NCERT Exemplar Solutions Class 10 Maths Coordinate Geometry — Exercises 7.3 and 7.4

NCERT Exemplar Problems, Class 10 Mathematics — Coordinate Geometry, Exercises 7.3 and 7.4. All 26 questions solved.

Exercise 7.3 is twenty short-answer questions and Exercise 7.4 is six long ones, and between them they cover everything the chapter can ask. The change from the first two exercises is that nothing is handed to you finished: you are given a condition and asked to produce the point, the ratio or the value that satisfies it.

That reverses the method. In Exercises 7.1 and 7.2 you substituted known coordinates into a formula and read off a number. Here you substitute an unknown and solve:

$$\text{unknown point } (x, y) \ \longrightarrow \ \text{impose the condition} \ \longrightarrow \ \text{solve the resulting equation}$$

The three standard set-ups, which between them handle about twenty of these twenty-six:

• a point on an axis is $(x, 0)$ or $(0, y)$ — one unknown, so one equation suffices
• an unknown ratio is $k : 1$ rather than $m_1 : m_2$ — one unknown instead of two, and the answer $k : 1$ can be cleared of fractions at the end
• an equidistant point gives $PA^2 = PB^2$, squared from the start so the roots never appear

A note on the current syllabus. The rationalised Class 10 textbook runs Introduction → Distance Formula → Section Formula → Summary. There is no “Area of a Triangle” section, so Questions 9, 16 and 17 of Exercise 7.3 and Questions 2 and 4 of Exercise 7.4 test material the book no longer carries. They are all still worth doing — the formula is straightforward and it reappears in Class 11 — but if you are revising strictly to the current syllabus, those five are the ones to leave until last. Where a collinearity question can be done without the area formula, this page does it without.

Key insight. When a question asks for a ratio, always write it $k : 1$. You then have a single unknown, a single linear equation, and no simultaneous system — and a negative $k$ is not an error but a message, telling you the division is external.

Exercise 7.3

Question 1

Name the type of triangle formed by the points $A(-5, 6)$, $B(-4, -2)$ and $C(7, 5)$.

Solution. Classification needs all three sides, and then a check for a right angle. Compute the squares and hold off on the roots — the comparisons are easier before the square roots go in.

$$AB^2 = (-4+5)^2 + (-2-6)^2 = 1 + 64 = 65$$

$$BC^2 = (7+4)^2 + (5+2)^2 = 121 + 49 = 170$$

$$CA^2 = (-5-7)^2 + (6-5)^2 = 144 + 1 = 145$$

All three different, so no two sides are equal: the triangle is scalene.

Is it right angled? That would need the two smaller squares to sum to the largest: $65 + 145 = 210$, and the largest is $170$. Not equal, so no right angle. (The other two pairings, $65 + 170 = 235$ and $145 + 170 = 315$, are nowhere near $145$ or $65$ either.)

And it is genuinely a triangle — the points are not collinear, since $\sqrt{65} + \sqrt{145} \approx 8.06 + 12.04 = 20.10$, which exceeds $\sqrt{170} \approx 13.04$.

Question 2

Find the points on the $x$-axis which are at a distance of $2\sqrt{5}$ from the point $(7, -4)$. How many such points are there?

Solution. A point on the $x$-axis has $y = 0$, so write it $(x, 0)$ — one unknown. Impose the distance condition, squared:

$$(x – 7)^2 + (0 + 4)^2 = \left(2\sqrt5\right)^2 = 20$$

$$(x-7)^2 + 16 = 20 \implies (x-7)^2 = 4 \implies x – 7 = \pm 2$$

So $x = 9$ or $x = 5$, giving the points $(9, 0)$ and $(5, 0)$.

There are two, and the geometry says why. The circle of radius $2\sqrt5 \approx 4.47$ centred at $(7,-4)$ reaches $4.47$ units in every direction, and the $x$-axis is only $4$ units above the centre. Since $4 < 4.47$, the axis cuts the circle, and a line that cuts a circle cuts it in exactly two places.

Had the distance been exactly $4$, there would be one point; had it been less than $4$, none.

Question 3

What type of a quadrilateral do the points $A(2, -2)$, $B(7, 3)$, $C(11, -1)$ and $D(6, -6)$ taken in that order, form?

Solution. Four sides and two diagonals settle every case. Sides first:

$$AB = \sqrt{25 + 25} = 5\sqrt2, \quad BC = \sqrt{16 + 16} = 4\sqrt2, \quad CD = \sqrt{25 + 25} = 5\sqrt2, \quad DA = \sqrt{16 + 16} = 4\sqrt2$$

Opposite sides are equal in pairs, so $ABCD$ is at least a parallelogram. Now the diagonals:

$$AC = \sqrt{81 + 1} = \sqrt{82}, \qquad BD = \sqrt{1 + 81} = \sqrt{82}$$

Equal. A parallelogram with equal diagonals has right angles at every vertex, so $ABCD$ is a rectangle.

Not a square, because adjacent sides $5\sqrt2$ and $4\sqrt2$ differ.

For extra security, confirm the diagonals bisect each other: the mid-point of $AC$ is $\left(\frac{13}{2}, -\frac{3}{2}\right)$ and the mid-point of $BD$ is $\left(\frac{13}{2}, -\frac{3}{2}\right)$. The same point.

