NCERT Exemplar Solutions Class 10 Maths Coordinate Geometry — Exercise 7.2

NCERT Exemplar Problems, Class 10 Mathematics — Coordinate Geometry, Exercise 7.2. All 12 true-or-false questions solved.

Twelve statements to be judged true or false. Guessing gets you half of them, so the marks are all in the second half of the instruction — justify your answer — and that is what this page is really about.

The justification is almost always one of four calculations, and knowing which one the statement calls for is most of the work:

• a claim about equal distances (perpendicular bisector, equidistant point, isosceles, circle) → compute the two distances and compare
• a claim about collinearity or “cannot be a triangle” → show the two shorter distances add to the longest
• a claim about a parallelogram or rectangle → show the diagonals share a mid-point, and for a rectangle that they are also equal
• a claim about a point lying on a segment in some ratio → run the section formula forwards and see whether it lands on the point

A note on the current syllabus. The rationalised Class 10 textbook has no “Area of a Triangle” section — the chapter is Introduction, Distance Formula, Section Formula, Summary, and nothing else. Most published solutions test collinearity by setting the area of the triangle to zero. That route is closed to you now, so the collinearity questions here (3, 5 and 10) are done the way the textbook’s own Example 3 does it: three points are collinear exactly when the two shorter of the three distances between them sum to the longest.

Key insight. A false statement is only worth marks if you can name the number that kills it. “False, because the points are not collinear” earns nothing; “False, because the area is $21$, not zero” — or better, “because $AB + BC \ne AC$” — earns everything. Aim every justification at a specific quantity.

State whether the following statements are true or false. Justify your answer.

Question 1

$\triangle ABC$ with vertices $A(-2, 0)$, $B(2, 0)$ and $C(0, 2)$ is similar to $\triangle DEF$ with vertices $D(-4, 0)$, $E(4, 0)$ and $F(0, 4)$.

Solution. Similar triangles need their three pairs of corresponding sides in the same ratio, so compute all six lengths before saying anything.

For $\triangle ABC$:

$$AB = \sqrt{(2+2)^2 + 0} = 4, \qquad BC = \sqrt{(0-2)^2 + (2-0)^2} = 2\sqrt{2}, \qquad CA = \sqrt{(-2-0)^2 + (0-2)^2} = 2\sqrt{2}$$

For $\triangle DEF$:

$$DE = \sqrt{(4+4)^2 + 0} = 8, \qquad EF = \sqrt{(0-4)^2 + (4-0)^2} = 4\sqrt{2}, \qquad FD = \sqrt{(-4-0)^2 + (0-4)^2} = 4\sqrt{2}$$

Now the ratios, taken in the order the naming $\triangle ABC \sim \triangle DEF$ demands:

$$\frac{AB}{DE} = \frac{4}{8} = \frac{1}{2}, \qquad \frac{BC}{EF} = \frac{2\sqrt2}{4\sqrt2} = \frac{1}{2}, \qquad \frac{CA}{FD} = \frac{2\sqrt2}{4\sqrt2} = \frac{1}{2}$$

All three equal, so the triangles are similar by the SSS similarity criterion.

The underlying reason is that $D$, $E$, $F$ are just $A$, $B$, $C$ with both coordinates doubled — the second triangle is an enlargement of the first about the origin with scale factor $2$, and an enlargement always produces a similar figure.

Question 2

Point $P(-4, 2)$ lies on the line segment joining the points $A(-4, 6)$ and $B(-4, -6)$.

Solution. All three points have $x = -4$, so all three lie on the vertical line $x = -4$. Being on the line is necessary but not sufficient — the claim is about the segment, so $P$ must also lie between $A$ and $B$.

On the line $x = -4$ the position is decided entirely by the $y$-coordinate. $A$ has $y = 6$, $B$ has $y = -6$, and $P$ has $y = 2$. Since

$$-6 < 2 < 6$$

$P$ is between them, and so lies on the segment.

The formal check is the same one that catches the fakes: $P$ lies on the segment exactly when $AP + PB = AB$. Here

$$AP = |6 – 2| = 4, \qquad PB = |2 – (-6)| = 8, \qquad AB = |6 – (-6)| = 12$$

and $4 + 8 = 12$. Confirmed.

Question 3

The points $(0, 5)$, $(0, -9)$ and $(3, 6)$ are collinear.

