 7)$ on the school ground, and we found its distance from the corner flag with Pythagoras: $\sqrt{24^2+7^2}=25$ m. The same trick works for any two flags, not only for flags measured from the corner. If one flag is at $(3, 2)$ and another at $(11, 8)$, then the second is $8$ m further along and $6$ m further up, and the straight line between them is the hypotenuse of a right-angled triangle with sides $8$ and $6$. Its length is $\sqrt{64+36}=10$ m.

In this section we turn that picture into a formula, the distance formula, and prove it. Then we use it for three jobs a Class 10 student meets again and again: finding a missing coordinate when a distance is given, deciding whether three points lie on one straight line, and deciding whether a triangle or a quadrilateral is isosceles, right-angled, a square, a rectangle or a rhombus. Everything is done with distances — no protractor, no ruler.

Points, axes and distances along a line

You met the Cartesian plane in Class 9. Two perpendicular number lines cross at a point called the origin; the horizontal one is the $x$-axis and the vertical one is the $y$-axis. Any point is described by an ordered pair of numbers, and the order matters: $(3, 5)$ and $(5, 3)$ are different points.

📘 Definition 7.1 — Coordinates of a point

The point $P(x, y)$ has abscissa (or $x$-coordinate) $x$, its signed distance from the $y$-axis, and ordinate (or $y$-coordinate) $y$, its signed distance from the $x$-axis. The two axes meet at the origin $O(0, 0)$ and split the plane into four quadrants: I $(+,+)$, II $(-,+)$, III $(-,-)$ and IV $(+,-)$. A point on the $x$-axis has the form $(x, 0)$; a point on the $y$-axis has the form $(0, y)$.

The easiest distances need no formula. If two points have the same ordinate, they lie on a horizontal line and the distance is just the difference of the abscissas, taken positive: the distance from $(-4, 5)$ to $(7, 5)$ is $|7-(-4)|=11$. Likewise two points with the same abscissa lie on a vertical line and the distance is the difference of the ordinates. Notice the double minus in $7-(-4)$: distance along a number line is always "larger minus smaller", never $7-4$.

The distance formula

Now take two points that are not on a horizontal or a vertical line. Draw the horizontal line through one and the vertical line through the other; they meet in a corner point, and a right-angled triangle appears with the segment as its hypotenuse. Its legs are horizontal and vertical, so their lengths are the easy distances above.

📐 Theorem 7.1 — The Distance Formula

The distance between the points $P(x_1, y_1)$ and $Q(x_2, y_2)$ is $$PQ=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}.$$ In particular, the distance of $P(x, y)$ from the origin is $OP=\sqrt{x^2+y^2}$.

Proof. Through $P$ draw a horizontal line and through $Q$ a vertical line; let them meet at $R$. The coordinates of $R$ are $(x_2, y_1)$: it is in the same column as $Q$ and in the same row as $P$. Then $\triangle PRQ$ has a right angle at $R$.

The side $PR$ is horizontal, so $PR=|x_2-x_1|$. The side $RQ$ is vertical, so $RQ=|y_2-y_1|$. By Pythagoras in $\triangle PRQ$,

$$PQ^2=PR^2+RQ^2=|x_2-x_1|^2+|y_2-y_1|^2=(x_2-x_1)^2+(y_2-y_1)^2,$$

because the square of a number and of its absolute value are equal. Taking the positive square root gives the formula. If $P$ and $Q$ are on one horizontal or vertical line, the triangle collapses to a segment, and the formula still gives the correct length (one of the two brackets is $0$). For the origin, put $(x_1, y_1)=(0, 0)$. $\blacksquare$

Because the brackets are squared, it makes no difference which point you call $P$ and which $Q$: $(x_2-x_1)^2=(x_1-x_2)^2$. For the same reason a negative difference causes no trouble — $(-3)^2=9$.

⚠️ Common mistakes

1. Squaring the coordinates separately. $(x_2-x_1)^2$ is not $x_2^2-x_1^2$. For $(1, 2)$ and $(4, 6)$: $(4-1)^2=9$, but $4^2-1^2=15$.

2. Forgetting the square root. $(x_2-x_1)^2+(y_2-y_1)^2$ is the square of the distance. For $(1, 2)$ and $(4, 6)$ it is $25$, and the distance is $5$.