Question 4

Find the value of $a$, if the distance between the points $A(-3, -14)$ and $B(a, -5)$ is $9$ units.

Solution.

$$(a + 3)^2 + (-5 + 14)^2 = 81$$

$$(a+3)^2 + 81 = 81 \implies (a+3)^2 = 0 \implies a = -3$$

Only one value, and that is the interesting part. The two points already differ by $9$ in the $y$-direction alone — from $-14$ up to $-5$ — so the entire permitted distance is used up vertically and nothing is left over for a horizontal offset. $A$ and $B$ must lie on the same vertical line.

This is worth flagging because distance questions almost always produce a $\pm$ pair, and a student who expects two answers will hunt for a second root that does not exist. The signal is $(a+3)^2 = 0$: a perfect square equal to zero has a repeated root, not two roots.

Question 5

Find a point which is equidistant from the points $A(-5, 4)$ and $B(-1, 6)$? How many such points are there?

Solution. Let the point be $P(x, y)$ and impose $PA^2 = PB^2$:

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

$$x^2 + 10x + 25 + y^2 – 8y + 16 = x^2 + 2x + 1 + y^2 – 12y + 36$$

The $x^2$ and $y^2$ terms cancel — they always do, which is why this method is reliable — leaving

$$10x – 8y + 41 = 2x – 12y + 37 \implies 8x + 4y + 4 = 0 \implies 2x + y + 1 = 0$$

That is not a point but a line: the perpendicular bisector of $AB$. So there are infinitely many such points, one for every solution of $2x + y + 1 = 0$.

A convenient one to name is the mid-point of $AB$:

$$\left(\frac{-5-1}{2}, \frac{4+6}{2}\right) = (-3, 5)$$

and indeed $2(-3) + 5 + 1 = 0$, so it lies on the line as it must.

The question asks “how many”, which is the whole point of it. Do not stop at one answer and assume it is the answer.

Question 6

Find the coordinates of the point $Q$ on the $x$-axis which lies on the perpendicular bisector of the line segment joining the points $A(-5, -2)$ and $B(4, -2)$. Name the type of triangle formed by the points $Q$, $A$ and $B$.

Solution. $A$ and $B$ share the $y$-coordinate $-2$, so $AB$ is horizontal and its perpendicular bisector is the vertical line through the mid-point of $AB$:

$$x = \frac{-5 + 4}{2} = -\frac{1}{2}$$

A point on the $x$-axis has $y = 0$, so $Q = \left(-\dfrac{1}{2}, 0\right)$.

Now the triangle. $Q$ is on the perpendicular bisector of $AB$, which means $QA = QB$ — so the triangle is isosceles before any arithmetic is done. Confirming:

$$QA = \sqrt{\left(-\tfrac12 + 5\right)^2 + 4} = \sqrt{\tfrac{81}{4} + 4} = \sqrt{\tfrac{97}{4}} = \frac{\sqrt{97}}{2}$$

$$QB = \sqrt{\left(-\tfrac12 – 4\right)^2 + 4} = \sqrt{\tfrac{81}{4} + 4} = \frac{\sqrt{97}}{2}$$

Equal, as promised. The third side is $AB = 4 – (-5) = 9$. Since $\frac{\sqrt{97}}{2} \approx 4.92 \ne 9$, the triangle is isosceles and not equilateral.

Question 7

Find the value of $m$ if the points $(5, 1)$, $(-2, -3)$ and $(8, 2m)$ are collinear.

Solution. Two points fix a line, so find the line through the first two and then force the third onto it.

From $(5, 1)$ to $(-2, -3)$ the rise is $-3 – 1 = -4$ and the run is $-2 – 5 = -7$, so the gradient is

$$\frac{-4}{-7} = \frac{4}{7}$$

The line through $(5, 1)$ with that gradient is $y – 1 = \frac{4}{7}(x – 5)$. Substituting $x = 8$:

$$y = 1 + \frac{4}{7}(3) = 1 + \frac{12}{7} = \frac{19}{7}$$

The third point’s $y$-coordinate is $2m$, so

$$2m = \frac{19}{7} \implies m = \frac{19}{14}$$

Do not stop at $\frac{19}{7}$. The unknown is $m$, and the point’s ordinate is $2m$ — the same unwrapping step that Exercise 7.1 tested twice.

Most solutions do this one by setting the area of the triangle to zero. The gradient route above uses only the equation of a line, which is what the current textbook gives you.

Question 8

If the point $A(2, -4)$ is equidistant from $P(3, 8)$ and $Q(-10, y)$, find the values of $y$. Also find distance $PQ$.

Solution. Equidistant means $AP = AQ$, so square both from the start:

$$AP^2 = (3-2)^2 + (8+4)^2 = 1 + 144 = 145$$

$$AQ^2 = (-10-2)^2 + (y+4)^2 = 144 + (y+4)^2$$

Setting them equal:

$$144 + (y+4)^2 = 145 \implies (y+4)^2 = 1 \implies y + 4 = \pm 1$$

So $y = -3$ or $y = -5$. Two positions for $Q$, and the question asks for $PQ$, so there are two answers there as well.

