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01. Real Numbers
- Real Numbers and Fundamental Theorem of Arithmetic
- Divisibility and Euclid’s Division Lemma
- Finding HCF and LCM Using Fundamental Theorem of Arithmetic
- Miscellaneous Questions
- Chapter Notes – Real Numbers
- NCERT Solutions – Real Numbers Exercise 1.1 – 1.4
- Revision Notes Real Numbers
- R S Aggarwal Real Numbers
- R D Sharma Real Numbers
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02. Polynomials
- Introduction to Polynomials and Its Zeroes
- Finding Zeroes of Quadratic Polynomial
- Relationship b/w the Zeroes and coefficients of a Polynomial
- Cubic Polynomial
- Division and Division Algorithm for Polynomials
- Geometrical Meaning of Zeroes of Polynomial
- Chapter Notes – Polynomials
- NCERT Solutions – Polynomials Exercise 2.1 – 2.4
- Revision Notes Polynomials
- R S Aggarwal Polynomials
- R D Sharma Polynomials
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03. Linear Equation
- Solution of linear Equation in Two Variable by Graphical and Algebraic Methods
- Some Basic Questions on Solving Pair of Linear Equations
- Some Basics Problems on Numbers and Ages
- Problems on Money Matters, Time, Distance, Speed, Time and Work
- Chapter Notes – Linear Equation
- NCERT Solutions – Linear Equation Exercise 3.1 – 3.7
- Revision Notes Linear Equation
- R S Aggarwal Linear Equation
- R D Sharma Linear Equation
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04. Quadratic Equation
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05. Arithmetic Progressions
- Introduction to Arithmetic Progression
- Miscellaneous Questions Based on nth Term Formula of A.P.
- Middle term (s) of Finite A.P. and Arithmetic Mean
- Selection of the Terms and Sum of nth Terms of A.P.
- Miscellaneous Questions Based on Sum of nth Term of A.P.
- Word Problems related to nth term and sum of nth terms of A.P.
- Chapter Notes – Arithmetic Progressions
- NCERT Solutions – Arithmetic Progressions Exercise 5.1 – 5.4
- Revision Notes Arithmetic Progressions
- R S Aggarwal Arithmetic Progressions
- R D Sharma Arithmetic Progressions
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06. Some Applications of Trigonometry
- Angle of Elevation and Depression; Problems involving Elevation and Double Elevation
- Miscellaneous Problems Involving Double Elevation, Depression and Elevation
- Miscellaneous Questions (Level-1, 2, 3, 4)
- Chapter Notes – Some Applications of Trigonometry
- NCERT Solutions – Some Applications of Trigonometry
- Revision Notes Some Applications of Trigonometry
- R D Sharma Some Applications of Trigonometry
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07. Coordinate Geometry
- Introduction to Terms related to Coordinate Geometry
- Locating Coordinates of Point on the Axes
- Finding vertices of a Geometrical Fig. and Distance b/w Two Points
- Distance b/w Two Points Formula Based Examples
- Finding the type of Triangle and Quadrilateral
- Find the Collinear or Non-collinear points and Missing Vertex of a Geometrical Figure
- Section Formula and Corollary
- Finding the Section Ratio and Involving Equation of Line
- Proof Related to Mid-points and Centroid of a Triangle
- Finding Area of Triangle and Quadrilateral
- Collinearity
- Examples Based on If Areas are Given
- Chapter Notes – Coordinate Geometry
- NCERT Solutions – Coordinate Geometry Exercise 7.1 – 7.4
- Revision Notes Coordinate Geometry
- R S Aggarwal Coordinate Geometry
- R D Sharma Coordinate Geometry
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08. Triangles
- Introduction, Similarity of Triangles, Properties of Triangles, Thales’s Theorem
- Questions Based on Thales’s Theorem, Converse Thales’s Theorem
- Questions Based on Converse Thales’s Theorem, Questions Based On Mid-points and its Converse
- Questions Based on Internal Bisectors of an Angle, Similar Triangles, AAA criterion for Similarity
- SSS & SAS Criterion of Similarity, Question Based on Criteria for Similarity
- Questions Based on Criteria for Similarity Conti.
- Questions Based On Area of Similar Triangles
- Questions Based On Area of Similar Triangles cont., Pythagoras Theorem Level-1
- Pythagoras Theorem Level-1 cont.
