Rs Aggarwal 2019 Solutions for Class 10 Math Chapter 6 Coordinate Geometry are provided here with simple step-by-step explanations. These solutions for Coordinate Geometry are extremely popular among Class 10 students for Math Coordinate Geometry Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rs Aggarwal 2019 Book of Class 10 Math Chapter 6 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rs Aggarwal 2019 Solutions. All Rs Aggarwal 2019 Solutions for class Class 10 Math are prepared by experts and are 100% accurate.

Page No 307:

Question 1:

Find the distance between the points:

(i) A(9, 3) and B(15, 11)
(ii) A(7, −4) and B(−5, 1)
(iii) A(−6, −4) and B(9, −12)
(iv) A(1, −3) and B(4, −6)
(v) P(a + b, ab) and Q(ab, a + b)
(vi) P(a sin α, a cos α) and Q(a cos α, −a sin α)

Answer:

  (i) A(9, 3) and B(15, 11)
The given points are A(9, 3) and B(15, 11).
Then (x1 = 9, y1 = 3) and (x2 = 15, y2 = 11)
AB = x2-x12+y2-y12      =15-92+11-32      =15-92+11-32      =62+82      =36+64      =100      = 10 units

(ii) A(7, −4) and B(−5, 1)
The given points are A(7, −4) and B(−5, 1).
Then (x1= 7, y1 = −4) and (x2 = −5, y2 = 1)
AB = x2-x12+y2-y12       =-5-72+1--42       =-5-72+1+42       =-122+52       =144+25       =169       = 13 units

(iii) A(−6, −4) and B(9, −12)
The given points are A(−6, −4) and B(9, −12).
Then (x1 = −6, y1 = −4) and (x2 = 9, y2 = −12)
AB = x2-x12+y2-y12      =9--62+-12--42      =9+62+-12+42      =152+-82      =225+64      =289      = 17 units

(iv) A(1, −3) and B(4, −6)
The given points are A(1, −3) and B(4, −6).
Then (x1 = 1, y1 = −3) and (x2 = 4, y2 = −6)
AB = x2-x12+y2-y12      =4-12+-6--32      =4-12+-6+32      =32+-32      =9+9      =18      =9×2      = 32 units

(v) P(a + b, ab) and Q(ab, a + b)
The given points are P(a + b, a − b) and Q(a − b, a + b).
Then (x1 = a + b, y1 = a − b) and (x2 = a − b, y2 = a + b)
PQ = x2-x12+y2-y12       =a-b-a+b2+a+b-a-b2       =a-b-a-b2+a+b-a+b2       =-2b2+2b2       =4b2+4b2       =8b2       =4×2b2       = 22b units

(vi) P(a sin α, a cos α) and Q(a cos α, −a sin α)
The given points are P(a sin α, a cos α) and Q(a cos α, −a sin α).
Then (x1 = a sin α, y1 = a cos α) and (x2 = a cos α, y2 = −a sin α)
PQ = x2-x12+y2-y12       =a cos α-a sin α2+-a sinα-a cos α2       =a2cos2α +a2 sin2 α - 2a2cos α×sin α+a2 sin2 α+a2cos2α+2a2cos α× sin α       =2a2cos2α +2a2 sin2 α       =2a2cos2α + sin2 α       =2a21                  From the identity cos2α + sin2 α = 1       =2a2       = 2a units

Page No 307:

Question 2:

Find the distance of each of the following points from the origin:

(i) A(5, −12)
(ii) B(−5, 5)
(iii) C(−, −6)4

Answer:

(i) A(5, −12)
Let O(0, 0) be the origin.
 OA = 5-02+-12-02      = 52+-122      = 25+144      = 169      =13 units

(ii) B(−5, 5)
Let O(0, 0) be the origin.
OB = -5-02+5-02       = -52+52       = 25+25       = 50       =25×2       =52 units

(iii) C(−4, −6)
Let O(0,0) be the origin.
OC = -4-02+-6-02       = -42+-62       = 16+36       = 52       =4×13       =213 units

Page No 307:

Question 3:

Find all possible values of x for which the distance between the points A(x, −1) and B(5, 3) is 5 units.

Answer:

Given AB = 5 units
Therefore, (AB)2 = 25 units
5-x2+3--12 = 255-x2+3+12 = 255-x2+42 = 255-x2+16 = 255-x2 = 25-165-x2 = 95-x =± 95-x= ±35-x = 3  or   5-x=-3x=2 or 8
Therefore, x = 2 or 8.

Page No 307:

Question 4:

Find all possible values of y for which the distance between the points A2,-3 and B10,y is 10 units.