With $y = -3$, $Q = (-10, -3)$:

$$PQ = \sqrt{(-10-3)^2 + (-3-8)^2} = \sqrt{169 + 121} = \sqrt{290}$$

With $y = -5$, $Q = (-10, -5)$:

$$PQ = \sqrt{(-10-3)^2 + (-5-8)^2} = \sqrt{169 + 169} = \sqrt{338} = 13\sqrt{2}$$

Both are valid; the question has two solutions and both must be reported. Keep the pairing straight — $y = -3$ goes with $\sqrt{290}$, and $y = -5$ with $13\sqrt2$. The printed answer lists the two $y$ values and the two distances as separate lists, in that same order.

Question 9

Find the area of the triangle whose vertices are $(-8, 4)$, $(-6, 6)$ and $(-3, 9)$.

Solution.

$$\operatorname{ar} = \tfrac12\Big|(-8)(6 – 9) + (-6)(9 – 4) + (-3)(4 – 6)\Big| = \tfrac12\Big|24 – 30 + 6\Big| = \tfrac12\left|0\right| = 0$$

The area is zero, which means the three points are collinear — there is no triangle at all.

The question does not warn you, and the honest answer is to say so rather than to report “$0$ square units” as if a flat triangle were a triangle. Check it directly: from $(-8,4)$ to $(-6,6)$ the gradient is $\frac{2}{2} = 1$, and from $(-6,6)$ to $(-3,9)$ it is $\frac{3}{3} = 1$. The same gradient through a shared point means one line, $y = x + 12$, and all three points satisfy it.

This question uses the area formula, which is not in the current Class 10 textbook. The collinearity can be established from the gradients, as above, or from $AB + BC = 2\sqrt2 + 3\sqrt2 = 5\sqrt2 = AC$.

Question 10

In what ratio does the $x$-axis divide the line segment joining the points $(-4, -6)$ and $(-1, 7)$? Find the coordinates of the point of division.

Solution. Let the ratio be $k : 1$, so that the point of division is

$$\left(\frac{k(-1) + 1(-4)}{k+1},\ \frac{k(7) + 1(-6)}{k+1}\right)$$

The point lies on the $x$-axis, so its $y$-coordinate is zero:

$$\frac{7k – 6}{k + 1} = 0 \implies 7k – 6 = 0 \implies k = \frac{6}{7}$$

The ratio is $\frac{6}{7} : 1$, which clears to $6 : 7$. It is positive, confirming the division is internal — as it must be, since the two points lie on opposite sides of the $x$-axis and the segment therefore crosses it.

Now the $x$-coordinate, with $k = \frac{6}{7}$:

$$x = \frac{\frac{6}{7}(-1) – 4}{\frac{6}{7} + 1} = \frac{-\frac{6}{7} – 4}{\frac{13}{7}} = \frac{-\frac{34}{7}}{\frac{13}{7}} = -\frac{34}{13}$$

Question 11

Find the ratio in which the point $P\left(\dfrac{3}{4}, \dfrac{5}{12}\right)$ divides the line segment joining the points $A\left(\dfrac{1}{2}, \dfrac{3}{2}\right)$ and $B(2, -5)$.

Solution. The fractions look forbidding but change nothing. Let the ratio be $k : 1$ and use the $x$-coordinates alone, since one equation determines one unknown:

$$\frac{2k + \frac{1}{2}}{k + 1} = \frac{3}{4}$$

Cross-multiplying:

$$4\left(2k + \tfrac12\right) = 3(k+1) \implies 8k + 2 = 3k + 3 \implies 5k = 1 \implies k = \frac{1}{5}$$

So the ratio is $\frac{1}{5} : 1$, that is $1 : 5$.

Always verify with the coordinate you did not use — it is free, and it catches arithmetic slips that would otherwise go unnoticed:

$$y = \frac{\frac15(-5) + \frac32}{\frac15 + 1} = \frac{-1 + \frac32}{\frac65} = \frac{\frac12}{\frac65} = \frac{1}{2} \cdot \frac{5}{6} = \frac{5}{12} \ \checkmark$$

That matches $P$’s $y$-coordinate, so the answer is right — and, incidentally, it confirms that $P$ really does lie on the segment. Had the check failed, $P$ would have been off the line altogether.

Question 12

If $P(9a – 2, -b)$ divides line segment joining $A(3a + 1, -3)$ and $B(8a, 5)$ in the ratio $3 : 1$, find the values of $a$ and $b$.

Solution. The ratio is given, so run the section formula forwards and match coordinates. With $m_1 : m_2 = 3 : 1$:

$$x_P = \frac{3(8a) + 1(3a+1)}{4} = \frac{24a + 3a + 1}{4} = \frac{27a + 1}{4}$$

Setting this equal to the given $x_P = 9a – 2$:

$$\frac{27a + 1}{4} = 9a – 2 \implies 27a + 1 = 36a – 8 \implies 9a = 9 \implies a = 1$$

Now the $y$-coordinate, which does not involve $a$ at all:

$$y_P = \frac{3(5) + 1(-3)}{4} = \frac{12}{4} = 3$$

and the given $y_P$ is $-b$, so $-b = 3$ and $b = -3$.