- Miscellaneous Questions
- Chapter Notes – Triangles
- NCERT Solutions – Triangles Exercise 8.1 – 8.6
- Revision Notes Triangles
- R S Aggarwal Triangles
- R D Sharma Triangles
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09. Circles
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11. Introduction to Trigonometry
- Introduction, Trigonometric Ratios, Sums when Sides of Triangle are given, Miscellaneous Questions
- Trigonometric Ratios for Some Specific Angles, Trigonometric Ratio for 30 & 60, Value Table, Word Problems
- Complimentary angles in Inverse Multiplication, Miscellaneous Problems
- Miscellaneous Questions
- Chapter Notes – Introduction to Trigonometry
- NCERT Solutions – Introduction to Trigonometry
- Revision Notes Introduction to Trigonometry
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12. Surface Areas and Volumes
- Introduction, Surface Area of Cube, Cuboid, Cylinder, Hollow cylinder, Cone, Sphere, Type-1 : Surface Area and Volume of A Solid
- Volume Related Questions, Based On Embarkment
- Based on the Rate of Flowing, Frustum of Cone (A) Surface Area (B) Volume
- Quantity, Percentage, Very Short Answer Types Questions
- Chapter Notes – Surface Areas and Volumes
- NCERT Solutions – Surface Areas and Volumes
- Revision Notes Surface Areas and Volumes
- R S Aggarwal Surface Areas and Volumes
- R D Sharma Surface Areas and Volumes
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13. Statistics
- Introduction and Mean of Ungrouped Data
- Mean of Grouped Data
- Median of Ungrouped and grouped Data
- Mode of Ungrouped and Grouped Data
- Mean, Median and Mode & Cumulative Frequency Curve-Less Than Type
- Cumulative Frequency Curve (Ogive)-More Than Type
- Miscellaneous Questions
- Chapter Notes – Statistics
- NCERT Solutions – Statistics
- Revision Notes Statistics
- R S Aggarwal Statistics
- R D Sharma Statistics
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14. Probability
- Activities, Understanding term Probability, Experiments
- Calculations
- Event, Favorable Outcomes, Probability, Some important experiments- Tossing a coin, Rolling dice, Drawing a card
- Complement of an event, Elementary Events, Equally Likely events, Some problems of Probability
- Chapter Notes – Probability
- NCERT Solutions – Probability
- Revision Notes Probability
- R S Aggarwal Probability
- R D Sharma Probability
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15. Construction
- Introduction, Construction to Divide a Line Segment in Certain Ratio, Construction of Similar Triangles
- Construction of Different Angles, Construction of Angle Bisector
- Construction of Tangent of Circle at a Given Point
- Chapter Notes – Construction
- NCERT Solutions – Construction
- R S Aggarwal Construction
- R D Sharma Construction
NCERT Solutions – Coordinate Geometry Exercise 7.1 – 7.4
Q.1 Find the distance between the following pairs of points :
(i) (2, 3), (4, 1) (ii) (–5, 7), (–1, 3) (iii) (a, b),(–a,–b)
Sol. (i) Let P(2, 3) and Q (4, 1) be the given points.
Here, x1=2,y1=3 and x2=4,y2=1
Therefore, PQ=(x2−x1)2+(y2−y1)2−−−−−−−−−−−−−−−−−−√
⇒PQ=(4−2)2+(1−3)2−−−−−−−−−−−−−−−√ =(2)2+(−2)2−−−−−−−−−−√
⇒PQ=4+4−−−−√=8–√=22–√
(ii) Let P(– 5, 7) and Q(–1, 3) be the given points.
Here x1=−5,y1=7 and x2=−1,y2=3
Therefore, PQ=(x2−x1)2+(y2−y1)2−−−−−−−−−−−−−−−−−−√
⇒PQ=(−1+5)2+(3−7)2−−−−−−−−−−−−−−−−√ =(4)2+(−4)2−−−−−−−−−−√
⇒PQ=16+16−−−−−−√=32−−√=16×2−−−−−√=42–√
(iii) Let P(a, b) and Q(–a, –b) be the given points.
Here, x1=a,y1=b and x2=−a,y2=−b
Therefore, PQ=(x2−x1)2+(y2−y1)2−−−−−−−−−−−−−−−−−−√
⇒PQ=(−a−a)2+(−b−b)2−−−−−−−−−−−−−−−−−−√ =(−2a)2+(−2b)2−−−−−−−−−−−−−√
⇒PQ=4a2+4b2−−−−−−−−√=2a2+b2−−−−−−√
Q.2 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?
Sol. Let P(0, 0) and Q(36, 15) be the given points.
Here x1=0,y1=0 and x2=36,y2=15
Therefore, PQ=(x2−x1)2+(y2−y1)2−−−−−−−−−−−−−−−−−−√
⇒PQ=(36−0)2+(15−0)2−−−−−−−−−−−−−−−−−√ =1296+225−−−−−−−−−√=1521−−−−√=39
In fact, the positions of towns A and B are given by (0, 0) and (36, 15) respectively and so the distance between them = 39 km as calculated above.
Q.3 Determine if the points (1, 5), (2, 3) and (–2, –11) are collinear.