3. Mixing the coordinates. Subtract abscissa from abscissa and ordinate from ordinate. Subtracting $x_2-y_1$ is meaningless.

4. Splitting the square root. $\sqrt{9+16}=5$, not $\sqrt9+\sqrt{16}=7$. A root of a sum is not the sum of the roots.

✏️ Example 1 — Using the distance formula

Find the distance between (a) $A(2, 3)$ and $B(6, 6)$; (b) $P(-3, -1)$ and $Q(4, -2)$; (c) $O(0, 0)$ and $R(-5, 12)$; (d) $S(-4, 5)$ and $T(7, 5)$.

💡 Show solution▶

(a) $AB=\sqrt{(6-2)^2+(6-3)^2}=\sqrt{16+9}=\sqrt{25}=5$ units.

(b) $PQ=\sqrt{(4-(-3))^2+(-2-(-1))^2}=\sqrt{7^2+(-1)^2}=\sqrt{49+1}=\sqrt{50}=5\sqrt2\approx7.07$ units.

(d) Same ordinate, so $ST=|7-(-4)|=11$ units. The formula agrees: $\sqrt{11^2+0^2}=11$ ✓.

Check In (b), swap the points: $\sqrt{(-3-4)^2+(-1-(-2))^2}=\sqrt{49+1}$, the same ✓.

Drag point $A$ or point $B$ with your finger or mouse (or type new coordinates). The blue leg is the change in $x$, the red leg is the change in $y$, and the segment $AB$ is the hypotenuse. The readouts show the formula at work. Press Play to cycle through some examples with whole-number distances, then drag to explore for yourself.

$A$: $x_1$$y_1$$B$: $x_2$$y_2$

$|x_2-x_1|$ and $|y_2-y_1|$—

$d^2=(\Delta x)^2+(\Delta y)^2$—

distance $d$—

✏️ Example 2 — Finding a missing coordinate

Find the values of $y$ for which the distance between $Q(-1, y)$ and $R(5, 3)$ is $10$ units.

💡 Show solution▶

Formula $QR=\sqrt{(5-(-1))^2+(3-y)^2}=10$.

Square both sides $36+(3-y)^2=100$, so $(3-y)^2=64$.

Take roots $3-y=8$ or $3-y=-8$, so $y=-5$ or $y=11$. Two values, because $Q$ can be $8$ units below or $8$ units above the level of $R$.

Check $y=-5$: $\sqrt{36+64}=10$ ✓. $y=11$: $\sqrt{36+(-8)^2}=10$ ✓.

Collinear points and the shape of a figure

📘 Definition 7.2 — Collinear points

Three or more points are collinear if they all lie on one straight line.

Look at three points $A$, $B$, $C$ on a line, with $B$ between $A$ and $C$. Then walking from $A$ to $B$ and on to $C$ is the same as walking straight from $A$ to $C$: $AB+BC=AC$. If the points do not lie on a line, they form a triangle, and in a triangle two sides together are always longer than the third. So the distance formula settles collinearity.

📐 Theorem 7.2 — Test for collinear points

Three distinct points $A$, $B$, $C$ are collinear exactly when the largest of the three distances $AB$, $BC$, $CA$ equals the sum of the other two. Otherwise they are the vertices of a triangle.

Use exact values (square roots left as roots) when you add distances; a calculator value such as $7.62$ can hide a tiny difference that decides the answer, as the next example shows. The same distance idea identifies shapes:

🧭 Problem-solving strategy — what shape is it?

Figure	Distances to compute	Conclusion
Triangle	$AB, BC, CA$	all equal: equilateral · exactly two equal: isosceles · all different: scalene
Triangle	squares of the sides	$AB^2+BC^2=CA^2$ (largest side) means a right angle at $B$
Quadrilateral $ABCD$	four sides	all four equal: rhombus (or square) · opposite sides equal: parallelogram
Quadrilateral $ABCD$	sides and both diagonals $AC$, $BD$	parallelogram + equal diagonals: rectangle · rhombus + equal diagonals: square

Always compare squared distances: it avoids square roots and the arithmetic is exact. Take the square roots only for the final answer.