The sign is the trap. The point is named $P(9a-2, -b)$, not $P(9a-2, b)$, so finding that the ordinate is $3$ gives $b = -3$, not $b = 3$.

Question 13

If $(a, b)$ is the mid-point of the line segment joining the points $A(10, -6)$ and $B(k, 4)$ and $a – 2b = 18$, find the value of $k$ and the distance $AB$.

Solution. Write down what the mid-point is:

$$a = \frac{10 + k}{2}, \qquad b = \frac{-6 + 4}{2} = -1$$

Note that $b$ is fully determined straight away — it does not involve $k$ — which is what makes the extra condition usable. Substituting into $a – 2b = 18$:

$$a – 2(-1) = 18 \implies a + 2 = 18 \implies a = 16$$

Then from $a = \frac{10+k}{2}$:

$$\frac{10 + k}{2} = 16 \implies 10 + k = 32 \implies k = 22$$

So $B = (22, 4)$, and

$$AB = \sqrt{(22 – 10)^2 + (4 + 6)^2} = \sqrt{144 + 100} = \sqrt{244} = 2\sqrt{61}$$

$\sqrt{244}$ must be simplified: $244 = 4 \times 61$, and $61$ is prime, so $2\sqrt{61}$ is as far as it goes.

Question 14

The centre of a circle is $(2a, a – 7)$. Find the values of $a$ if the circle passes through the point $(11, -9)$ and has diameter $10\sqrt{2}$ units.

Solution. The diameter is $10\sqrt2$, so the radius is $5\sqrt2$ and

$$r^2 = \left(5\sqrt2\right)^2 = 50$$

Halving the diameter is the step most often skipped; using $10\sqrt2$ as the radius gives $r^2 = 200$ and no sensible answer.

The given point is on the circle, so its distance from the centre is $r$:

$$(2a – 11)^2 + \big[(a – 7) – (-9)\big]^2 = 50$$

$$(2a-11)^2 + (a + 2)^2 = 50$$

$$4a^2 – 44a + 121 + a^2 + 4a + 4 = 50$$

$$5a^2 – 40a + 125 = 50 \implies 5a^2 – 40a + 75 = 0 \implies a^2 – 8a + 15 = 0$$

$$(a – 3)(a – 5) = 0 \implies a = 3 \text{ or } a = 5$$

Both are genuine. With $a = 3$ the centre is $(6, -4)$; with $a = 5$ it is $(10, -2)$. Check the second: $(11-10)^2 + (-9+2)^2 = 1 + 49 = 50$. Correct. Two different circles of the same radius pass through the given point with a centre of the required form, and nothing in the question rules either out.

Question 15

The line segment joining the points $A(3, 2)$ and $B(5, 1)$ is divided at the point $P$ in the ratio $1 : 2$ and it lies on the line $3x – 18y + k = 0$. Find the value of $k$.

Solution. Find $P$ first — the ratio is given, so this is a direct substitution:

$$P = \left(\frac{1(5) + 2(3)}{3},\ \frac{1(1) + 2(2)}{3}\right) = \left(\frac{11}{3},\ \frac{5}{3}\right)$$

$P$ lies on the given line, so its coordinates satisfy the equation:

$$3\left(\frac{11}{3}\right) – 18\left(\frac{5}{3}\right) + k = 0$$

$$11 – 30 + k = 0 \implies k = 19$$

The coefficients $3$ and $18$ were chosen to cancel the thirds, which is a small sign that the arithmetic is on track — if the fractions are not clearing, check the section formula weights.

Question 16

If $D\left(\dfrac{-1}{2}, \dfrac{5}{2}\right)$, $E(7, 3)$ and $F\left(\dfrac{7}{2}, \dfrac{7}{2}\right)$ are the midpoints of sides of $\triangle ABC$, find the area of the $\triangle ABC$.

Solution. You could find $A$, $B$ and $C$ and then take the area, but there is no need. The triangle joining the mid-points of the sides — the medial triangle — has exactly one quarter the area of the original.

The reason is the mid-point theorem: each side of the medial triangle joins the mid-points of two sides of $\triangle ABC$, so it is parallel to the third side and half its length. All three sides are halved, so the medial triangle is similar to $\triangle ABC$ with scale factor $\frac12$, and areas scale as the square of the linear factor:

$$\frac{\operatorname{ar}(\triangle DEF)}{\operatorname{ar}(\triangle ABC)} = \left(\frac12\right)^2 = \frac14$$

So compute $\operatorname{ar}(\triangle DEF)$ and multiply by four:

$$\operatorname{ar}(\triangle DEF) = \frac12\left|-\frac12\left(3 – \frac72\right) + 7\left(\frac72 – \frac52\right) + \frac72\left(\frac52 – 3\right)\right|$$

$$= \frac12\left|-\frac12\left(-\frac12\right) + 7(1) + \frac72\left(-\frac12\right)\right| = \frac12\left|\frac14 + 7 – \frac74\right| = \frac12 \cdot \frac{22}{4} = \frac{11}{4}$$

Therefore

$$\operatorname{ar}(\triangle ABC) = 4 \times \frac{11}{4} = 11 \text{ square units}$$

This question uses the area formula, which is not in the current Class 10 textbook. The quarter-area relationship itself comes from the mid-point theorem and the similarity results of Chapter 6, both of which are still on the syllabus.