Sol. Let A (1, 5), B(2, 3) and C(–2, –11) be the given points. Then, we have
AB=(2−1)2+(3−5)2−−−−−−−−−−−−−−−√=12+(−2)2−−−−−−−−−√ =1+4−−−−√=5–√
BC=(−2−2)2+(−11−3)2−−−−−−−−−−−−−−−−−−−√
=(−4)2+(−14)2−−−−−−−−−−−−√ =16+196−−−−−−−√
=212−−−√=4×53−−−−−√=253−−√
and, AC=(−2−1)2+(−11−5)2−−−−−−−−−−−−−−−−−−−√
=(−3)2+(−16)2−−−−−−−−−−−−√
=9+256−−−−−−√=265−−−√
Clearly, BC≠AB+AC,AB≠BC+AC and AC≠BC
Hence, A,B and C are not collinear.
Q.4 Check whether (5, –2), (6, 4) and (7, – 2) are the vertices of an isosceles triangle.
Sol. Let A(5, –2), B(6, 4) and C(7, –2) are the given point. Then,
AB=(6−5)2+(4+2)2−−−−−−−−−−−−−−−√
=1+36−−−−−√=37−−√
BC=(7−6)2+(−2−4)2−−−−−−−−−−−−−−−−√
=1+36−−−−−√=37−−√
Clearly, AB = BC
Therefore, ΔABC is an isosceles triangle.
Q.5 In a classroom, four friends are seated at the points A, B, C and D as shown in figure. 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, and why?
Sol. Clearly from the figure, the coordinates of points A, B, C and D are (3, 4), (6, 7), (9, 4) and (6, 1).
By using distance formula, we get
AB=(6−3)2+(7−4)2−−−−−−−−−−−−−−−√
=9+9−−−−√=18−−√=32–√
BC=(9−6)2+(4−7)2−−−−−−−−−−−−−−−√
=9+9−−−−√=18−−√=32–√
CD=(6−9)2+(1−4)2−−−−−−−−−−−−−−−√
=9+9−−−−√=18−−√=32–√
and, DA=(3−6)2+(4−1)2−−−−−−−−−−−−−−−√
=9+9−−−−√=18−−√=32–√
⇒AB=BC=CD=DA=32–√
Also, AC=(9−3)2+(4−4)2−−−−−−−−−−−−−−−√
=36+0−−−−−√=36−−√=6
and, BD=(6−6)2+(1−7)2−−−−−−−−−−−−−−−√
=0+36−−−−−√=36−−√=6
⇒AC=BD=6
Thus, the four sides are equal and the diagonals are also equal. Therefore, ABCD is a square.
Hence, Champa is correct.
(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)
Sol. (i) Let A(–1,–2), B(1, 0), C(–1, 2) and D(–3, 0) be the given points. Then
AB=(1+1)2+(0+2)2−−−−−−−−−−−−−−−√=4+4−−−−√=8–√
BC=(−1−1)2+(2−0)2−−−−−−−−−−−−−−−−√=4+4−−−−√=8–√
CD=(−3+1)2+(0−2)2−−−−−−−−−−−−−−−−√=4+4−−−−√=8–√
DA=(−1+3)2+(−2−0)2−−−−−−−−−−−−−−−−−−√=4+4−−−−√=8–√
AC=(−1+1)2+(2+2)2−−−−−−−−−−−−−−−−√=0+16−−−−−√=4
and BD=(−3−1)2+(0−0)2−−−−−−−−−−−−−−−−√=16+0−−−−−√=4
Clearly, four sides AB, BC, CD and DA are equal. Also, diagonals AC and BD are equal.
Therefore, the quadrilateral ABCD is a square.
(ii) Let A(–3, 5), B(3, 1), C(0, 3) and D(–1,– 4) are the given points. Plot these points as shown.
Clearly, the points A, C and B are collinear. So, no quadrilateral is formed by these points.
(iii) Let A(4, 5), B(7, 6), C(4, 3) and D(1, 2) be the given points. Then,
AB=(7−4)2+(6−5)2−−−−−−−−−−−−−−−√=9+1−−−−√=10−−√
BC=(4−7)2+(3−6)2−−−−−−−−−−−−−−−√=9+9−−−−√=18−−√
CD=(1−4)2+(2−3)2−−−−−−−−−−−−−−−√=9+1−−−−√=10−−√
DA=(4−1)2+(5−2)2−−−−−−−−−−−−−−−√=9+9−−−−√=18−−√
AC=(4−4)2+(3−5)2−−−−−−−−−−−−−−−√=0+4−−−−√=4–√=2
and, BD=(1−7)2+(2−6)2−−−−−−−−−−−−−−−√=36+16−−−−−−√=52−−√
Clearly, AB = CD, BC = DA and AC≠BD
Therefore, the quadrilateral ABCD is a parallelogram.
Q.7 Find the point on the x-axis which is equidistant from (2, – 5) and (– 2, 9).
Sol. Since the point on x – axis have its ordinate = 0, so (x, 0) is any point on the x – axis.
Since P(x, 0) is equidistant from A(2, –5) and B(–2, 9)
PA = PB ⇒PA2=PB2
⇒(x−2)2+(0+5)2=(x+2)2+(0−9)2
⇒x2−4x+4+25=x2+4x+4+81
⇒−4x−4x=81−25
⇒−8x=56
⇒x=56−8=−7
Therefore, the point equidistant from given points on the axis is (–7, 0).