Answer:

The given points are A2,-3 and B10,y.
AB=2-102+-3-y2          =-82+-3-y2          =64+9+y2+6y
AB=1064+9+y2+6y=1073+y2+6y=100        Squaring both sidesy2+6y-27=0
y2+9y-3y-27=0yy+9-3y+9=0y+9y-3=0y+9=0 or y-3=0
y=-9 or y=3
Hence, the possible values of y are -9 and 3.

Page No 307:

Question 5:

Find the values of x for which the distance between the points P(x, 4) and Q(9, 10) is 10 units.

Answer:

The given points are P(x, 4) and Q(9, 10).
PQ=x-92+4-102          =x-92+-62          =x2-18x+81+36           =x2-18x+117
PQ=10x2-18x+117=10x2-18x+117=100        Squaring both sidesx2-18x+17=
x2-17x-x+17=0xx-17-1x-17=0x-17x-1=0x-17=0 or x-1=0
x=17 or x=1
Hence, the values of x are 1 and 17.

Page No 307:

Question 6:

If the point A(x, 2) is equidistant from the points B(8, − 2) and C(2, − 2), find the value of x. Also, find the length of AB.

Answer:

As per the question
AB=ACx-82+2+22=x-22+2+22
Squaring both sides, we get
x-82+42=x-22+42x2-16x+64+16=x2+4-4x+1616x-4x=64-4x=6012=5
Now,
AB=x-82+2+22     =5-82+2+22                x=2     =-32+42     =9+16=25=5
Hence, x = 5 and AB = 5 units.

Page No 307:

Question 7:

If the point A(0, 2) is equidistant from the points B(3, p) and C(p, 5), find the value of p. Also, find the length of AB.

Answer:

As per the question
AB=AC0-32+2-p2=0-p2+2-52-32+2-p2=-p2+-32
Squaring both sides, we get
-32+2-p2=-p2+-329+4+p2-4p=p2+94p=4p=1
Now,
AB=0-32+2-p2     =-32+2-12                p=1     =9+1     =10 units
Hence, p = 1 and AB10 units.



Page No 308:

Question 8:

Find the coordinates of the point on x-axis which is equidistant from the points (–2, 5) and (2, –3).

Answer:

Let the point on the x - axis be (x, 0).
We have A-2,5 and B2,-3AX=BXAX2=BX2    .....1Using distance formula:d=x2-x12+y2-y12For AXAX=x--22+0-52AX2=x+22+-52AX2=x2+4x+29     .....2
And BX=x-22+0--32BX2=x-22+32BX2=x2-4x+13    .....3From 1,2 and 3x2+4x+29=x2-4x+13x=-2
Point on the x - axis is (-2, 0).

Page No 308:

Question 9:

Find points on the x-axis, each of which is at a distance of 10 units from the point A(11, −8).

Answer:

Let P (x, 0) be the point on the x-axis. Then as per the question, we have
AP=10x-112+0+82=10x-112+82=100                     Squaring both sidesx-112=100-64=36
x-11=±6x=11±6x=11-6, 11+6x=5, 17
Hence, the points on the x-axis are (5, 0) and (17, 0).

Page No 308:

Question 10:

Find the point on the y-axis which is equidistant from the points A(6, 5) and B(− 4, 3).

Answer:

Let P (0, y) be a point on the y-axis. Then as per the question, we have
AP=BP0-62+y-52=0+42+y-3262+y-52=42+y-3262+y-52=42+y-32                     Squaring both sides
36+y2-10y+25=16+y2-6y+94y=36y=9
Hence, the point on the y-axis is (0, 9).

Page No 308:

Question 11:

If the point P(x, y) is equidistant from the points A(5, 1) and B(− 1, 5), prove that 3x = 2y.

Answer:

As per the question, we have
AP=BPx-52+y-12=x+12+y-52x-52+y-12=x+12+y-52        Squaring both sidesx2-10x+25+y2-2y+1=x2+2x+1+y2-10y+25
-10x-2y=2x-10y8y=12x3x=2y
Hence, 3x = 2y.

Page No 308:

Question 12:

If P(x, y) is a point equidistant from the points A(6, −1) and B(2, 3), show that xy = 3.

Answer:

The given points are A(6, −1) and B(2, 3). The point P(x, y) is equidistant from the points A and B. So, PA = PB.
Also, (PA)2 = (PB)2
6-x2+-1-y2 = 2-x2+3-y2x2-12x+36+y2+2y+1 = x2-4x+4+y2-6y+9x2+y2-12x+2y+37 = x2+y2-4x-6y+13x2+y2-12x+2y-x2-y2+4x+6y =13-37-8x+8y = -24-8x-y = -24x-y =-24-8x-y =3

Hence proved.