Question 17

The points $A(2, 9)$, $B(a, 5)$ and $C(5, 5)$ are the vertices of a triangle $ABC$ right angled at $B$. Find the values of $a$ and hence the area of $\triangle ABC$.

Solution. The right angle is at $B$, so $BA \perp BC$ and Pythagoras applies with $AC$ as the hypotenuse:

$$BA^2 + BC^2 = AC^2$$

$$\big[(2-a)^2 + 16\big] + \big[(5-a)^2 + 0\big] = \big[9 + 16\big]$$

$$4 – 4a + a^2 + 16 + 25 – 10a + a^2 = 25$$

$$2a^2 – 14a + 20 = 0 \implies a^2 – 7a + 10 = 0 \implies (a-2)(a-5) = 0$$

So $a = 2$ or $a = 5$.

The second root must be rejected, and the question does not tell you so. If $a = 5$ then $B = (5, 5)$, which is the point $C$. Two vertices coinciding is not a triangle — there is no angle at $B$ to be right, and the “area” would be zero. So $a = 2$.

Then $A(2,9)$, $B(2,5)$, $C(5,5)$. $AB$ is vertical of length $4$ and $BC$ is horizontal of length $3$, meeting at right angles at $B$ exactly as required:

$$\operatorname{ar}(\triangle ABC) = \frac12 \times 4 \times 3 = 6 \text{ square units}$$

The printed answer gives $a = 2$ without mentioning the other root. It is worth knowing it is there, because a quadratic that produces a degenerate figure is a standard exam trick and the marks are in spotting it.

This question uses the area formula, which is not in the current Class 10 textbook — though here the triangle has perpendicular legs, so $\frac12 \times \text{base} \times \text{height}$ does the job without it.

Question 18

Find the coordinates of the point $R$ on the line segment joining the points $P(-1, 3)$ and $Q(2, 5)$ such that $PR = \dfrac{3}{5}PQ$.

Solution. Convert the length relation into a ratio before doing anything else. If $PR = \frac35 PQ$ then $R$ is three-fifths of the way from $P$ to $Q$, so

$$RQ = PQ – PR = \frac25 PQ, \qquad \text{hence} \qquad PR : RQ = \frac35 : \frac25 = 3 : 2$$

Now apply the section formula with $m_1 : m_2 = 3 : 2$:

$$R = \left(\frac{3(2) + 2(-1)}{5},\ \frac{3(5) + 2(3)}{5}\right) = \left(\frac{6 – 2}{5},\ \frac{15 + 6}{5}\right) = \left(\frac{4}{5},\ \frac{21}{5}\right)$$

The commonest error is using the ratio $3 : 5$ straight from the fraction in the question. But $\frac35$ compares $PR$ to the whole segment $PQ$, whereas the section formula wants $PR$ compared to the remaining part $RQ$. Whenever a question gives a fraction of the total, subtract to find the other piece first.

Question 19

Find the values of $k$ if the points $A(k + 1, 2k)$, $B(3k, 2k + 3)$ and $C(5k – 1, 5k)$ are collinear.

Solution. Collinearity means the gradient from $A$ to $B$ equals the gradient from $B$ to $C$. Written without dividing — which avoids trouble when a denominator turns out to be zero:

$$(y_B – y_A)(x_C – x_B) = (y_C – y_B)(x_B – x_A)$$

The four differences:

$$y_B – y_A = 3, \qquad x_C – x_B = 2k – 1, \qquad y_C – y_B = 3k – 3, \qquad x_B – x_A = 2k – 1$$

So

$$3(2k – 1) = (3k – 3)(2k – 1)$$

$$(2k-1)\big[3 – (3k – 3)\big] = 0 \implies (2k – 1)(6 – 3k) = 0$$

giving $k = \dfrac{1}{2}$ or $k = 2$.

Both work, and they work for different reasons — which is why factorising rather than dividing mattered.

With $k = 2$: $A(3, 4)$, $B(6, 7)$, $C(9, 10)$, all on the line $y = x + 1$.

With $k = \frac12$: $A\left(\frac32, 1\right)$, $B\left(\frac32, 4\right)$, $C\left(\frac32, \frac52\right)$ — every $x$-coordinate is $\frac32$, so all three sit on the vertical line $x = \frac32$. Vertical lines have no gradient, so dividing by $x_B – x_A$ at the start would have thrown this solution away.

This question is usually set with the area formula, which is not in the current Class 10 textbook. The gradient form above uses only the equation of a line.

Question 20

Find the ratio in which the line $2x + 3y – 5 = 0$ divides the line segment joining the points $(8, -9)$ and $(2, 1)$. Also find the coordinates of the point of division.