Q.8 Find the value of y for which the distance between the points P(2, –3) and Q (10, y) is 10 units.
Sol. P(2, –3) and Q(10, y) are given points such that PQ = 10 units.
But, PQ=(10−2)2+(y+3)2−−−−−−−−−−−−−−−−√
⇒10=64+y2+6y+9−−−−−−−−−−−−−√
⇒100=73+y2+6y
⇒y2+6y−27=0⇒(y+9)(y−3)=0
⇒ y=−9 or 3
Thus, the possible values of y are – 9 or 3.
Q.9 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.
Sol. Since the point Q(0, 1) is equidistant from P(5, –3) and R(x, 6) therefore,
QP = QR ⇒QP2=QR2
⇒(5−0)2+(−3−1)2=(x−0)2+(6−1)2
⇒25+16=x2+25
⇒x2=16⇒x=±4
Thus, R is (4, 6) or (– 4, 6).
Now, QR = Distance between Q (0, 1) and R (4, 6)
=(4−0)2+(6−1)2−−−−−−−−−−−−−−−√=16+25−−−−−−√=4–√1
Also, QR = Distance between Q(0, 1) and R(– 4, 6)
=(−4−0)2+(6−1)2−−−−−−−−−−−−−−−−√=16+25−−−−−−√=41−−√
and, PR = Distance between P(5, –3) and R( 4, 6)
=(4−5)2+(6+3)2−−−−−−−−−−−−−−−√=1+81−−−−−√=82−−√
Also, PR = Distance between P(5, –3) and R(– 4, 6)
=(4−5)2+(6+3)2−−−−−−−−−−−−−−−√=81+81−−−−−−√=92–√
Q.10 Find a relation between x and y such that the point (x, y) is equidistant from the points (3, 6) and (–3, 4).
Sol. Let the point P(x, y) be equidistant from the points A(3, 6) and B(–3, 4)
i.e., PA = PB
⇒PA2=PB2
⇒(x−3)2+(y−6)2=(x+3)2+(y−4)2
⇒x2−6x+9+y2−12y+36=x2+6x+9+y2−8y+16
⇒−6x−6x−12y+8y+36−16=0
⇒−12x−4y+20=0
⇒3x+y−5=0
This is the required relation.
Q.1 Find the coordinates of the point which divides join of (–1, 7) and (4, –3) in the ratio 2 : 3.
Sol. Let P(x, y) be the required point.
By Section formula
P(x,y)=[mx2+nx1m+n,my2+ny1m+n]
Where, m = 2 and n = 3
Then, x=2×4+3×−12+3
and, y=2×−3+3×72+3
⇒x=8−35
and y=−6+215
⇒ x = 1 and y = 3
So, the coordinates of P are (1, 3).
Q.2 Find the coordinates of the points of trisection of the line segment joining (4,–1) and (–2,–3).
Sol. Let A(4,–1) and B(–2,–3) be the points of trisection of P and Q.
Then, AP = PQ= QB = k (say).
Therefore, PB = PQ + QB = 2k
and, AQ = AP + PQ = 2k
⇒ AP : PB = k : 2k = 1 : 2
and AQ : QB = 2k : k = 2 : 1
So, P divides AB internally in the ratio 1 : 2, while Q divides AB internally in the ratio 2 : 1.
Thus, the coordinates of
P are (1×−2+2×41+2,1×−3+2×−11+2)=(−2+83,−3−23)=(2,−53)
and, the coordinates of Q are (2×−2+1×42+1,2×−3+1×−12+1)=(−4+43,−6−13)=(0,−73)
Hence, the two points of trisection are (2,−53) and (0,−73).
Q.3 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 figure. Niharika runs 14 th the distance AD on the 2nd line and posts a green flag. Preet runs 15th 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?
Sol. Clearly from the figure, the position of green flag posted by Niharika is given by P(2,14×100),i.e., P(2, 25)
and that of red flag posted by Preet is given by Q(8,15×100) i.e., Q(8, 20).
Now, PQ=(8−2)2+(20−25)2−−−−−−−−−−−−−−−−−√
=(6)2+(−5)2−−−−−−−−−−√
=36+25−−−−−−√=61−−√
Therefore, the distance between the flags = 61−−√ metres
Let M be the position of the blue flag posted by Rashmi in the halfway of line segment PQ.
Therefore, M is the given by (2+82,25+202) or (102,452),i.e., (5, 22.5) .
Thus, the blue flag is on the fifth line at a distance 22.5m above it.
Q.4 Find the ratio in which the line segment joining the points of (–3, 10) and (6, – 8) is divided by (–1, 6).
Sol. Let the point P(–1, 6) divide the line joining A(–3, 10) and B(6, –8) in the ratio k : 1. Then, the
coordinates of P are (6k−3k+1,−8k+10k+1).
But, the coordinates of P are given as (–1, 6).
