NCERT Exemplar Solutions Class 10 Maths Coordinate Geometry — Exercise 7.1

NCERT Exemplar Problems, Class 10 Mathematics — Coordinate Geometry, Exercise 7.1. All 20 multiple-choice questions solved.

Twenty questions running on three formulae and almost nothing else. The whole exercise is a drill in choosing which of the three to reach for, and in reading the question closely enough to know what is actually being asked.

The distance between $P(x_1, y_1)$ and $Q(x_2, y_2)$:

$$PQ = \sqrt{(x_2 – x_1)^2 + (y_2 – y_1)^2}$$

The point dividing the segment from $A(x_1, y_1)$ to $B(x_2, y_2)$ internally in the ratio $m_1 : m_2$:

$$\left(\frac{m_1 x_2 + m_2 x_1}{m_1 + m_2},\ \frac{m_1 y_2 + m_2 y_1}{m_1 + m_2}\right), \qquad \text{and the mid-point } \left(\frac{x_1 + x_2}{2},\ \frac{y_1 + y_2}{2}\right)$$

And the area of the triangle with vertices $A(x_1, y_1)$, $B(x_2, y_2)$, $C(x_3, y_3)$:

$$\operatorname{ar}(\triangle ABC) = \tfrac{1}{2}\left|x_1(y_2 – y_3) + x_2(y_3 – y_1) + x_3(y_1 – y_2)\right|$$

A note on the current syllabus. That third formula is no longer in the Class 10 textbook. The rationalised chapter runs Introduction → Distance Formula → Section Formula → Summary, and there is no “Area of a Triangle” section at all. Questions 7 and 18 here therefore test material the book no longer carries, and Question 20 — a collinearity test, which most solution sets do with the area formula — has to be argued a different way. The textbook’s own route for collinearity is the distance sum: three points are collinear exactly when the two shorter distances between them add to the longest. Worth knowing, because it is the method your textbook example uses and the area route is not available to you.

Two small mercies: the Introduction to this chapter was edited to match the cuts, which is more than can be said for Chapters 1 and 6, and everything removed is still ordinary knowledge you can use — it is simply not examinable from this book.

Key insight. Four of these twenty are decided not by a calculation but by one word in the question — segment rather than line, interior rather than on or inside, the vertex order inside a name like $AOBC$. Do the arithmetic second. Read the sentence first.

Choose the correct answer from the given four options:

Question 1

The distance of the point $P(2, 3)$ from the $x$-axis is

Solution. The distance of a point from the $x$-axis is the length of the perpendicular dropped onto it, and that perpendicular is vertical — it changes only the $y$-coordinate, running from $(2, 3)$ down to $(2, 0)$. So the distance is $|y| = 3$.

The trap is answering $2$. The $x$-coordinate measures how far the point is from the $y$-axis, not from the $x$-axis; each coordinate names the distance from the other axis.

Question 2

The distance between the points $A(0, 6)$ and $B(0, -2)$ is

Solution. Both points have $x = 0$, so both sit on the $y$-axis and the segment joining them is vertical. The distance formula collapses to the difference of the $y$-coordinates:

$$AB = \sqrt{(0-0)^2 + (-2-6)^2} = \sqrt{64} = 8$$

Read as a picture: from $6$ down to $0$ is six units, and from $0$ down to $-2$ is two more. Eight.

Question 3

The distance of the point $P(-6, 8)$ from the origin is

Solution. Distance from the origin is the special case of the distance formula with $(x_2, y_2) = (0, 0)$:

$$OP = \sqrt{(-6)^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10$$

The squares kill the minus sign, which is the point of the question — $(-6, 8)$ and $(6, 8)$ are the same distance from the origin.

Question 4

The distance between the points $(0, 5)$ and $(-5, 0)$ is

Solution.

$$\sqrt{(-5-0)^2 + (0-5)^2} = \sqrt{25 + 25} = \sqrt{50} = 5\sqrt{2}$$

Both points are five units from the origin, one up the $y$-axis and one along the negative $x$-axis, so they are the ends of the hypotenuse of a right isosceles triangle with legs $5$ — and the hypotenuse of such a triangle is always $\sqrt{2}$ times a leg.

Answering $5$ means having found one leg and stopped; answering $10$ means having added the legs instead of using Pythagoras.