Solution. Look at the $x$-coordinates before computing anything. The first two points have $x = 0$, so both lie on the $y$-axis. If all three were collinear, the line through them would have to be the $y$-axis itself — a line is fixed by two points, and the only line through $(0,5)$ and $(0,-9)$ is $x = 0$.

But the third point is $(3, 6)$, and $3 \ne 0$. So it is not on the $y$-axis, and the three are not collinear.

If you prefer the distance argument, name the points $A(0,5)$, $B(0,-9)$, $C(3,6)$:

$$AB = 14, \qquad AC = \sqrt{9 + 1} = \sqrt{10} \approx 3.16, \qquad BC = \sqrt{9 + 225} = \sqrt{234} \approx 15.30$$

The longest is $BC$, and $AB + AC = 14 + 3.16 = 17.16 \ne 15.30$. Not collinear.

Question 4

Point $P(0, 2)$ is the point of intersection of $y$-axis and perpendicular bisector of line segment joining the points $A(-1, 1)$ and $B(3, 3)$.

Solution. $P$ is certainly on the $y$-axis. The question is whether it is on the perpendicular bisector of $AB$, and a point is on that bisector exactly when it is equidistant from $A$ and $B$. So measure both:

$$PA = \sqrt{(0+1)^2 + (2-1)^2} = \sqrt{2}, \qquad PB = \sqrt{(0-3)^2 + (2-3)^2} = \sqrt{10}$$

These are not equal, so $P$ is not on the perpendicular bisector, and the statement is false.

Where does the bisector actually meet the $y$-axis? Take $(0, y)$ and force the distances equal:

$$1 + (y-1)^2 = 9 + (y-3)^2 \implies y^2 – 2y + 2 = y^2 – 6y + 18 \implies 4y = 16 \implies y = 4$$

So the correct point is $(0, 4)$, not $(0, 2)$. It is worth seeing where $(0,2)$ came from: it is the mid-point of $AB$ shifted onto the axis, or a half-remembered version of it. The mid-point of $AB$ is $(1, 2)$, which does lie on the bisector — but it is not on the $y$-axis.

Question 5

Points $A(3, 1)$, $B(12, -2)$ and $C(0, 2)$ cannot be the vertices of a triangle.

Solution. Three points fail to make a triangle exactly when they are collinear, so this is a collinearity test wearing different clothes.

$$AB = \sqrt{81 + 9} = \sqrt{90} = 3\sqrt{10}, \qquad BC = \sqrt{144 + 16} = \sqrt{160} = 4\sqrt{10}, \qquad CA = \sqrt{9 + 1} = \sqrt{10}$$

Now check whether the two shorter add to the longest:

$$CA + AB = \sqrt{10} + 3\sqrt{10} = 4\sqrt{10} = BC$$

They do. So $A$ lies between $C$ and $B$ on a straight line, the three points are collinear, and there is no triangle.

Seeing it another way: the line through $C(0,2)$ and $A(3,1)$ drops one unit for every three across, so its equation is $y = 2 – \tfrac{x}{3}$. At $x = 12$ that gives $y = 2 – 4 = -2$, which is exactly $B$. Confirmed.

The verdict is subtle to state. The statement says they cannot be the vertices of a triangle, and that is true — so the answer here is True, even though the underlying geometric fact is a failure.

Question 6

Points $A(4, 3)$, $B(6, 4)$, $C(5, -6)$ and $D(-3, 5)$ are the vertices of a parallelogram.

Solution. In a quadrilateral named $ABCD$ the diagonals are $AC$ and $BD$, and the shape is a parallelogram exactly when those two diagonals bisect each other — that is, when they share a mid-point.

$$\text{mid-point of } AC = \left(\frac{4+5}{2}, \frac{3-6}{2}\right) = \left(\frac{9}{2}, -\frac{3}{2}\right)$$

$$\text{mid-point of } BD = \left(\frac{6-3}{2}, \frac{4+5}{2}\right) = \left(\frac{3}{2}, \frac{9}{2}\right)$$

Nowhere near each other. The diagonals do not bisect each other, so $ABCD$ is not a parallelogram.

The side lengths tell the same story if you would rather check those: $AB = \sqrt{4+1} = \sqrt5$ while $CD = \sqrt{64 + 121} = \sqrt{185}$, and opposite sides of a parallelogram are equal.

The mid-point test is the better one to reach for, because it needs two short calculations rather than four, and because it is the test that also settles rectangles, rhombuses and squares once you add a second condition.