Therefore, 6k−3k+1=−1 and −8k+10k+1=6
⇒6k−3=−k−1 and −8k+10=6k+6
⇒6k+k=−1+3 and −8k−6k=6−10
⇒7k=2 and −14k=−4
⇒k=27
Hence, the point P divides AB in the ratio 2 : 7.
Q.5 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.
Sol. Let the required ratio be k : 1. Then, the coordinates of the point P of division are (−4k+1k+1,5k−5k+1).
But it is a point on x-axis on which y – coordinate of every point is zero.
Therefore, 5k−5k+1=0
⇒5k−5=0
⇒ 5k = 5
⇒ k = 1
Thus, the required ratio is 1 : 1 and the point of division P is given by
(−4×1+11+1,0),i.e., (−4+12,0) i.e., (−32,0)
Q.6 If (1, 2), (4, y), (x, 6) and (3, 5) are the vertices of a parallelogram taken in order, find
x and y.
Sol. Let A(1, 2), B(4, y), C(x, 6) and D(3, 5) be the vertices of a parallelogram ABCD.
Since the diagonals of a parallelogram bisect each other,
Therefore, x+12=3+42
⇒ x + 1 = 7
⇒ x = 6
and, 5+y2=6+22
⇒ 5 + y = 8
⇒ y = 3
Hence, x = 6 and y = 3
Q.7 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).
Sol. Let AB be a diameter of the circle having its centre at C(2, –3) such that the coordinates of end B are (1, 4).
Let the coordinates of A be (x, y).
Since C is the mid – point of AB , therefore
x+12=2⇒x+1=4⇒x=3
and, y+42=−3⇒y+4=−6⇒y=−10
Hence, the coordinates of A are (3, –10) .
Q.8 If A and B are (–2, –2) and (2, – 4) respectively, find the coordinates of P such that AP = 37 AB and P lies on the line segment AB.
Sol. We have, AP=37AB
⇒ABAP=73
⇒AP+PBAP=3+43
⇒1+PBAP=1+43⇒PBAP=43
⇒APPB=34⇒AP:PB=3:4
Let P(x, y) be the point which divides the join of A(–2, – 2) and B(2, – 4) in the ratio 3 : 4.
Therefore, x=3×2+4×−23+4=6−87 =−27
and, y=3×−4+4×−23+4=−12−87=−207
Hence, the coordinates of the point P are (−27,−207).
Q.9 Find the coordinates of the points which divide the line segment joining A(– 2, 2) and B(2, 8) into four equal parts.
Sol. Let P1,P2 and P3 be the points that divide the line segment joining A(–2, 2) and B(2, 8) into four equal parts.
Since P2 divides the segment into two equal parts
Therefore, coordinates of P2 (i.e., mid – point) are
(−2+22,2+82), i.e., (0, 5).
Now, P1 divides the line segment AP2 into two equal parts.
Therefore, coordinates of P1(i.e., mid – point) are
(−2+02,2+52), i.e., (−1,72).
Again, P3 is the mid point of line segment P2B.
Therefore, coordinates of P3 are (0+22,5+82), i.e., (1,132).
Q.10 Find the area of a rhombus if its vertices are (3, 0), (4, 5), (– 1, 4) and (– 2,–1) taken in order.
Sol. Let A(3, 0), B(4, 5), C(–1, 4) and D(–2,– 1) be the vertices of the rhombus ABCD.
Diagonal AC=(−1−3)2+(4−0)2−−−−−−−−−−−−−−−−√ =16+16−−−−−−√=42–√ and, diagonal
BD=(−2−4)2+(−1−5)2−−−−−−−−−−−−−−−−−−√ =36+36−−−−−−√=62–√
Area of the rhombus ABCD
=12× (Product of lengths of diagonals )
=12×AC×BD
=12×42–√×62–√ sq. units
= 24 sq. units
Q.1 Find the area of the triangle whose vertices are :
(i) (2, 3), (–1, 0), (2, – 4) (ii) (– 5,–1), (3, – 5), (5, 2)
Sol. (i) LetA=(x1,y1)=(2,3),B=(x2,y2)=(−1,0)
and C=(x3,y3)=(2,−4)
Area of ΔABC
=12[x1(y2−y3)+x2(y3−y1)+x3(y1−y2)]
=12[2(0+4)+(−1)(−4−3)+2(3−0)]
=12(8+7+6)=212sq.units
(ii) Let A=(x1,y1)=(−5,−1),B=(x2,y2)=(3,−5)
and C=(x3,y3)=(5,2)
Area of ΔABC
=12[x1(y2−y3)+x2(y3−y1)+x3(y1−y2)]
=12[−5(−5−2)+3(2+1)+5(−1+5)]
=12(35+9+20)=12×64
= 32 sq. units
Q.2 In each of the following find the value of ‘k’ for which the points are collinear :
(i) (7, –2), (5, 1), (3, k) (ii) (8, 1), (k, – 4), (2, – 5)
Sol. (i) Let the given points be A=(x1,y1)=(7,−2),B=(x2,y2)=(5,1) and C=(x3,y3)=(3,k).
