Do the points (3, 2), (−2, −3) and (2, 3) form a triangle? If so, name the type of triangle formed.
Let P(3, 2), Q(−2, −3) and R(2, 3) be the given points. Three points form a triangle exactly when the sum of any two of the three distances between them exceeds the third (otherwise they are collinear, or two coincide). So the first step is always to find all three distances using the distance formula, \(d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\).
PQ:
QR:
PR:
Check the triangle condition: \(PQ+PR=5\sqrt2+\sqrt2=6\sqrt2\approx8.49\), and \(QR=2\sqrt{13}\approx7.21\). Since \(PQ+PR>QR\), and the same holds for the other two pairs, P, Q, R are not collinear — they do form a triangle.
To name the type, compare \(PQ^2+PR^2\) with \(QR^2\):
Since \(PQ^2+PR^2=QR^2\), by the converse of the Pythagoras Theorem the angle at P is \(90^{\circ}\). So P, Q, R form a right triangle, right-angled at P.
Show that the points (1, 7), (4, 2), (−1, −1) and (−4, 4) are the vertices of a square.
Let A(1, 7), B(4, 2), C(−1, −1) and D(−4, 4) be the given points. A quadrilateral is a square exactly when all four sides are equal and both diagonals are equal — a rhombus alone (equal sides only) would not guarantee right angles, so both checks are needed.
The four sides:
All four sides equal \(\sqrt{34}\).
The two diagonals:
Both diagonals equal \(\sqrt{68}\). Since \(AB=BC=CD=DA\) and \(AC=BD\), ABCD is a square.
Alternative solution (NCERT's own second method): find the four sides and just ONE diagonal, say AC, as above. Then check the converse of Pythagoras on triangle ADC:
Since \(AD^2+DC^2=AC^2\), the angle at D is \(90^{\circ}\). A quadrilateral with all four sides equal and one angle \(90^{\circ}\) is a square — so ABCD is a square, without needing to compute the second diagonal BD at all.
Ashima, Bharti and Camella are seated at A(3, 1), B(6, 4) and C(8, 6) respectively in a classroom (Fig. 7.6). Do you think they are seated in a line? Give reasons for your answer.
Three points lie on a line exactly when the distance between the two outer points equals the sum of the distances to the point in between — that is the same triangle-inequality idea as Example 1, run in reverse (equality instead of strict inequality means the “triangle” has collapsed onto a straight line).
Now check: \(AB+BC=3\sqrt2+2\sqrt2=5\sqrt2=AC\). Since \(AB+BC=AC\), the three points are collinear — so yes, Ashima, Bharti and Camella are seated in a line, with Bharti between Ashima and Camella.
Find a relation between x and y such that the point (x, y) is equidistant from the points (7, 1) and (3, 5).
“Equidistant from A and B” means the point P(x, y) satisfies \(PA=PB\), so \(PA^2=PB^2\) — squaring first avoids ever having to deal with square roots.
Let A(7, 1), B(3, 5). Then \(PA^2=PB^2\) gives:
Expand both sides:
The \(x^2\) and \(y^2\) terms cancel from both sides, leaving:
which is the required relation: \(\boldsymbol{x-y=2}\).
Remark: the graph of \(x-y=2\) is a straight line, and it is exactly the perpendicular bisector of AB — every point equidistant from two fixed points lies on their perpendicular bisector, which is the geometric meaning of what was just computed algebraically.
Find a point on the y-axis which is equidistant from the points A(6, 5) and B(−4, 3).
Every point on the y-axis has the form (0, y), so let the required point be P(0, y). It must satisfy \(PA^2=PB^2\).
Expand:
So the required point is (0, 9).
Check:
Both distances come out equal to \(\sqrt{52}\), confirming (0, 9) is equidistant from A and B. (By the Remark in Example 4, this is exactly the point where the y-axis meets the perpendicular bisector of AB.)
Find the distance between the following pairs of points:
(i) (2, 3), (4, 1)
(ii) (−5, 7), (−1, 3)
(iii) (a, b), (−a, −b)
All three parts use the distance formula \(d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\) directly.
(i)
(ii)
(iii) Here the two points are (a, b) and (−a, −b):
Find the distance between the points (0, 0) and (36, 15). Can you now find the distance between the two towns A and B discussed in Section 7.2.
Using the distance formula:
So the distance is 39 units.
In Section 7.2, town A was placed at the origin (0, 0) and town B, being 36 km east and 15 km north of A, was at (36, 15) — exactly this pair of points, with 1 km as one unit. So the distance between towns A and B is 39 km.
Determine if the points (1, 5), (2, 3) and (−2, −11) are collinear.