Question 5

$AOBC$ is a rectangle whose three vertices are vertices $A(0, 3)$, $O(0, 0)$ and $B(5, 0)$. The length of its diagonal is

Solution. The name carries the information. In a quadrilateral $AOBC$ the vertices are visited in that order, so the sides are $AO$, $OB$, $BC$, $CA$ and the diagonals are $AB$ and $OC$ — the pairs of vertices that are not adjacent in the name.

So $AB$ is a diagonal, not a side, and

$$AB = \sqrt{(5-0)^2 + (0-3)^2} = \sqrt{25 + 9} = \sqrt{34}$$

The fourth vertex $C$ is $(5, 3)$, and $OC = \sqrt{25 + 9} = \sqrt{34}$ as well, as it must be in a rectangle.

The wrong answers $5$ and $3$ are the two sides $OB$ and $AO$ — which is exactly what you get by treating $AOBC$ as though the letters were in the order $ABCO$.

Question 6

The perimeter of a triangle with vertices $(0, 4)$, $(0, 0)$ and $(3, 0)$ is

Solution. Two of the vertices lie on the axes and the third is the origin, so two sides run along the axes and the triangle is right angled at $(0,0)$:

• vertical leg from $(0,0)$ to $(0,4)$: length $4$
• horizontal leg from $(0,0)$ to $(3,0)$: length $3$
• hypotenuse from $(0,4)$ to $(3,0)$: $\sqrt{9 + 16} = 5$

Perimeter $= 4 + 3 + 5 = 12$.

Option (A) is the hypotenuse alone. Option (D) is what you get by computing $\sqrt{9+16}$ as $\sqrt{9} + \sqrt{16}$ somewhere and mislaying a term — a square root never distributes over a sum.

Question 7

The area of a triangle with vertices $A(3, 0)$, $B(7, 0)$ and $C(8, 4)$ is

Solution. Both $A$ and $B$ lie on the $x$-axis, so take $AB$ as the base. Its length is $7 – 3 = 4$, and the height is the perpendicular distance from $C$ to the $x$-axis, which is $4$:

$$\operatorname{ar}(\triangle ABC) = \tfrac{1}{2} \times 4 \times 4 = 8$$

By the coordinate formula, for a check:

$$\tfrac{1}{2}\left|3(0-4) + 7(4-0) + 8(0-0)\right| = \tfrac{1}{2}\left|-12 + 28\right| = 8$$

Notice that $C$ is not above the segment $AB$ at all — it sits out at $x = 8$, past $B$. That makes no difference. The area depends on the perpendicular distance to the line $AB$, not on whether the foot of that perpendicular lands inside the segment.

Option (B), $28$, is the value of $\left|3(0-4) + 7(4-0)\right|$ with the halving forgotten and a sign lost; (A) is that value halved.

This question uses the area formula, which is not in the current Class 10 textbook.

Question 8

The points $(-4, 0)$, $(4, 0)$, $(0, 3)$ are the vertices of a

Solution. All three side lengths, so that nothing is guessed:

$$\text{from } (-4,0) \text{ to } (4,0): \ 8, \qquad \text{from } (4,0) \text{ to } (0,3): \ \sqrt{16+9} = 5, \qquad \text{from } (0,3) \text{ to } (-4,0): \ \sqrt{16+9} = 5$$

Two sides equal and the third different: isosceles, which also rules out scalene and equilateral.

Is it right angled? Only if $5^2 + 5^2 = 8^2$, i.e. $50 = 64$, which is false. So (A) is out. The answer is forced.

The picture makes it obvious once drawn: the third vertex sits on the $y$-axis, exactly above the mid-point of the base, so the triangle is symmetric about the $y$-axis and the two slant sides must be equal.

Question 9

The point which divides the line segment joining the points $(7, -6)$ and $(3, 4)$ in ratio $1 : 2$ internally lies in the

Solution. With $A(7, -6)$, $B(3, 4)$ and $m_1 : m_2 = 1 : 2$:

$$x = \frac{1 \cdot 3 + 2 \cdot 7}{1 + 2} = \frac{17}{3}, \qquad y = \frac{1 \cdot 4 + 2 \cdot (-6)}{1 + 2} = \frac{-8}{3}$$

The point is $\left(\frac{17}{3}, -\frac{8}{3}\right)$: $x$ positive, $y$ negative, so it is in the fourth quadrant.