These points lie on a line if
Area (ΔABC)=0
⇒x1(y2−y3)+x2(y3−y1)+x3(y1−y2)=0
⇒7(1−k)+5(k+2)+3(−2−1)=0
⇒7−7k+5k+10−9=0
⇒8−2k=0
⇒ 2k = 8
⇒ k = 4
Hence, the given points are collinear for k = 4
(ii) Let the given points be A=(x1,y1)=(8,1),B=(x2,y2)=(k,−4)andC=(x3,y3)=(2,−5).
If the given points are collinear, then
⇒x1(y2−y3)+x2(y3−y1)+x3(y1−y2)=0
⇒8(−4+5)+k(−5−1)+2(1+4)=0
⇒8−6k+10=0
⇒ – 6k = – 18
⇒ k = 3
Hence, the given points are collinear for k = 3.
Q.3 Find the area of the triangle formed by joining the mid-points of the sides of the triangle whose vertices are (0, –1), (2, 1), and (0, 3). Find the ratio of this area to the area of the given triangle.
Sol. Let A=(x1,y1)=(0,−1),B=(x2,y2)=(2,1)
and C=(x3,y3)=(0,3) be the vertices of ΔABC.
Area (ΔPQR)
=12[1(0−1)+1(1−2)+0(2−0)]
=12(−1−1+0)=−1
= 1 sq. unit (numerically)
Area (ΔABC)
= 12[x1(y2−y3)+x2(y3−y1)+x3(y1−y2)]
=12[0(1−3)+2(3+1)+0(−1−1)]
=12(0+8+0)=4 sq. units
Let P(0+22,3+12),i.e., (1, 2), Q(2+02,1−12), i.e., (1, 0) and R(0+02,3−12), i.e., (0, 1) are the vertices
of ΔPQR formed by joining the mid-points of the sides of ΔABC.
Ratio of the area (ΔPQR) to the area (ΔABC) = 1 : 4.
Q.4 Find the area of the quadrilateral whose vertices, taken in order, are (– 4, –2), (– 3, –5), (3, –2) and (2, 3).
Sol. Let A(– 4, –2), B (– 3, –5), C (3, –2) and D (2, 3) be the vertices of the quadrilateral ABCD.
= Area of Δ ABC + Area of Δ ACD
=12[−4(−5+2)−3(−2+2)+3(−2+5)]
+12[−4(−2−3)+3(3+2)+2(−2+2)]
=12(12−0+9)+12(20+15+0)
=12(21+35)=12×56=28 sq. units
Q.5 You have studied in class IX (Chapter 9 Example 3) that a median of a triangle divides it into two triangles of equal areas. Verify this result for Δ ABC whose vertices are A(4, –6), B(3, –2) and C(5, 2).
Sol. Since AD is the median of ΔABC, therefore, D is the mid-point of BC. Coordinates of D are
(3+52,2+22),i.e., (4, 0).
Area of ΔADC
=12[4(0−2)+4(2+6)+5(−6−0)]
=12(−8+32−30)=12×−6=−3
= 3sq. units (numerically)
=12[4(−2−0)+3(0+6)+4(−6+2)]
=12(−8+18−16)=12(−6)=−3
= 3 sq. units (numerically)
Clearly, area (ΔADC) = area (ΔABD)
Hence, the median of the triangle divides it into two triangles of equal areas.
Q.1 Determine the ratio in which the line 2x + y – 4 = 0 divides the line segment joining the
points A(2, –2) and B(3, 7).
Sol. Suppose the line 2x + y – 4 = 0 divides the line segment joining A(2, –2) and B(3, 7) in the ratio k : 1 at
point C. Then, the coordinates of C are (3k+2k+1,7k−2k+1)
But C lies on 2x + y – 4 = 0 therefore,
2(3k+2k+1)+(7k−2k+1)−4=0
⇒6k+4+7k−2−4k−4=0
⇒ 9k – 2 = 0
⇒ 9k = 2 ⇒ k=29
So, the required ratio is 2 : 9 internally.
Q.2 Find a relation between x and y if the points (x, y), (1, 2) and (7, 0) are collinear.
Sol. The points A(x, y), B(1, 2) and C(7, 0) will be are collinear.
x(2 – 0) + 1(0 – y) + 7(y – 2) = 0
⇒ 2x – y + 7y – 14 = 0
⇒ 2x + 6y – 14 = 0
⇒ x + 3y – 7 = 0
which is the relation between x and y.
Q.3 Find the centre of a circle passing through the points (6, –6), (3, –7) and (3, 3).
Sol. Let P(x, y) be the centre of the circle passing through the points A(6, –6) B(3, –7) and C(3, 3).
Then, AP = BP = CP.