Let A(1, 5), B(2, 3), C(−2, −11). Find all three distances and check whether the largest equals the sum of the other two.
Now check: does \(AB+BC\) equal AC, or does any one distance equal the sum of the other two? Numerically, \(AB\approx2.236\), \(BC\approx14.318\), \(AC\approx16.279\). Testing every combination:
\(AB+BC\approx2.236+14.318=16.554\neq AC(\approx16.279)\)
\(AB+AC\approx18.515\neq BC\), and \(BC+AC\approx30.597\neq AB\).
None of the three equalities holds, so the points do not satisfy the collinearity condition. A, B and C are NOT collinear — they form a (very thin, but genuine) triangle.
Check whether (5, −2), (6, 4) and (7, −2) are the vertices of an isosceles triangle.
Let A(5, −2), B(6, 4), C(7, −2). An isosceles triangle needs at least two equal sides, so find all three.
Since \(AB=BC=\sqrt{37}\) while \(AC=2\) is different, two sides are equal and the third is not. Yes, A, B, C are the vertices of an isosceles triangle.
In a classroom, 4 friends are seated at the points A, B, C and D as shown in Fig. 7.8. Champa and Chameli walk into the class and after observing for a few minutes Champa asks Chameli, “Don’t you think ABCD is a square?” Chameli disagrees. Using distance formula, find which of them is correct.
Reading the coordinates off Fig. 7.8: A(3, 4), B(6, 7), C(9, 4), D(6, 1). ABCD is a square exactly when all four sides are equal AND both diagonals are equal (equal sides alone would only prove a rhombus).
The four sides:
All four sides equal \(\sqrt{18}=3\sqrt2\).
The two diagonals:
Both diagonals equal 6. Since all four sides are equal AND both diagonals are equal, ABCD IS a square — so Champa is correct, and Chameli is wrong.
Name the type of quadrilateral formed, if any, by the following points, and give reasons for your answer:
(i) (−1, −2), (1, 0), (−1, 2), (−3, 0)
(ii) (−3, 5), (3, 1), (0, 3), (−1, −4)
(iii) (4, 5), (7, 6), (4, 3), (1, 2)
(i) Let A(−1, −2), B(1, 0), C(−1, 2), D(−3, 0).
All four sides equal \(2\sqrt2\). Now the diagonals:
Both diagonals equal 4 and are also equal to each other. All sides equal AND diagonals equal ⇒ ABCD is a square.
(ii) Let A(−3, 5), B(3, 1), C(0, 3), D(−1, −4). Before computing all six sides/diagonals, check whether any three of these four points are collinear — if they are, no proper quadrilateral is formed at all.
Since \(AC+CB=\sqrt{13}+\sqrt{13}=2\sqrt{13}=AB\), the points A, C, B are collinear, with C lying exactly on the segment AB (in fact C is the midpoint of AB). So A, B, C, D do NOT form a proper quadrilateral; three of the four given points lie on one straight line.
(iii) Let A(4, 5), B(7, 6), C(4, 3), D(1, 2).
Opposite sides come out equal in pairs: \(AB=CD=\sqrt{10}\) and \(BC=DA=3\sqrt2\). Now the diagonals:
The diagonals (\(2\sqrt5\) and \(6\)) are NOT equal, so this is not a rectangle — but opposite sides are equal, which is enough to conclude ABCD is a parallelogram (a general one, not a rhombus, rectangle or square).
Find the point on the x-axis which is equidistant from (2, −5) and (−2, 9).
Every point on the x-axis has the form (x, 0). Let P(x, 0) be equidistant from A(2, −5) and B(−2, 9), so \(PA^2=PB^2\).
Expand both sides:
So the required point is (−7, 0).
Find the values of y for which the distance between the points P(2, −3) and Q(10, y) is 10 units.
By the distance formula, \(PQ^2=100\):
So \((y+3)^2=36\), giving \(y+3=\pm6\):
So y = 3 or y = −9. (Both are genuinely valid — there is no reason to prefer one over the other; the point Q simply sits on a circle of radius 10 about P, and that circle meets the vertical line x = 10 at two points.)
If Q(0, 1) is equidistant from P(5, −3) and R(x, 6), find the values of x. Also find the distances QR and PR.
Step 1 — find x. \(QP^2=QR^2\):
So \(x^2=16\), giving x = 4 or x = −4.
Step 2 — find QR. Since \(QR^2\) only involves \(x^2\) (which is 16 either way), QR is the same for both values of x:
QR = \(\sqrt{41}\) for either value of x.
Step 3 — find PR. Unlike QR, this genuinely depends on the sign of x, so both cases must be worked out separately.