✏️ Example 3 — Collinear or not?

Are the points (a) $A(1, -1)$, $B(3, 3)$, $C(5, 7)$ collinear? (b) $P(1, 1)$, $Q(2, 3)$, $R(4, 8)$ collinear?

💡 Show solution▶

(a) Distances $AB=\sqrt{4+16}=\sqrt{20}=2\sqrt5$, $\;BC=\sqrt{4+16}=2\sqrt5$, $\;AC=\sqrt{16+64}=\sqrt{80}=4\sqrt5$.

(a) Test $AB+BC=2\sqrt5+2\sqrt5=4\sqrt5=AC$. The largest distance is the sum of the other two, so the points are collinear, with $B$ the middle point (it is also halfway, since $AB=BC$).

(b) Distances $PQ=\sqrt{1+4}=\sqrt5$, $\;QR=\sqrt{4+25}=\sqrt{29}$, $\;PR=\sqrt{9+49}=\sqrt{58}$.

(b) Test Is $\sqrt5+\sqrt{29}=\sqrt{58}$? Decimals: $2.236+5.385=7.621$ against $7.616$ — painfully close. Square the left side: $(\sqrt5+\sqrt{29})^2=5+29+2\sqrt{145}\approx34+24.08=58.08\ne58$. So the points are not collinear; they form a very thin triangle.

✏️ Example 4 — Is it a square?

Show that $A(1, 2)$, $B(5, 3)$, $C(4, 7)$ and $D(0, 6)$ are the vertices of a square.

💡 Show solution▶

Sides $AB^2=4^2+1^2=17$, $\;BC^2=(-1)^2+4^2=17$, $\;CD^2=(-4)^2+(-1)^2=17$, $\;DA^2=1^2+(-4)^2=17$. All four sides are equal, so $ABCD$ is a rhombus.

Diagonals $AC^2=3^2+5^2=34$ and $BD^2=(-5)^2+3^2=34$. The diagonals are equal.

Conclusion A rhombus with equal diagonals is a square. (Also $AB^2+BC^2=34=AC^2$, so the angle at $B$ is a right angle.)

Why both? Equal sides alone give only a rhombus — a pushed-over square. Equal diagonals are what force the corners to be right angles.

✏️ Example 5 — A point equidistant from two points

Find the point on the $y$-axis that is equidistant from $C(5, -2)$ and $D(-3, 2)$.

💡 Show solution▶

Set up A point on the $y$-axis is $Q(0, y)$. We need $QC=QD$, which is the same as $QC^2=QD^2$.

Equation $QC^2=(5-0)^2+(-2-y)^2=25+(y+2)^2$ and $QD^2=(-3-0)^2+(2-y)^2=9+(y-2)^2$. So $25+y^2+4y+4=9+y^2-4y+4$.

Solve The $y^2$ cancel: $29+4y=13-4y$, so $8y=-16$ and $y=-2$. The point is $Q(0, -2)$.

Check $QC^2=25+0=25$ and $QD^2=9+(-4)^2=25$. Both distances are $5$ ✓.

The point $P(x, 0)$ slides along the $x$-axis. The blue segment is $PA$ and the red segment is $PB$. Drag the slider (or press Play) and watch the two lengths: when they are equal, $P$ is equidistant from $A$ and $B$ and the verdict turns green. Choose a pair of points, or snap to the exact answer.

Points:$x$:0

$PA$—

$PB$—

exact answer—

✏️ Example 6 — The circular park

On a town map (1 unit = 100 m) a circular park has its centre at $C(4, 5)$ and radius $5$ units. Which of the benches $P(7, 9)$, $Q(1, 1)$, $R(9, 8)$ and $S(2, 3)$ lie inside the park, on its boundary, or outside?

💡 Show solution▶

Idea A point is inside a circle if its distance from the centre is less than the radius, on it if equal, outside if greater. The radius squared is $25$.

Distances squared $CP^2=3^2+4^2=25$, $\;CQ^2=(-3)^2+(-4)^2=25$, $\;CR^2=5^2+3^2=34$, $\;CS^2=(-2)^2+(-2)^2=8$.