Solution. Let the ratio be $k : 1$. The point of division is

$$\left(\frac{2k + 8}{k+1},\ \frac{k – 9}{k+1}\right)$$

It lies on the given line, so substitute into $2x + 3y – 5 = 0$:

$$2\left(\frac{2k+8}{k+1}\right) + 3\left(\frac{k-9}{k+1}\right) – 5 = 0$$

Multiply through by $(k+1)$, which is safe since $k = -1$ would make the point undefined:

$$2(2k + 8) + 3(k – 9) – 5(k+1) = 0$$

$$4k + 16 + 3k – 27 – 5k – 5 = 0 \implies 2k – 16 = 0 \implies k = 8$$

So the ratio is $8 : 1$, positive and therefore an internal division. The point:

$$\left(\frac{2(8) + 8}{9},\ \frac{8 – 9}{9}\right) = \left(\frac{24}{9},\ \frac{-1}{9}\right) = \left(\frac{8}{3},\ -\frac{1}{9}\right)$$

Confirm it is on the line: $2\left(\frac83\right) + 3\left(-\frac19\right) – 5 = \frac{16}{3} – \frac13 – 5 = 5 – 5 = 0$. Correct.

Exercise 7.4

Question 21

If $(-4, 3)$ and $(4, 3)$ are two vertices of an equilateral triangle, find the coordinates of the third vertex, given that the origin lies in the interior of the triangle.

(Exercise 7.4, Question 1.)

Solution. Call the given vertices $A(-4, 3)$ and $B(4, 3)$. They share the $y$-coordinate $3$, so $AB$ is horizontal of length $8$, and its perpendicular bisector is the $y$-axis. The third vertex of an equilateral triangle lies on the perpendicular bisector of any side, so $C = (0, y)$ for some $y$.

All sides equal, so $CA = AB = 8$:

$$(0 + 4)^2 + (y – 3)^2 = 64 \implies 16 + (y-3)^2 = 64 \implies (y-3)^2 = 48$$

$$y – 3 = \pm 4\sqrt{3} \implies y = 3 + 4\sqrt3 \ \text{ or } \ y = 3 – 4\sqrt3$$

Two equilateral triangles exist — one with the third vertex above $AB$ and one below. The condition about the origin decides between them.

$4\sqrt3 \approx 6.93$, so the two candidates are $y \approx 9.93$ and $y \approx -3.93$.

• If $C = (0, 3 + 4\sqrt3)$, all three vertices have $y \ge 3$, so the whole triangle sits above the line $y = 3$ and the origin is nowhere near it.
• If $C = (0, 3 – 4\sqrt3) \approx (0, -3.93)$, the triangle spans $y$ from $-3.93$ up to $3$. The origin is on the axis of symmetry $x = 0$ and its height $0$ lies between $-3.93$ and $3$, so it is inside.

Question 22

$A(6, 1)$, $B(8, 2)$ and $C(9, 4)$ are three vertices of a parallelogram $ABCD$. If $E$ is the midpoint of $DC$, find the area of $\triangle ADE$.

(Exercise 7.4, Question 2.)

Solution. Find $D$ first, using the fact that the diagonals $AC$ and $BD$ of parallelogram $ABCD$ bisect each other — or equivalently $D = A – B + C$:

$$D = (6 – 8 + 9,\ 1 – 2 + 4) = (7, 3)$$

Check: the mid-point of $AC$ is $\left(\frac{6+9}{2}, \frac{1+4}{2}\right) = \left(\frac{15}{2}, \frac52\right)$, and the mid-point of $BD$ is $\left(\frac{8+7}{2}, \frac{2+3}{2}\right) = \left(\frac{15}{2}, \frac52\right)$. The same point, so $D$ is right.

Now $E$, the mid-point of $DC$ with $D(7,3)$ and $C(9,4)$:

$$E = \left(\frac{7+9}{2},\ \frac{3+4}{2}\right) = \left(8,\ \frac{7}{2}\right)$$

And the area of $\triangle ADE$ with $A(6,1)$, $D(7,3)$, $E\left(8, \frac72\right)$:

$$\operatorname{ar} = \frac12\left|6\left(3 – \frac72\right) + 7\left(\frac72 – 1\right) + 8(1 – 3)\right|$$

$$= \frac12\left|6\left(-\frac12\right) + 7\left(\frac52\right) – 16\right| = \frac12\left|-3 + \frac{35}{2} – 16\right| = \frac12\left|\frac{-6 + 35 – 32}{2}\right| = \frac12 \cdot \frac{3}{2} = \frac{3}{4}$$

A sanity check on the size: $\triangle ADC$ is half the parallelogram, and $E$ is the mid-point of $DC$, so $\triangle ADE$ is half of $\triangle ADC$ and therefore a quarter of the parallelogram. The parallelogram’s area is $3$, and a quarter of $3$ is $\frac34$. Consistent.

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

Question 23

The points $A(x_1, y_1)$, $B(x_2, y_2)$ and $C(x_3, y_3)$ are the vertices of $\triangle ABC$.

(Exercise 7.4, Question 3.)

Solution. This question builds the centroid from scratch in four steps, and the point of it is the pattern that emerges: every answer after the first is the same point.

Question 23 (i)

The median from $A$ meets $BC$ at $D$. Find the coordinates of the point $D$.

A median runs from a vertex to the mid-point of the opposite side, so $D$ is simply the mid-point of $BC$:

$$D = \left(\frac{x_2 + x_3}{2},\ \frac{y_2 + y_3}{2}\right)$$

Question 23 (ii)

Find the coordinates of the point $P$ on $AD$ such that $AP : PD = 2 : 1$.