You need not finish the arithmetic. The ratio $1:2$ means the point is nearer $A(7,-6)$ than $B(3,4)$ — one-third of the way along — and $A$ itself is in the fourth quadrant. A point a third of the way from $A$ towards $B$ has not yet crossed the $x$-axis, because the crossing happens at $\tfrac{6}{10} = \tfrac{3}{5}$ of the way.

The commonest error here is swapping the weights and computing $\frac{2 \cdot 3 + 1 \cdot 7}{3} = \frac{13}{3}$ with $y = \frac{2\cdot 4 – 6}{3} = \frac{2}{3}$, which lands in the first quadrant. In $m_1 : m_2$, the first weight multiplies the second point.

Question 10

The point which lies on the perpendicular bisector of the line segment joining the points $A(-2, -5)$ and $B(2, 5)$ is

Solution. The perpendicular bisector of a segment is exactly the set of points equidistant from its two ends, and it certainly passes through the mid-point. So find the mid-point first:

$$\left(\frac{-2 + 2}{2}, \frac{-5 + 5}{2}\right) = (0, 0)$$

The mid-point is the origin, which is option (A). No further work is needed — the mid-point always lies on the perpendicular bisector.

As a check, $OA = \sqrt{4 + 25} = \sqrt{29}$ and $OB = \sqrt{4 + 25} = \sqrt{29}$. Equal, as required.

The other three fail: for $(0, 2)$, the distances are $\sqrt{4 + 49} = \sqrt{53}$ and $\sqrt{4 + 9} = \sqrt{13}$.

Question 11

The fourth vertex $D$ of a parallelogram $ABCD$ whose three vertices are $A(-2, 3)$, $B(6, 7)$ and $C(8, 3)$ is

Solution. In the parallelogram $ABCD$ the diagonals are $AC$ and $BD$, and the diagonals of a parallelogram bisect each other — so $AC$ and $BD$ share a mid-point.

$$\text{mid-point of } AC = \left(\frac{-2 + 8}{2}, \frac{3 + 3}{2}\right) = (3, 3)$$

Let $D = (x, y)$. Then the mid-point of $BD$ must also be $(3, 3)$:

$$\frac{6 + x}{2} = 3 \implies x = 0, \qquad \frac{7 + y}{2} = 3 \implies y = -1$$

So $D = (0, -1)$.

The shortcut worth memorising: for a parallelogram named in order $ABCD$,

$$D = A – B + C$$

which here gives $(-2 – 6 + 8,\ 3 – 7 + 3) = (0, -1)$ in one line. It follows from the same bisection fact, and it is the result the book’s Sample Question 4 proves in general.

Option (A), $(0, 1)$, is what the sign slip $7 + y = 6$ read as $y = 1$ produces.

Question 12

If the point $P(2, 1)$ lies on the line segment joining points $A(4, 2)$ and $B(8, 4)$, then

Solution. Measure all three lengths:

$$AP = \sqrt{(2-4)^2 + (1-2)^2} = \sqrt{5}, \qquad PB = \sqrt{(8-2)^2 + (4-1)^2} = \sqrt{45} = 3\sqrt{5}$$

$$AB = \sqrt{(8-4)^2 + (4-2)^2} = \sqrt{20} = 2\sqrt{5}$$

So $AP = \sqrt{5}$ and $AB = 2\sqrt{5}$, giving $AP = \tfrac{1}{2}AB$ — option (D). Check the others against the numbers: $\tfrac{1}{3}AB = \tfrac{2\sqrt 5}{3} \ne \sqrt 5$, so (A) fails; $AP \ne PB$, so (B) fails; $PB = 3\sqrt 5 \ne \tfrac{2\sqrt 5}{3}$, so (C) fails.

A misprint in the question. $P$ does not lie on the line segment $AB$, and the question says it does. $P$ lies on the line $AB$ — all three points satisfy $y = \tfrac{x}{2}$ — but going from $A(4,2)$ towards $B(8,4)$ the $x$-coordinate increases from $4$ to $8$, and $P$ has $x = 2$. So $P$ sits on $BA$ produced, beyond $A$ and away from $B$.

You can see it in the numbers you have just computed: if $P$ were between $A$ and $B$ then $AP + PB$ would equal $AB$, and here $\sqrt{5} + 3\sqrt{5} = 4\sqrt{5}$, which is twice $AB$. That is the signature of a point outside the segment.