Question 7

A circle has its centre at the origin and a point $P(5, 0)$ lies on it. The point $Q(6, 8)$ lies outside the circle.

Solution. The radius is the distance from the centre to any point on the circle:

$$r = OP = \sqrt{5^2 + 0^2} = 5$$

A point lies outside the circle when its distance from the centre exceeds the radius. For $Q$:

$$OQ = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10$$

And $10 > 5$, so $Q$ is outside.

Compare this with Question 16 of Exercise 7.1, where the distance came out exactly equal to the radius and the point was therefore on the circle rather than outside it. The three cases are worth keeping distinct in your head: distance less than $r$ means inside, equal means on, greater means outside.

Question 8

The point $A(2, 7)$ lies on the perpendicular bisector of line segment joining the points $P(6, 5)$ and $Q(0, -4)$.

Solution. Again the test is equal distances:

$$AP = \sqrt{(2-6)^2 + (7-5)^2} = \sqrt{16 + 4} = \sqrt{20} = 2\sqrt{5}$$

$$AQ = \sqrt{(2-0)^2 + (7+4)^2} = \sqrt{4 + 121} = \sqrt{125} = 5\sqrt{5}$$

$2\sqrt5 \ne 5\sqrt5$, so $A$ is not equidistant from $P$ and $Q$, and therefore does not lie on the perpendicular bisector.

Both lengths landing on a multiple of $\sqrt5$ is a coincidence of the numbers, not a sign of anything. What matters is that $\sqrt{20}$ and $\sqrt{125}$ differ, and they differ by a lot — the point is two and a half times as far from $Q$ as from $P$.

Question 9

Point $P(5, -3)$ is one of the two points of trisection of the line segment joining the points $A(7, -2)$ and $B(1, -5)$.

Solution. The two points of trisection divide $AB$ in the ratios $1:2$ and $2:1$. Compute the first:

$$\left(\frac{1 \cdot 1 + 2 \cdot 7}{1 + 2}, \frac{1 \cdot (-5) + 2 \cdot (-2)}{1 + 2}\right) = \left(\frac{15}{3}, \frac{-9}{3}\right) = (5, -3)$$

That is $P$ exactly. So yes, $P$ is a point of trisection — the one nearer $A$.

For completeness the other one, dividing in $2:1$, is

$$\left(\frac{2 \cdot 1 + 1 \cdot 7}{3}, \frac{2 \cdot (-5) + 1 \cdot (-2)}{3}\right) = (3, -4)$$

Notice how much easier it is to run the section formula forwards — plug in the ratio, get a point, compare — than to start from $P$ and solve for the ratio. Whenever a statement names a specific ratio, test it that way round.

Question 10

Points $A(-6, 10)$, $B(-4, 6)$ and $C(3, -8)$ are collinear such that $AB = \frac{2}{9}AC$.

Solution. Two claims here, and both need checking.

Collinear? Compute the three distances:

$$AB = \sqrt{4 + 16} = \sqrt{20} = 2\sqrt{5}, \qquad BC = \sqrt{49 + 196} = \sqrt{245} = 7\sqrt{5}, \qquad AC = \sqrt{81 + 324} = \sqrt{405} = 9\sqrt{5}$$

$$AB + BC = 2\sqrt5 + 7\sqrt5 = 9\sqrt5 = AC$$

The two shorter sum to the longest, so the points are collinear with $B$ between $A$ and $C$.

The ratio? From those same lengths,

$$\frac{AB}{AC} = \frac{2\sqrt5}{9\sqrt5} = \frac{2}{9}$$

so $AB = \tfrac{2}{9}AC$ as claimed. Both parts hold.

Equivalently — and this is how the answer key puts it — $B$ divides $AC$ in the ratio $2 : 7$. Check with the section formula:

$$\left(\frac{2 \cdot 3 + 7 \cdot (-6)}{9}, \frac{2 \cdot (-8) + 7 \cdot 10}{9}\right) = \left(\frac{6 – 42}{9}, \frac{-16 + 70}{9}\right) = (-4, 6) = B$$

These are not new numbers, incidentally. This is the textbook’s own Example 7 in Chapter 7 — the same three points and the same $2:7$ ratio — so of all twelve statements here, this is the one most likely to appear on a paper.

Question 11

The point $P(-2, 4)$ lies on a circle of radius $6$ and centre $C(3, 5)$.