Verdict $P$ and $Q$ are on the boundary ($CP=CQ=5$); $R$ is outside ($CR=\sqrt{34}\approx5.83$); $S$ is inside ($CS=\sqrt8\approx2.83$).

Real distance The bench $R$ is $5.83$ units $=583$ m from the centre, so about $83$ m beyond the edge of the park.

Exercises

Find the distance between the points. (Leave answers as simple surds, and give a decimal where it helps.)

1 $(6, 8)$ and the origin
2 $(-8, -15)$ and the origin
3 $(2, -3)$ and $(5, 1)$
4 $(-4, -1)$ and $(2, 5)$
5 $(0, 7)$ and $(24, 0)$
6 $(a, b)$ and $(-a, -b)$
7 $(1.5, 2)$ and $(4.5, 6)$
8 $(3, -6)$ and $(3, 9)$

Check: type two points — the squared distance and the distance

$x_1$$y_1$$x_2$$y_2$

$\Delta x,\ \Delta y$—

$d^2$—

$d$—

💡 Show solutions▶

1 $\sqrt{6^2+8^2}=\sqrt{100}=10$.

2 $\sqrt{(-8)^2+(-15)^2}=\sqrt{64+225}=\sqrt{289}=17$.

3 $\sqrt{(5-2)^2+(1-(-3))^2}=\sqrt{9+16}=5$.

4 $\sqrt{6^2+6^2}=\sqrt{72}=6\sqrt2\approx8.49$.

5 $\sqrt{(24-0)^2+(0-7)^2}=\sqrt{576+49}=\sqrt{625}=25$.

6 $\sqrt{(-a-a)^2+(-b-b)^2}=\sqrt{4a^2+4b^2}=2\sqrt{a^2+b^2}$.

7 $\sqrt{3^2+4^2}=5$.

8 Same abscissa, so the distance is $|9-(-6)|=15$.

Missing coordinates and equidistant points.

1 Find $x$ if the distance between $(x, 3)$ and $(7, -1)$ is $5$.
2 Find $y$ if the distance between $(-3, y)$ and $(5, 2)$ is $10$.
3 Find the point on the $x$-axis that is equidistant from $(-1, 3)$ and $(5, 3)$.
4 Find the point on the $y$-axis that is equidistant from $A(6, 5)$ and $B(-2, 1)$.
5 Find the relation between $x$ and $y$ if $P(x, y)$ is equidistant from $(3, 6)$ and $(-3, 4)$.
6 Find the points on the $x$-axis whose distance from $(4, 12)$ is $13$.
7 If $P(x, y)$ is equidistant from $A(5, 1)$ and $B(-1, 5)$, show that $3x=2y$.
8 Which of $(6, -3)$, $(4, 5)$, $(-5, -5)$ and $(0, -6)$ is nearest to the origin?

💡 Show solutions▶

1 $(7-x)^2+(-1-3)^2=25$ gives $(7-x)^2=9$, so $7-x=\pm3$ and $x=4$ or $x=10$.

2 $(5+3)^2+(2-y)^2=100$ gives $(2-y)^2=36$, so $2-y=\pm6$ and $y=-4$ or $y=8$.

3 $P(x, 0)$: $(x+1)^2+9=(x-5)^2+9$, so $2x+1=-10x+25$ and $x=2$. $P(2, 0)$. (It lies halfway between the two points' abscissas, because they have equal ordinates.)

4 $Q(0, y)$: $36+(y-5)^2=4+(y-1)^2$ gives $61-10y=5-2y$, so $8y=56$ and $y=7$. $Q(0, 7)$. Check: $QA^2=36+4=40=QB^2=4+36$ ✓.

5 $(x-3)^2+(y-6)^2=(x+3)^2+(y-4)^2$ gives $45-6x-12y=25+6x-8y$, so $12x+4y=20$, i.e. $3x+y=5$. Check with $(1, 2)$: both squared distances are $20$ ✓.

6 $P(x, 0)$: $(x-4)^2+144=169$ gives $(x-4)^2=25$, so $x=9$ or $x=-1$. The points are $(9, 0)$ and $(-1, 0)$.

7 $(x-5)^2+(y-1)^2=(x+1)^2+(y-5)^2$ gives $26-10x-2y=26+2x-10y$, so $8y=12x$, i.e. $2y=3x$ ✓.