None of this changes the answer, because option (D) is a statement about lengths and it is true either way. But the sentence as printed is false, and it is worth knowing that $AP + PB = AB$ is the test for “between”.

Question 13

If $P\left(\frac{a}{3}, 4\right)$ is the mid-point of the line segment joining the points $Q(-6, 5)$ and $R(-2, 3)$, then the value of $a$ is

Solution. The mid-point of $QR$ is

$$\left(\frac{-6 + (-2)}{2}, \frac{5 + 3}{2}\right) = (-4, 4)$$

The $y$-coordinates already agree — the given point has $y = 4$ and so does the mid-point, which is the question telling you your arithmetic is on track. Matching the $x$-coordinates:

$$\frac{a}{3} = -4 \implies a = -12$$

Option (A) is the trap, and it is the whole point of the question. The mid-point’s $x$-coordinate is $-4$, but that is the value of $\tfrac{a}{3}$, not of $a$. Multiply back.

Question 14

The perpendicular bisector of the line segment joining the points $A(1, 5)$ and $B(4, 6)$ cuts the $y$-axis at

Solution. Use the characterisation rather than the geometry: a point is on the perpendicular bisector of $AB$ exactly when it is equidistant from $A$ and $B$.

A point on the $y$-axis has the form $(0, y)$. Set its two distances equal — squared, to be rid of the roots:

$$(0-1)^2 + (y-5)^2 = (0-4)^2 + (y-6)^2$$

$$1 + y^2 – 10y + 25 = 16 + y^2 – 12y + 36$$

$$26 – 10y = 52 – 12y \implies 2y = 26 \implies y = 13$$

So the point is $(0, 13)$.

Option (D), $(13, 0)$, is the same number put on the wrong axis — a point on the $y$-axis has its $x$-coordinate zero. Getting $(0,-13)$ means a sign slip when moving $-12y$ across; getting $(0,12)$ usually means expanding $(y-6)^2$ as $y^2 – 12y + 26$.

Squaring both sides is safe here even though squaring is normally a dangerous move, because both distances are lengths and so both sides were non-negative to begin with.

Question 15

The coordinates of the point which is equidistant from the three vertices of the $\triangle AOB$ as shown in the Fig. 7.1 is

Solution. The figure gives $O(0,0)$, $A(0, 2y)$ on the $y$-axis and $B(2x, 0)$ on the $x$-axis, so $\triangle AOB$ is right angled at $O$ and $AB$ is its hypotenuse.

For a right-angled triangle, the point equidistant from all three vertices — the circumcentre — is the mid-point of the hypotenuse. So the answer is the mid-point of $AB$:

$$\left(\frac{0 + 2x}{2}, \frac{2y + 0}{2}\right) = (x, y)$$

If you do not want to quote that fact, prove it here in three lines. Let $P = (x, y)$; then

$$PA = \sqrt{x^2 + (y – 2y)^2} = \sqrt{x^2 + y^2}, \quad PB = \sqrt{(x – 2x)^2 + y^2} = \sqrt{x^2 + y^2}, \quad PO = \sqrt{x^2 + y^2}$$

All three equal. That is also why the coordinates in the figure were given as $2x$ and $2y$ rather than $x$ and $y$: the doubling is there purely so that the halving in the mid-point formula leaves a clean answer.

Option (C) is the mid-point of $OB$ paired with the mid-point of… nothing in particular — it is what you get by halving the answer a second time.

Question 16

A circle drawn with origin as the centre passes through $\left(\frac{13}{2}, 0\right)$. The point which does not lie in the interior of the circle is

Solution. The circle is centred at the origin and passes through $\left(\tfrac{13}{2}, 0\right)$, which is $\tfrac{13}{2}$ units from the centre. So the radius is $r = \tfrac{13}{2} = 6.5$.

A point lies in the interior exactly when its distance from the centre is strictly less than $r$. Compute all four:

$$\text{(A)}\ \sqrt{\tfrac{9}{16} + 1} = \sqrt{\tfrac{25}{16}} = \tfrac{5}{4} = 1.25 \quad < 6.5$$

$$\text{(B)}\ \sqrt{4 + \tfrac{49}{9}} = \sqrt{\tfrac{85}{9}} = \tfrac{\sqrt{85}}{3} \approx 3.07 \quad < 6.5$$

$$\text{(C)}\ \sqrt{25 + \tfrac{1}{4}} = \sqrt{\tfrac{101}{4}} = \tfrac{\sqrt{101}}{2} \approx 5.02 \quad < 6.5$$

$$\text{(D)}\ \sqrt{36 + \tfrac{25}{4}} = \sqrt{\tfrac{169}{4}} = \tfrac{13}{2} = 6.5 \quad = r$$

The first three are inside. The fourth is at a distance exactly equal to the radius, so it lies on the circle.