Solution. For $P$ to lie on the circle, its distance from the centre must equal the radius exactly.

$$PC = \sqrt{(-2-3)^2 + (4-5)^2} = \sqrt{25 + 1} = \sqrt{26} \approx 5.10$$

But the radius is $6$, and $\sqrt{26} \ne 6$ — indeed $\sqrt{26} < \sqrt{36} = 6$. So $P$ does not lie on the circle. It lies inside it, a little under a unit short of the boundary.

Note how narrowly this one fails. $\sqrt{26}$ and $6$ are close enough that an approximate calculation might not separate them, which is why the comparison should be made by squaring rather than by taking roots: $26$ against $36$ is unmistakable, $5.10$ against $6$ invites doubt.

Question 12

The points $A(-1, -2)$, $B(4, 3)$, $C(2, 5)$ and $D(-3, 0)$ in that order form a rectangle.

Solution. A rectangle is a parallelogram with equal diagonals, so there are two things to establish.

Do the diagonals bisect each other? The diagonals of $ABCD$ are $AC$ and $BD$:

$$\text{mid-point of } AC = \left(\frac{-1+2}{2}, \frac{-2+5}{2}\right) = \left(\frac{1}{2}, \frac{3}{2}\right)$$

$$\text{mid-point of } BD = \left(\frac{4-3}{2}, \frac{3+0}{2}\right) = \left(\frac{1}{2}, \frac{3}{2}\right)$$

Identical, so $ABCD$ is a parallelogram.

Are the diagonals equal?

$$AC = \sqrt{(2+1)^2 + (5+2)^2} = \sqrt{9 + 49} = \sqrt{58}, \qquad BD = \sqrt{(-3-4)^2 + (0-3)^2} = \sqrt{49 + 9} = \sqrt{58}$$

Equal. A parallelogram with equal diagonals is a rectangle, so the statement is true.

Is it perhaps a square? The sides settle that:

$$AB = \sqrt{25 + 25} = 5\sqrt{2}, \qquad BC = \sqrt{4 + 4} = 2\sqrt{2}$$

Adjacent sides differ, so it is a rectangle and not a square. The order matters too — “in that order” is doing real work, because the same four points listed as $A$, $B$, $D$, $C$ would trace a crossed figure and not a rectangle at all.

Common mistakes

• Questions 2 and 9 — confusing “on the line” with “on the segment”. Three points can share a line and still not have the middle one between the other two. The test that settles it is $AP + PB = AB$, and it costs three distance calculations.
• Questions 3, 5 and 10 — reaching for the area formula. It is no longer in the Class 10 textbook, so build the habit now of testing collinearity by the distance sum instead. It is also the more informative test, because it tells you which point is in the middle — information the area formula throws away.
• Questions 4 and 8 — checking the wrong property of the perpendicular bisector. Students often try to find its equation, or its gradient, and get lost. A point is on the perpendicular bisector of $PQ$ if and only if it is equidistant from $P$ and $Q$: two distances, one comparison, done.
• Questions 7 and 11 — comparing roots instead of squares. $\sqrt{26}$ against $6$ is an invitation to a decimal slip. Square both sides first: $26$ against $36$ leaves no room for doubt, and the inequality direction is preserved because both quantities are positive.
• Question 6 — testing sides when the diagonals are quicker. Equal opposite sides do confirm a parallelogram, but that is four distance calculations against two mid-points. The mid-point test is also the one that extends cleanly to rectangles, as Question 12 shows.
• Question 5 — answering the geometry instead of the statement. The points are collinear, which feels like a negative result, but the statement claimed they “cannot be the vertices of a triangle” — so the verdict is True. Read what was asserted, then decide, in that order.
• Question 1 — declaring similarity from a single pair of sides. Two sides in the ratio $1:2$ prove nothing on their own. All three pairs are needed for SSS, and they must be paired according to the vertex correspondence named in the statement.

Practise next

• Exercise 7.3 — twenty short-answer questions. Everything you have just verified here gets run in reverse: instead of testing whether a given point satisfies a condition, you are given the condition and have to produce the point.
• Exercise 7.1 — worth a second look if any verdict here surprised you, particularly Question 16 on interiors and Question 12 on segments. The same two distinctions are being tested in both exercises.
• Exercise 6.2 — for the similarity criteria that Question 1 leans on. Knowing SSS, SAS and AAA apart is assumed here and examined there.