If x = −4:
If x = 4:
So PR = \(9\sqrt2\) when x = −4, and PR = \(\sqrt{82}\) when x = 4.
Find a relation between x and y such that the point (x, y) is equidistant from the point (3, 6) and (−3, 4).
Let P(x, y) satisfy \(PA^2=PB^2\) with A(3, 6), B(−3, 4).
Expand both sides:
Collect like terms — subtract the right side from the left:
So the required relation is \(\boldsymbol{3x+y=5}\) (equivalently \(3x+y-5=0\)).
Find the coordinates of the point which divides the line segment joining the points (4, −3) and (8, 5) in the ratio 3 : 1 internally.
Using the section formula, the point P(x, y) dividing A\((x_1,y_1)\) to B\((x_2,y_2)\) internally in the ratio \(m_1:m_2\) is:
Here A(4, −3), B(8, 5), \(m_1:m_2=3:1\):
So the required point is (7, 3).
In what ratio does the point (−4, 6) divide the line segment joining the points A(−6, 10) and B(3, −8)?
Let \((-4,6)\) divide AB internally in the ratio \(m_1:m_2\). By the section formula:
Using the x-coordinate:
So \(m_1:m_2=2:7\). (The y-coordinate should give the same ratio — checking: \(\dfrac{-8m_1+10m_2}{m_1+m_2}=\dfrac{-8(2)+10(7)}{9}=\dfrac{-16+70}{9}=\dfrac{54}{9}=6\), which matches, confirming the ratio.)
So (−4, 6) divides AB in the ratio 2 : 7.
Alternatively (NCERT's own second method): write the ratio as \(k:1\). Then:
So \(k:1=2:7\), the same answer as before — this form is often quicker when only ONE unknown (k) needs to be solved for, instead of two (\(m_1,m_2\)).
Note: the ratio can also be found by computing the distances AP and PB directly and taking their ratio, provided A, P, B are known to be collinear.
Find the coordinates of the points of trisection (i.e., points dividing in three equal parts) of the line segment joining the points A(2, −2) and B(−7, 4).
Let P and Q be the points of trisection of AB, so \(AP=PQ=QB\). Then P divides AB internally in the ratio 1 : 2 (one part out of the three from A), and Q divides AB internally in the ratio 2 : 1 (two parts out of the three from A).
Point P (ratio 1 : 2):
So P = (−1, 0).
Point Q (ratio 2 : 1):
So Q = (−4, 2).
Note: Q could also have been found as the midpoint of P and B, using the simpler midpoint formula \(\left(\dfrac{x_1+x_2}{2},\dfrac{y_1+y_2}{2}\right)\), since Q is exactly halfway between P and B.
Find the ratio in which the y-axis divides the line segment joining the points (5, −6) and (−1, −4). Also find the point of intersection.
Let the y-axis divide the segment joining A(5, −6) and B(−1, −4) in the ratio \(k:1\). By the section formula, the dividing point is:
Every point on the y-axis has x-coordinate 0, so:
So the ratio is 5 : 1. Substituting \(k=5\) into the y-coordinate:
So the point of intersection is \(\left(0,\ \dfrac{-13}{3}\right)\).
If the points A(6, 1), B(8, 2), C(9, 4) and D(p, 3) are the vertices of a parallelogram, taken in order, find the value of p.
The key fact about a parallelogram is that its diagonals bisect each other — so the midpoint of one diagonal must equal the midpoint of the other. Taken in order, ABCD has diagonals AC and BD.
Equating the midpoints:
Matching x-coordinates:
(The y-coordinates already agree on both sides without needing p, which is an automatic consistency check that the parallelogram condition is being applied correctly.)
So p = 7.
Find the coordinates of the point which divides the join of (−1, 7) and (4, −3) in the ratio 2 : 3.
Using the section formula with A(−1, 7), B(4, −3), ratio \(m_1:m_2=2:3\):
So the required point is (1, 3).
Find the coordinates of the points of trisection of the line segment joining (4, −1) and (−2, −3).
Let P and Q trisect the segment joining A(4, −1) and B(−2, −3), so P divides AB in the ratio 1 : 2 and Q divides AB in the ratio 2 : 1 (exactly as in Example 8).
Point P (ratio 1 : 2):
So P = \(\left(2,\ \dfrac{-5}{3}\right)\).
Point Q (ratio 2 : 1):
So Q = \(\left(0,\ \dfrac{-7}{3}\right)\).