And a point on the circle is not in the interior. That is the whole question, and it is a good one: “does not lie in the interior” is not the same as “lies outside”. Had the question asked which point lies outside the circle, the honest answer would have been none of them.

Notice too that (D) is a rigged number — $36 + \tfrac{25}{4} = \tfrac{169}{4}$ is designed to give the exact radius, so if you are getting a messy decimal near $6.5$ you have slipped.

Question 17

A line intersects the $y$-axis and $x$-axis at the points $P$ and $Q$, respectively. If $(2, -5)$ is the mid-point of $PQ$, then the coordinates of $P$ and $Q$ are, respectively

Solution. $P$ is on the $y$-axis, so $P = (0, b)$. $Q$ is on the $x$-axis, so $Q = (a, 0)$. Their mid-point is

$$\left(\frac{0 + a}{2}, \frac{b + 0}{2}\right) = \left(\frac{a}{2}, \frac{b}{2}\right)$$

Setting this equal to $(2, -5)$:

$$\frac{a}{2} = 2 \implies a = 4, \qquad \frac{b}{2} = -5 \implies b = -10$$

So $P = (0, -10)$ and $Q = (4, 0)$.

Option (A) is the given mid-point’s own coordinates redistributed onto the two axes — the same halving-forgotten error as Question 13, and it is why that error is worth naming. When a point is the mid-point, the ends are twice as far out, not the same distance.

Note also the order the question asks for: $P$ first, and $P$ is the one on the $y$-axis. Option (B) has the right shape and the wrong signs; (C) has both coordinates doubled onto the wrong axes.

Question 18

The area of a triangle with vertices $(a, b + c)$, $(b, c + a)$ and $(c, a + b)$ is

Solution. Put the three points into the area formula and let the algebra run:

$$\operatorname{ar} = \tfrac{1}{2}\Big|a\big[(c+a) – (a+b)\big] + b\big[(a+b) – (b+c)\big] + c\big[(b+c) – (c+a)\big]\Big|$$

Each bracket simplifies to a difference of two of the letters:

$$= \tfrac{1}{2}\Big|a(c – b) + b(a – c) + c(b – a)\Big| = \tfrac{1}{2}\Big|ac – ab + ab – bc + bc – ac\Big| = \tfrac{1}{2}\left|0\right| = 0$$

Every term cancels against another. The area is $0$, which means the three points are collinear for any values of $a$, $b$ and $c$ whatsoever.

That is not an accident of the algebra, and it is worth seeing why. Each point has the form $(t,\ s – t)$ where $s = a + b + c$ is the same for all three — the first point is $(a, s-a)$, the second $(b, s-b)$, the third $(c, s-c)$. Every such point satisfies

$$x + y = s$$

and $x + y = s$ is the equation of a straight line. All three points sit on one line, so there is no triangle at all.

Since the answer is $0$ for all $a, b, c$, options (A), (C) and (D) can each be killed by a single example: take $a = 1, b = 2, c = 3$ and they give $36$, $6$ and $6$, none of which is the area.

This question uses the area formula, which is not in the current Class 10 textbook. The collinearity it establishes can be shown instead by the line $x + y = a+b+c$, as above.

Question 19

If the distance between the points $(4, p)$ and $(1, 0)$ is $5$, then the value of $p$ is

Solution.

$$\sqrt{(4-1)^2 + (p-0)^2} = 5 \implies 9 + p^2 = 25 \implies p^2 = 16 \implies p = \pm 4$$

Both values work, and the wording of the options is the whole question. Nothing in the problem says $p$ is positive, and $p^2 = 16$ has two real solutions.

Geometrically: the points at distance $5$ from $(1, 0)$ form a circle of radius $5$ about $(1,0)$, and the vertical line $x = 4$ cuts that circle twice — once above the $x$-axis at $(4, 4)$ and once below at $(4, -4)$. A line meeting a circle meets it twice unless it is tangent, and $x = 4$ is not tangent here because $4$ is strictly between $1 – 5$ and $1 + 5$.