Apply the section formula to the segment from $A(x_1, y_1)$ to $D$, with $m_1 : m_2 = 2 : 1$:

$$x_P = \frac{2 \cdot \frac{x_2 + x_3}{2} + 1 \cdot x_1}{2 + 1} = \frac{(x_2 + x_3) + x_1}{3} = \frac{x_1 + x_2 + x_3}{3}$$

The $2$ in the numerator cancels the $2$ in the mid-point’s denominator, which is the whole reason the answer comes out symmetric. The $y$-coordinate goes the same way:

$$y_P = \frac{y_1 + y_2 + y_3}{3}$$

Question 23 (iii)

Find the coordinates of points $Q$ and $R$ on medians $BE$ and $CF$, respectively such that $BQ : QE = 2 : 1$ and $CR : RF = 2 : 1$.

Nothing new is required. $E$ is the mid-point of $CA$ and $F$ is the mid-point of $AB$, so repeating part (ii) with the letters rotated:

$$x_Q = \frac{2 \cdot \frac{x_3 + x_1}{2} + x_2}{3} = \frac{x_1 + x_2 + x_3}{3}, \qquad x_R = \frac{2 \cdot \frac{x_1 + x_2}{2} + x_3}{3} = \frac{x_1 + x_2 + x_3}{3}$$

and the same for the $y$-coordinates. The expression is symmetric in the three vertices, so relabelling cannot change it.

Question 23 (iv)

What are the coordinates of the centroid of the triangle $ABC$?

Parts (ii) and (iii) have just shown that $P$, $Q$ and $R$ are the same point. Each lies on a different median, so that single point lies on all three medians at once — the three medians are concurrent, and their common point is the centroid.

It also sits two-thirds of the way down each median from the vertex, since each of the three ratios was $2 : 1$. That is the standard fact about the centroid, and this question has just proved it rather than quoted it.

Question 24

If the points $A(1, -2)$, $B(2, 3)$, $C(a, 2)$ and $D(-4, -3)$ form a parallelogram, find the value of $a$ and height of the parallelogram taking $AB$ as base.

(Exercise 7.4, Question 4.)

Solution. In parallelogram $ABCD$ the diagonals $AC$ and $BD$ bisect each other, so their mid-points coincide:

$$\left(\frac{1 + a}{2},\ \frac{-2 + 2}{2}\right) = \left(\frac{2 – 4}{2},\ \frac{3 – 3}{2}\right)$$

The $y$-coordinates already agree — both are $0$ — so the $x$-coordinates give the answer:

$$\frac{1 + a}{2} = -1 \implies a = -3$$

So $C = (-3, 2)$.

Now the height. Taking $AB$ as base, the height is the perpendicular distance from the opposite side to $AB$ — equivalently from $C$ to the line $AB$. The cleanest route is through the area, because the area of a parallelogram is base times height:

$$\text{area} = AB \times h \implies h = \frac{\text{area}}{AB}$$

The diagonal $AC$ cuts the parallelogram into two congruent triangles, so

$$\text{area} = 2 \times \operatorname{ar}(\triangle ABC) = 2 \times \frac12\Big|1(3 – 2) + 2(2 + 2) + (-3)(-2 – 3)\Big|$$

$$= \left|1 + 8 + 15\right| = 24$$

And the base:

$$AB = \sqrt{(2-1)^2 + (3+2)^2} = \sqrt{1 + 25} = \sqrt{26}$$

Therefore

$$h = \frac{24}{\sqrt{26}} = \frac{24\sqrt{26}}{26} = \frac{12\sqrt{26}}{13} \approx 4.71$$

Rationalising the denominator is what turns $\frac{24}{\sqrt{26}}$ into the printed form $\frac{12\sqrt{26}}{13}$ — multiply above and below by $\sqrt{26}$, then cancel the factor $2$.

The area formula used here is not in the current Class 10 textbook, though the area of a parallelogram as base times height is Class 9 material.

Question 25

Students of a school are standing in rows and columns in their playground for a drill practice. $A$, $B$, $C$ and $D$ are the positions of four students as shown in figure 7.4. Is it possible to place Jaspal in the drill in such a way that he is equidistant from each of the four students $A$, $B$, $C$ and $D$? If so, what should be his position?

(Exercise 7.4, Question 5.)

Solution. Read the positions off the grid, taking the column as the $x$-coordinate and the row as the $y$-coordinate:

$$A(3, 5), \qquad B(7, 9), \qquad C(11, 5), \qquad D(7, 1)$$

A point equidistant from all four must in particular be equidistant from $A$ and $C$, and from $B$ and $D$ — so it lies on both perpendicular bisectors. Find them.

$A$ and $C$ share the row $y = 5$, so $AC$ is horizontal and its perpendicular bisector is the vertical line through the mid-point $\left(\frac{3+11}{2}, 5\right) = (7,5)$, namely $x = 7$.

$B$ and $D$ share the column $x = 7$, so $BD$ is vertical and its perpendicular bisector is the horizontal line through $\left(7, \frac{9+1}{2}\right) = (7,5)$, namely $y = 5$.