8 Squared distances: $45$, $41$, $50$, $36$. The smallest is $36$, so $(0, -6)$ is nearest (distance $6$).

III

Collinearity and shapes.

1 Show that $(0, -3)$, $(2, 1)$ and $(4, 5)$ are collinear.
2 Show that $(1, 2)$, $(3, 5)$ and $(7, 10)$ are not collinear.
3 Show that $(-2, 0)$, $(2, 0)$ and $(0, 4)$ form an isosceles triangle that is not equilateral.
4 Show that $A(0, 0)$, $B(4, 3)$ and $C(-2, 11)$ form a right-angled triangle. Which angle is the right angle?
5 Show that $A(0, 0)$, $B(4, 3)$, $C(9, 3)$, $D(5, 0)$ form a rhombus but not a square.
6 Show that $A(0, 0)$, $B(3, 4)$, $C(-5, 10)$, $D(-8, 6)$ form a rectangle. Is it a square?
7 Find the point equidistant from $O(0, 0)$, $B(6, 0)$ and $C(0, 8)$, and the common distance.

💡 Show solutions▶

1 Distances: $\sqrt{4+16}=2\sqrt5$, $\;\sqrt{4+16}=2\sqrt5$, $\;\sqrt{16+64}=4\sqrt5$. Since $2\sqrt5+2\sqrt5=4\sqrt5$, the points are collinear.

2 Distances: $\sqrt{13}$, $\sqrt{41}$ and $10$. $(\sqrt{13}+\sqrt{41})^2=54+2\sqrt{533}\approx100.17\ne100$, so the largest distance is not the sum of the others: not collinear.

3 With $P(-2, 0)$, $Q(2, 0)$, $R(0, 4)$: $PR^2=4+16=20$, $QR^2=4+16=20$, $PQ^2=16$. Two sides equal, the third different: isosceles, not equilateral. (It is not right-angled either, since $20+16\ne20$ and $20+20\ne16$.)

4 $AB^2=25$, $BC^2=36+64=100$, $AC^2=4+121=125$. Since $25+100=125$, $AB^2+BC^2=AC^2$: the right angle is at $B$. (The triangle is not isosceles.)

5 $AB^2=25$, $BC^2=25$, $CD^2=16+9=25$, $DA^2=25$: all sides equal, a rhombus. Diagonals: $AC^2=81+9=90$ and $BD^2=1+9=10$ are unequal, so it is not a square.

6 $AB^2=25$, $BC^2=64+36=100$, $CD^2=9+16=25$, $DA^2=64+36=100$: opposite sides equal (a parallelogram). Diagonals: $AC^2=25+100=125$ and $BD^2=121+4=125$, equal. So it is a rectangle. Adjacent sides are $5$ and $10$, so it is not a square.

7 Let the point be $(x, y)$. Equal distances from $O$ and $B$: $x^2+y^2=(x-6)^2+y^2$ gives $x=3$. From $O$ and $C$: $x^2+y^2=x^2+(y-8)^2$ gives $y=4$. The point is $(3, 4)$ and its distance from each vertex is $5$ (the midpoint of the hypotenuse of a right triangle).

Applications.

1 Road map. On a map (1 unit = 10 km), town $A$ is at $(2, 3)$ and town $B$ is at $(8, 11)$. How far apart are the towns by air?
2 Throw to the keeper. Taking the batter as the origin (units in metres), a fielder stands at $(-12, 9)$ and the wicket-keeper at $(12, 16)$. How far must the fielder throw?
3 Bus route. A bus goes from the depot $P(0, 0)$ to stop $Q(6, 8)$, then to stop $R(6, 0)$ and back to the depot, all in a straight line each time (units in km). How long is the round trip?
4 Drone. A drone flies in a straight line from $(3, 4)$ to $(15, 9)$ over a field (units in metres, ignoring height). How far does it fly?
5 Football pitch. The corners of a field are $(0, 0)$, $(8, 0)$, $(8, 6)$ and $(0, 6)$ (units in metres). A ball lies at $(4, 3)$. How far is it from each corner?
6 Wi-Fi range. A router at $(2, -1)$ (units in metres) reaches phones up to $5$ m away. Which of the phones at $(5, 3)$, $(4, 0)$, $(3, -7)$ and $(6, 2)$ can connect?