Now, AP = BP
⇒AP2=BP2
⇒(x−6)2+(y+6)2=(x−3)2+(y+7)2
⇒x2−12x+36+y2+12y+36 =x2−6x+9+y2+14y+49
⇒−12x+6x+12y−14y+72−58 = 0
⇒−6x−2y+14=0
⇒3x+y−7=0 ….(1)
and, BP = CP
⇒BP2=CP2
⇒(x−3)2+(y+7)2=(x−3)2+(y−3)2
⇒x2−6x+9+y2+14y+49 =x2−6x+9+y2−6y+9
⇒−6x+6x+14y+6y+58−18 = 0
⇒20y+40=0
⇒y=−4020=−2 …(2)
Putting y = – 2 in (1), we get
3x – 2 – 7 = 0
⇒ 3x = 9
⇒ x = 3
Thus, the centre of the circle is (3, –2).
Q.4 The two opposite vertices of a square are (–1, 2) and (3, 2). Find the coordinates of the other two vertices.
Sol. Let ABCD be a square and let A(–1, 2) and C(3, 2) be the given angular points. Let B(x, y) be the unknown vertex.
Then, AB = BC
⇒AB2=BC2
⇒(x+1)2+(y−2)2=(x−3)2+(y−2)2
⇒x2+2x+1+y2−4y+4=x2−6x+9+y2−4y+4
⇒2x+1=−6x+9
⇒ 8x = 8
⇒ x = 1 …(1)
In Δ ABC, we have
AB2+BC2=AC2
⇒ (x+1)2+(y−2)2+(x−3)2+(y−2)2=(3+1)2+(2−2)2
⇒2x2+2y2+2x−4y−6x−4y+1+4+9+4=16
⇒2x2+2y2−4x−8y+2=0
⇒x2+y2−2x−4y+1=0 …(2)
Substituting the value of x from (1) into (2), we get
1+y2−2−4y+1=0
⇒y2−4y=0⇒y(y−4)=0
⇒y=0 or 4
Hence, the required vertices of the square are (1, 0) and (1, 4).
Q.5 The class X students of a secondary school in Krishinagar have been allotted a rectangular plot of land for their gardening activity. Saplings of Gulmohar are planted on the boundary at a distance of 1m from each other. There is a triangular grassy lawn in the plot as shown in the figure. The students are to sow seeds of flowering plants on the remaining area of the plot.
(i) Taking A as origin, find the coordinates of the vertices of the triangle.
(ii) What will be the coordinates of the vertices of ΔPQR if C is the origin ? Also calculate the area of the triangle in these cases. What do you observe ?
Sol. (i) Taking A as the origin, AD and AB as the coordinate axes. Clearly, the points P,Q and R are (4, 6),
(3, 2) and (6, 5) respectively.
(ii) Taking C as the origin, CB and CD as the coordinate axes. Clearly, the points P, Q and R are
given by (12, 2), (13, 6) and (10, 3) respectively,
We know that the area of the triangle whose vertices are (x1,y1),(x2,y2) and (x3,y3) is given by
= 12[x1(y2−y3)+x2(y3−y1)+x3(y1−y2)]
Therefore, area of Δ PQR in the 1st case
12[4(2−5)+3(5−6)+6(6−2)]
=12(4×−3+3×−1+6×4)
=12(−12−3+24)=92 sq. units
and, area of ΔPQR in the 2nd case
=12[12(6−3)+13(3−2)+10(2−6)]
=12(12×3+13×1+10×−4)
=12(36+13−40)=92 sq. units
Thus, we observe that the areas are the same in both the cases.
Q.6 The vertices of a ΔABC are A(4, 6), B(1, 5) and C(7, 2). A line is drawn to intersect sides AB and AC at D and E respectively such that ADAB=AEAC=14. Calculate the area of ΔADE and compare it with the area of ΔABC.
Sol. We know that if a line divides any two sides of a triangle in the same ratio, then the line is parallel to
the third side.
Therefore, DE || BC
Clearly, ΔADE∼ΔABC (AAA similarity)
We know that the ratio of the areas of two similar triangles is equal to the ratio of the squares of any
two corresponding sides.
Therefore, Area(ΔADE)Area(ΔABC)=AD2AB2
=(ADAB)2=(14)2=116 …(1)
Now, Area (ΔABC) =12[4(5−2)+1(2−6)+7(6−5)]
=12(12−4+7)
=152 sq. units …(2)
From (1) and (2),
Area (ΔADE)=116×Area(ΔABC)
= (116×152) sq. units
=1532 sq. units
Also, form (1), area (ΔADE) : area (ΔABC) = 1 : 16.