The two bisectors meet at $(7, 5)$ — which is the mid-point of both diagonals at once. That is the only candidate, so check it against all four:

$$\text{to } A(3,5): \ 4, \qquad \text{to } B(7,9): \ 4, \qquad \text{to } C(11,5): \ 4, \qquad \text{to } D(7,1): \ 4$$

All four distances are $4$. So yes, it is possible.

The reason it works so neatly is that $ABCD$ is a square. Each side is $\sqrt{4^2 + 4^2} = 4\sqrt2$, and both diagonals $AC$ and $BD$ have length $8$; a quadrilateral with four equal sides and equal diagonals is a square, and a square’s diagonals bisect each other, so their common mid-point is equidistant from all four vertices. Had the four students formed a general quadrilateral, no such point need have existed.

Position $(7, 5)$ means Jaspal stands in the 7th column and the 5th row.

Question 26

Ayush starts walking from his house to office. Instead of going to the office directly, he goes to a bank first, from there to his daughter’s school and then reaches the office. What is the extra distance travelled by Ayush in reaching his office? (Assume that all distances covered are in straight lines.)

If the house is situated at $(2, 4)$, bank at $(5, 8)$, school at $(13, 14)$ and office at $(13, 26)$ and coordinates are in km.

(Exercise 7.4, Question 6.)

Solution. Add up the three legs he actually walks, then subtract the direct distance he could have walked.

House to bank, from $(2,4)$ to $(5,8)$:

$$\sqrt{(5-2)^2 + (8-4)^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \text{ km}$$

Bank to school, from $(5,8)$ to $(13,14)$:

$$\sqrt{(13-5)^2 + (14-8)^2} = \sqrt{64 + 36} = \sqrt{100} = 10 \text{ km}$$

School to office, from $(13,14)$ to $(13,26)$:

$$\sqrt{0 + (26-14)^2} = 12 \text{ km}$$

Total walked: $5 + 10 + 12 = 27$ km.

House to office directly, from $(2,4)$ to $(13,26)$:

$$\sqrt{(13-2)^2 + (26-4)^2} = \sqrt{121 + 484} = \sqrt{605} = 11\sqrt{5} \approx 24.6 \text{ km}$$

$\sqrt{605} = \sqrt{121 \times 5} = 11\sqrt5$, and $\sqrt5 \approx 2.236$, so $11\sqrt5 \approx 24.60$.

Extra distance:

$$27 – 11\sqrt{5} \approx 27 – 24.6 = 2.4 \text{ km}$$

The first three legs coming out as $5$, $10$ and $12$ is the question being kind — those are exact whole numbers, and if any of yours is not, check that leg again before going on. The detour, for all that, costs only about a tenth of the journey.

Common mistakes

• Exercise 7.3 Question 4 — hunting for a second root that is not there. $(a+3)^2 = 0$ has one repeated root, not a $\pm$ pair, because the vertical separation of the two points already equals the whole permitted distance. Read a perfect square equal to zero as the signal it is.
• Exercise 7.3 Question 5 — stopping at one point. The set of points equidistant from $A$ and $B$ is a whole line, the perpendicular bisector. The question asks “how many”, and the answer is infinitely many; naming the mid-point alone answers only half of it.
• Exercise 7.3 Question 18 — using the fraction in the question as the ratio. $PR = \frac35 PQ$ compares $PR$ with the whole segment; the section formula needs $PR : RQ$, which is $3 : 2$. Subtract to find the second piece before substituting.
• Exercise 7.3 Question 17 — keeping both roots of the quadratic. $a = 5$ puts $B$ exactly on top of $C$, and a triangle with two coincident vertices is not a triangle. Whenever a geometric question ends in a quadratic, test each root against the figure before reporting it.
• Exercise 7.3 Questions 7, 12 and Exercise 7.1 Question 13 — solving for the expression instead of the letter. The ordinate was $2m$, not $m$; the ordinate was $-b$, not $b$. Finish by unwrapping whatever the unknown is buried inside.
• Exercise 7.3 Question 19 — dividing by $x_B – x_A$ to compare gradients. One of the two solutions here, $k = \frac12$, makes all three points share the $x$-coordinate $\frac32$ — a vertical line, where the gradient does not exist. Cross-multiply instead of dividing and both roots survive.
• Exercise 7.3 Question 14 — using the diameter as the radius. $10\sqrt2$ is the diameter, so $r = 5\sqrt2$ and $r^2 = 50$. Squaring the diameter instead gives $200$ and a quadratic with no useful roots.
• Exercise 7.4 Question 1 — ignoring the condition that seems decorative. “The origin lies in the interior” is not scene-setting; it is the only thing separating the two equilateral triangles that the algebra produces. A condition tacked onto the end of a question is usually the one that selects the answer.

Practise next

• Exercise 7.1 and Exercise 7.2 — worth a second pass now. Several of the short-answer techniques here, particularly the $k : 1$ ratio and the squared equidistance condition, make the earlier true-or-false justifications much quicker to write.
• Exercise 6.3 — for the similar-triangle results behind Question 16’s quarter-area relationship, which is the mid-point theorem in disguise.
• Exercise 8.1 — Introduction to Trigonometry, the next unit. It opens on right-angled triangles, and the distance formula you have used throughout this unit is Pythagoras wearing coordinates.