💡 Show solutions▶

1 $AB=\sqrt{6^2+8^2}=10$ units $=100$ km.

2 $\sqrt{(12+12)^2+(16-9)^2}=\sqrt{576+49}=25$ m.

3 $PQ=10$, $QR=8$ (same abscissa), $RP=6$ (same ordinate). Round trip $=10+8+6=24$ km.

4 $\sqrt{12^2+5^2}=13$ m.

5 Each distance is $\sqrt{4^2+3^2}=5$ m (e.g. to $(8, 6)$: $\sqrt{16+9}$). The ball is at the centre of the field, equidistant from all four corners.

6 Squared distances from $(2, -1)$: $(5, 3)$: $9+16=25$; $(4, 0)$: $4+1=5$; $(3, -7)$: $1+36=37$; $(6, 2)$: $16+9=25$. The range squared is $25$, so the phones at $(5, 3)$, $(4, 0)$ and $(6, 2)$ connect (the first and last are exactly at the edge); the phone at $(3, -7)$ does not.

🎯 Exam corner Board · 1 markBoard · 2 marksBoard · 3 marksBoard · Case studyJEE Main

How this is examined: the distance formula appears as a one-mark direct question, as a missing coordinate, as a collinearity or shape proof (three marks) and as a case study set on a map or a ground. Always state the formula first, show the substitution, and give the final answer with units.

E1. Board · 1 mark The distance of the point $(-7, 24)$ from the origin is (A) $25$ (B) $17$ (C) $31$ (D) $7$

💡 Show solution▶

Idea $OP=\sqrt{(-7)^2+24^2}=\sqrt{49+576}=\sqrt{625}=25$.

Traps (C) adds the two coordinates' sizes, $7+24=31$, instead of using Pythagoras. (D) is the distance from the $y$-axis only.

Answer (A) $25$.

E2. Board · 1 mark Assertion (A): The points $O(0, 0)$, $A(3, 4)$ and $B(6, 8)$ are collinear. Reason (R): $OA=5$, $AB=5$ and $OB=10$.
(A) Both A and R are true, and R is the correct explanation of A. (B) Both A and R are true, but R is not the correct explanation of A. (C) A is true but R is false. (D) A is false but R is true.

💡 Show solution▶

Check R $OA=\sqrt{9+16}=5$, $AB=\sqrt{9+16}=5$, $OB=\sqrt{36+64}=10$. R is true.

Link $OA+AB=5+5=10=OB$: the largest distance equals the sum of the other two, which is exactly the test for collinear points (Theorem 7.2). So A is true and R explains it.

Answer (A).

E3. Board · 2 marks Find the values of $x$ for which the distance between $A(-2, 3)$ and $B(x, -2)$ is $13$ units.

💡 Show solution▶

Equation $AB^2=(x+2)^2+(-2-3)^2=169$, so $(x+2)^2+25=169$ and $(x+2)^2=144$. (1 mark)

Solve $x+2=\pm12$, so $x=10$ or $x=-14$. (1 mark) Check: $\sqrt{12^2+5^2}=13$ for both ✓.

E4. Board · 3 marks Show that the points $P(2, -1)$, $Q(5, 2)$ and $R(2, 5)$ are the vertices of an isosceles right-angled triangle.

💡 Show solution▶

Sides $PQ^2=3^2+3^2=18$, $\;QR^2=(-3)^2+3^2=18$, $\;PR^2=0^2+6^2=36$. (1 mark)

Isosceles $PQ=QR=\sqrt{18}=3\sqrt2$, so the triangle is isosceles. (1 mark)

Right angle $PQ^2+QR^2=18+18=36=PR^2$, so the angle at $Q$ is $90^\circ$ (converse of Pythagoras). (1 mark)

E5. Board · Case study Three friends take part in a relay on a school ground. Taking a corner as the origin and $1$ unit $=1$ m, Aman stands at $A(2, 2)$, Bhavna at $B(14, 7)$ and Chitra at $C(2, 7)$.
(i) Find the distances $AB$ and $BC$. (2 marks)
(ii) Show that the three friends stand at the corners of a right-angled triangle. (1 mark)
(iii) A fourth friend, Dev, stands at $D(14, 2)$. Show that $CD=AB$. (1 mark)

💡 Show solution▶

(i) $AB=\sqrt{(14-2)^2+(7-2)^2}=\sqrt{144+25}=13$ m. $\;BC=\sqrt{(2-14)^2+0}=12$ m (same ordinate).