ALITER : (Using only Coordinate Geometry)
Since ADAB=14
⇒ABAD=41⇒AD+DBAD=1+31
⇒1+DBAD=1+31
⇒DBAD=31⇒ADDB=13
Since D divides AB in the ratio 1 : 3
Therefore, coordinates of D are
(1×1+3×41+3,1×5+3×64), i.e., (134,234)
From AEAC=14, we find that
AEEC=13
Since E divides AC in the ratio 1 : 3
Therefore, coordinates of E are
(1×7+3×41+3,1×2+3×61+3),i.e., (194,5)
Now, Area (ΔADE)
= [4(234−5)+134(5−6)+194(6−234)]
=12[4×34−134+194×14]
=132[48−52+19]
=1532 sq. units
Also find the area (ΔABC)
Area ΔABC=152 sq. units
Area (ΔADE) : Area (ΔABC)
=1532:152=132:12=1:16
Q.7 Let A(4, 2), B(6, 5) and C(1, 4) be the vertices of ΔABC.
(i) The median from A meets BC at D. Find the coordinates of the point D.
(ii) Find the coordinates of the point P on AD such that AP : PD = 2 : 1.
(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.
(iv) What do you observe ?
(Note : The point which is common to all the three medians is called centroid and this point divides each median in the ratio 2 : 1).
(v) If A(x1,y1),B(x2,y2) and C(x3,y3) are the vertices of ΔABC, find the coordinates of the centroid of the triangle.
Sol. Let A(4, 2), B(6, 5) and C(1, 4) be the vertices of ΔABC.
(i) Since AD is the median of ΔABC, Therefore, D is the mid-point of BC.
Its coordinates are (6+12,5+42),i.e., (72,92).
(ii) Since P divides AD in the ratio 2 : 1 , so its coordinates are P(2×72+1×42+1,2×92+1×22+1)
or P(7+43,9+23) i.e., P(113,113)
(iii) Since BE is the median of ΔABC, so E is the mid-point of AC and its coordinates are E(4+12,2+42) i.e., E(52,3)
Since Q divides BE in the ratio 2 : 1, so its coordinates are
Q(2×52+1×62+1,2×3+1×52+1) or Q(5+63,6+53), i.e., Q(113,113)
Since CF is the median of ΔABC, so F is the mid-point of AB. Therefore, its coordinates are
F(4+62,2+52), i.e., F(5,72).
Since R divides CF in the ratio 2 : 1, so its coordinates are
R(2×5+1×12+1,2×72+1×42+1) or R(10+13,7+43) or R(113,113)
(iv) We observe that the points P,Q and R coincide, the medians AD, BE and CF are concurrent at the point
(113,113). This point is known as the centroid of the triangle.
(v) Let A(x1,y1),B(x2,y2) and C(x3,y3) be the vertices of ΔABC whose medians are AD, BE, and CF respectively. So, D,E and F are respectively the mid-points of BC,CA and AB.
Coordinates of D are (x2+x32,y2+y32)
Coordinates of a point dividing AD in the ratio 2 : 1 are
(1.x1+2(x2+x32)1+21.y1+2(y2+y32)1+2) =(x1+x2+x33,y1+y2+y33)
The coordinates of E are (x1+x22,y1+y22).The coordinates of a point dividing BE in the ratio 2 : 1 are
(1.x2+2(x1+x32)1+2,1.y2+2(y1+y32))
=(x1+x2+x33,y1+y2+y33)
Similarly, the coordinates of a point dividing CF in the ratio 2 : 1 are
(x1+x2+x33,y1+y2+y33)
Thus, the point (x1+x2+x33,y1+y2+y33) is common to AD,BE and CF and divides them in the ratio 2 : 1
Therefore, the medians of a triangle are concurrent and the coordinates of the centroid are
(x1+x2+x33,y1+y2+y33)
Q.8 ABCD is a rectangle formed by joining the points A(– 1,–1), B(– 1, 4), C(5, 4) and D(5,–1). P, Q, R and S are the mid-points of AB, BC, CD and DA respectively. Is the quadrilateral PQRS a square ? a rectangle ? or a rhombus ? Justify your answer.
Sol. Various points are marked in the adjoining figure.
PQ=(2+1)2+(4−32)2−−−−−−−−−−−−−−−−√ = 32+(52)2−−−−−−−−√
= 9+254−−−−−√ =36+254−−−−−√=614−−√
QR=(5−2)2+(32−4)2−−−−−−−−−−−−−−−−√
= 32+(−52)2−−−−−−−−−√=9+254−−−−−√
=36+254−−−−−√=614−−√
RS=(2−5)2+(−1−32)2−−−−−−−−−−−−−−−−−√
= (−32)+(−52)2−−−−−−−−−−−−−√
=9+254−−−−−√=36+254−−−−−√=614−−√
and, SP=(−1−2)2+(32+1)2−−−−−−−−−−−−−−−−−√
=(−3)2+(52)2−−−−−−−−−−−√
=9+254−−−−−√=36+254−−−−−√=614−−√
⇒ PQ = QR = RS = SP
Now, PR=(5+1)2+(32−32)2−−−−−−−−−−−−−−−−√ = 36−−√=6
and SQ=(2−2)2+(4+1)2−−−−−−−−−−−−−−−√ =25−−√=5
⇒PR≠SQ
Since all the sides are equal but the diagonals are not equal,
therefore, PQRS is a rhombus.
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