Answering “$4$ only” is the standard reflex of taking the positive square root without asking whether the negative one is also admissible. It is admissible whenever the unknown is a coordinate rather than a length.

Question 20

If the points $A(1, 2)$, $O(0, 0)$ and $C(a, b)$ are collinear, then

Solution. The cleanest route uses the fact that the line passes through the origin. Two of the given points are $O(0,0)$ and $A(1, 2)$, so the line through them is

$$y = 2x$$

— it climbs two units for every one across, and it passes through $(0,0)$ by construction. For $C(a, b)$ to be on that same line, its coordinates must satisfy the equation:

$$b = 2a, \qquad \text{that is,} \qquad 2a = b$$

Alternatively, without writing down the line at all: if $O$, $A$, $C$ are collinear then $C$ is some multiple of $A$ as a displacement from the origin, say $C = k \cdot (1, 2) = (k, 2k)$. Then $a = k$ and $b = 2k = 2a$. Same conclusion.

Option (B), $a = 2b$, is the same relation with the letters exchanged, and it is the answer you get by matching “$2$” to the wrong coordinate. Test it: $a = 2b$ with $b = 1$ gives $C = (2, 1)$, and $(2,1)$ does not lie on $y = 2x$ since $1 \ne 4$. Whereas $2a = b$ with $a = 1$ gives $C = (1, 2)$, which is the point $A$ itself — trivially collinear, as it should be.

Most published solutions do this one with the area formula, setting $\operatorname{ar}(\triangle AOC) = 0$. That formula is not in the current textbook; the route above needs only the equation of a line, and the textbook’s own route — checking that the two shorter distances sum to the longest — would work too.

Common mistakes

• Question 1 — reading the coordinate off the wrong axis. The distance from the $x$-axis is $|y|$, and the distance from the $y$-axis is $|x|$. Each coordinate measures the distance from the other axis, because it counts how far you travel along its own.
• Question 5 — treating $AOBC$ as if it read $ABCO$. A quadrilateral’s name lists its vertices in order round the shape, so in $AOBC$ the sides are $AO$, $OB$, $BC$, $CA$ and the diagonals are $AB$ and $OC$. Both wrong options here, $5$ and $3$, are sides.
• Questions 13 and 17 — finding the mid-point and stopping. In Q13 the mid-point’s $x$-coordinate is $-4$, but that is $\tfrac{a}{3}$, so $a = -12$; in Q17 the mid-point is $(2,-5)$, but the endpoints on the axes are at $4$ and $-10$. Whenever the unknown is buried inside an expression, finish by unwrapping it.
• Question 16 — equating “not in the interior” with “outside”. The point in (D) is at a distance of exactly $6.5$ from the centre, which is the radius: it is on the circle. A circle divides the plane into three sets, not two, and questions about interiors are usually testing the boundary case.
• Question 19 — taking only the positive square root. $p^2 = 16$ gives $p = \pm 4$ because $p$ is a coordinate, and coordinates can be negative. Discard the negative root only when the unknown is a length, a radius or a count.
• Question 9 — putting the ratio weights on the wrong points. In the section formula for $m_1 : m_2$, the first weight $m_1$ multiplies the coordinates of the second point $B$. Swapping them here moves the answer from the fourth quadrant to the first — every option in the question is a quadrant, so the slip is invisible unless you check.
• Question 12 — trusting that a point said to be on a segment really is. $AP + PB = AB$ is the test. Here $\sqrt5 + 3\sqrt5 = 4\sqrt5$ against $AB = 2\sqrt5$, so $P$ is on the line but outside the segment, and the question’s own wording is wrong.

Practise next

• Exercise 7.2 — twelve true-or-false statements on the same three formulae, where the work is in the justification rather than the verdict. Several of them are Question 12 in disguise: a claim about a point lying on a segment, or inside a circle, that has to be tested rather than assumed.
• Exercise 7.3 — twenty short-answer questions, where these ideas run in reverse: instead of being handed a point and asked what is true of it, you are handed a condition and asked to find the point.
• Exercise 6.1 — worth revisiting if Question 8 was slow. Classifying a triangle from its side lengths is the same skill whether the lengths come from a figure or from the distance formula.