To conduct Sports Day activities, in your rectangular shaped school ground ABCD, lines have been drawn with chalk powder at a distance of 1 m each. 100 flower pots have been placed at a distance of 1 m from each other along AD, as shown in Fig. 7.12. Niharika runs \(\dfrac{1}{4}\)th the distance AD on the 2nd line and posts a green flag. Preet runs \(\dfrac{1}{5}\)th the distance AD on the eighth line and posts a red flag. What is the distance between both the flags? If Rashmi has to post a blue flag exactly halfway between the line segment joining the two flags, where should she post her flag?
Since 100 flower pots are placed 1 m apart along AD, \(AD=100\) m. Taking A as the origin with the “line” number as the x-coordinate and the distance along AD as the y-coordinate:
Green flag (Niharika, 2nd line, \(\tfrac14\) of AD):
Red flag (Preet, eighth line, \(\tfrac15\) of AD):
Distance between the flags — by the distance formula:
So the distance between the two flags is \(\sqrt{61}\) m.
Blue flag — exactly halfway between the green and red flags is their midpoint:
So Rashmi should post her blue flag at the point (5, 22.5), i.e. on the 5th line, 22.5 m from AD.
Find the ratio in which the line segment joining the points (−3, 10) and (6, −8) is divided by (−1, 6).
Let \((-1,6)\) divide the segment joining A(−3, 10) and B(6, −8) in the ratio \(m_1:m_2\). Using the x-coordinate of the section formula:
So the ratio is 2 : 7. (Checking the y-coordinate: \(\dfrac{2(-8)+7(10)}{9}=\dfrac{-16+70}{9}=\dfrac{54}{9}=6\), which matches.)
Find the ratio in which the line segment joining A(1, −5) and B(−4, 5) is divided by the x-axis. Also find the coordinates of the point of division.
Let the x-axis divide AB in the ratio \(k:1\). Every point on the x-axis has y-coordinate 0, so using the y-coordinate of the section formula:
So the ratio is 1 : 1 — the x-axis bisects AB.
Substituting \(k=1\) into the x-coordinate:
So the point of division is \(\left(\dfrac{-3}{2},\ 0\right)\).
If (1, 2), (4, y), (x, 6) and (3, 5) are the vertices of a parallelogram taken in order, find x and y.
As in Example 10, the diagonals of a parallelogram bisect each other. Taken in order, the diagonals here join the 1st & 3rd vertices, and the 2nd & 4th vertices.
Equating the midpoints (x-coordinates, then y-coordinates):
So x = 6 and y = 3.
Find the coordinates of a point A, where AB is the diameter of a circle whose centre is (2, −3) and B is (1, 4).
The centre of a circle is the midpoint of any diameter, so the centre is the midpoint of AB. Let A = (x, y).
Matching coordinates:
So A = (3, −10).
If A and B are (−2, −2) and (2, −4), respectively, find the coordinates of P such that \(AP=\dfrac{3}{7}AB\) and P lies on the line segment AB.
Since P lies on AB with \(AP=\tfrac37AB\), the remaining part \(PB=AB-AP=\tfrac47AB\). So P divides AB internally in the ratio \(AP:PB=3:4\).
Using the section formula with A(−2, −2), B(2, −4), ratio 3 : 4:
So \(P=\left(\dfrac{-2}{7},\ \dfrac{-20}{7}\right)\).
Find the coordinates of the points which divide the line segment joining A(−2, 2) and B(2, 8) into four equal parts.
Four equal parts need three dividing points, \(P_1,P_2,P_3\), splitting AB in ratios 1 : 3, 1 : 1 (the midpoint) and 3 : 1 respectively from A.
\(P_1\) (ratio 1 : 3):
\(P_2\) (ratio 1 : 1, the midpoint):
\(P_3\) (ratio 3 : 1):
So the three points are \(P_1=\left(-1,\ \dfrac72\right)\), \(P_2=(0,5)\), \(P_3=\left(1,\ \dfrac{13}{2}\right)\).
Find the area of a rhombus if its vertices are (3, 0), (4, 5), (−1, 4) and (−2, −1) taken in order. [Hint: Area of a rhombus \(=\dfrac12\) (product of its diagonals)]
Since the "Area of a Triangle" section is not part of this rationalised syllabus, the hint’s formula — half the product of the diagonals — is used instead, and the diagonal lengths come from the distance formula. Let A(3, 0), B(4, 5), C(−1, 4), D(−2, −1); the diagonals of ABCD taken in order are AC and BD.
By the hint,
So the area of the rhombus is 24 square units.
Cross-check that ABCD really is a rhombus (all four sides equal), confirming the hint’s formula applies:
All four sides equal \(\sqrt{26}\), confirming ABCD is genuinely a rhombus.
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