(ii) $AC=5$ m (same abscissa). $AC^2+BC^2=25+144=169=AB^2$, so the right angle is at $C$.

(iii) $CD=\sqrt{(14-2)^2+(2-7)^2}=\sqrt{144+25}=13$ m $=AB$. ($A, C, B, D$ are the corners of a $12\times5$ rectangle, whose diagonals $AB$ and $CD$ are equal.)

E6. JEE Main The point $P$ lies on the line $y=x$ and is equidistant from $A(0, 4)$ and $B(6, 2)$. Then $PA^2$ equals (A) $8$ (B) $10$ (C) $13$ (D) $18$

💡 Show solution▶

Idea Let $P(t, t)$. Then $PA^2=t^2+(t-4)^2=2t^2-8t+16$ and $PB^2=(t-6)^2+(t-2)^2=2t^2-16t+40$.

Solve $PA^2=PB^2$ gives $-8t+16=-16t+40$, so $8t=24$ and $t=3$. Then $P(3, 3)$ and $PA^2=9+1=10$.

Check $PB^2=(3-6)^2+(3-2)^2=9+1=10$ ✓.

Answer (B) $10$. (D) $18=3^2+3^2$ is $OP^2$, the distance from the origin, not from $A$.

🔬 Research–Teaching NexusMini project · optional, about 45 minutes

Research question. A delivery rider in a city of straight, parallel streets cannot go "as the crow flies": she must ride along the streets. How much longer is the street route than the straight-line distance, and does the extra distance depend on the direction of the trip?

Why it matters. Delivery companies, taxi apps and city planners all compare "straight-line distance" with "route distance". The straight-line distance (from the distance formula) is a quick estimate and a lower bound; the real route is always at least as long. Knowing how big the gap usually is helps in pricing a trip and in placing a warehouse.

What to do.

On squared paper (1 square = 100 m), mark a start point and six destinations in different directions, at least two of them on a horizontal or vertical line from the start.
For each destination, write the street distance (moving only along grid lines: $|\Delta x|+|\Delta y|$) and the straight-line distance (distance formula). Put them in a table with a third column "street ÷ straight".
Which destinations have a ratio of exactly $1$? Which have the biggest ratio? Where on the grid are they?
Check one straight-line distance on a real map app by measuring between two landmarks, and compare it with the app's walking distance.
Write a rule in words for when the street route is much longer than the straight line.

Tools: Squared paper or a spreadsheet, a calculator, a map app on a phone

Hand in: Your table, the grid drawing, the map-app comparison, and a short paragraph answering the research question.

Going further. Try a destination exactly diagonal from the start ($\Delta x=\Delta y$) and compute the ratio. Does any destination give a ratio larger than that? (Look for the largest value you can find, and explain why you cannot beat it.)

Teacher/parent note: the ratio is $\dfrac{|\Delta x|+|\Delta y|}{\sqrt{\Delta x^2+\Delta y^2}}$. It equals $1$ on horizontal and vertical trips, and is largest on the diagonal, where it is $\dfrac{2n}{n\sqrt2}=\sqrt2\approx1.414$ — the street route is about $41\%$ longer. For $(6, 8)$: street $14$, straight $10$, ratio $1.4$. It never exceeds $\sqrt2$ because $(|a|+|b|)^2\le2(a^2+b^2)$.

🧪 Self-check quiz

0/5 correct

Q1. The distance between $(2, 5)$ and $(-4, -3)$ is

Q2. The distance of the point $P(-5, 12)$ from the $y$-axis is

Q3. Three points satisfy $AB=7$, $BC=5$ and $AC=12$. Then

Q4. The triangle with vertices $(0, 0)$, $(6, 0)$ and $(3, 3)$ is

Q5. The point on the $x$-axis equidistant from $(2, 3)$ and $(6, 5)$ is

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