Page No 308:

Question 13:

Find the coordinates of the point equidistant from the three points A(5, 3), B(5, −5) and C(1, −5).

Answer:

Let the required point be P(x, y). Then AP = BP = CP
That is, (AP)2=(BP)2=(CP)2
This means (AP)2=(BP)
x-52+y-32=x-52+y+52  x2-10x+25+y2-6y+9=x2-10x+25+y2+10y+25  x2-10x+y2-6y+34=x2-10x+y2+10y+50  x2-10x+y2-6y-x2+10x-y2-10y = 50-34  -16y = 16  y =-1616=-1  And BP2=CP2 x-52+y+52=x-12+y+52 x2-10x+25+y2+10y+25=x2-2x+1+y2+10y+25 x2-10x+y2+10y+50=x2-2x+y2+10y+26 x2-10x+y2+10y-x2+2x-y2-10y = 26-50 -8x = -24 x =-24-8= 3
Hence, the required point is (3, −1).

Page No 308:

Question 14:

If the points A(4, 3) and B(x, 5) lie on a circle with centre O(2, 3), find the value of x.

Answer:

Given, the points A(4, 3) and B(x, 5) lie on a circle with centre O(2, 3).
Then OA = OB
Also (OA)2 = (OB)2
4-22+3-32 = x-22+5-3222+02 = x-22+224= x-22+4x-22 = 0x-2 = 0x=2
Therefore, x = 2.

Page No 308:

Question 15:

If the point C(− 2, 3) is equidistant from the points A(3, − 1) and B(x, 8), find the value of x. Also, find the distance BC.

Answer:

As per the question, we have
AC=BC-2-32+3+12=-2-x2+3-8252+42=x+22+-5225+16=x+22+25        Squaring both sides
25+16=x+22+25x+22=16x+2=±4x=-2±4=-2-4, -2+4=-6, 2
Now
BC=-2-x2+3-82      =-2-22+-5      =16+25=41 units
Hence, x = 2 or −6 and BC=41 units.

Page No 308:

Question 16:

If the point P(2, 2) is equidistant from the points A(−2, k) and B(−2k, −3), find k. Also find the length of AP.

Answer:

As per the question, we have
AP=BP2+22+2-k2=2+2k2+2+3242+2-k2=2+2k2+5216+4+k2-4k=4+4k2+8k+25        Squaring both sides
k2+4k+3=0k+1k+3=0k=-3, -1
Now for k=-1
AP=2+22+2-k2      =42+2+12                   =16+9=5 units
For k=-3
AP=2+22+2-k2      =42+2+32                   =16+25=41 units
Hence, k=-1,-3 ; AP=5 units for k=-1 and AP=41 units for k=-3.

Page No 308:

Question 17:

(i) If the point (x, y) is equidistant from the points (a + b, ba) and (a b, a + b), prove that bx = ay.

(ii) If the distances of P(x, y) from A(5, 1) and B(–1, 5) are equal then prove that 3x = 2y.

Answer:

(i) As per the question, we have
x-a-b2+y-b+a2=x-a+b2+y-a-b2x-a-b2+y-b+a2=x-a+b2+y-a-b2      Squaring both sidesx2+a+b2-2xa+b+y2+a-b2-2ya-b=x2+a-b2-2xa-b+y2+a+b2-2ya+b-xa+b-ya-b=-xa-b-ya+b
-xa-xb-ay+by=-xa+bx-ya-byby=bx
Hence, bx = ay.

(ii)
As per the question, we have
AP=BPx-52+y-12=x+12+y-52x-52+y-12=x+12+y-52        Squaring both sidesx2-10x+25+y2-2y+1=x2+2x+1+y2-10y+25
-10x-2y=2x-10y8y=12x3x=2y
Hence, 3x = 2y.

Page No 308:

Question 18:

Using the distance formula, show that the given points are collinear:
(i) (1, −1), (5, 2) and (9, 5)                                         (ii) (6, 9), (0, 1) and (−6, −7)
(iii) (−1, −1), (2, 3) and (8, 11)                                  (iv) (−2, 5), (0, 1) and (2, −3)

Answer:

(i)
Let A(1, −1), B(5, 2) and C(9, 5) be the given points. Then
AB=5-12+2+12=42+32=25=5 unitsBC=9-52+5-22=42+32=25=5 unitsAC=9-12+5+12=82+62=100=10 units
AB+BC=5+5 units=10 units=AC
Hence, the given points are collinear.

(ii)
Let A(6, 9), B(0, 1) and C(−6, −7) be the given points. Then
AB=0-62+1-92=-62+-82=100=10 unitsBC=-6-02+-7-12=-62+-82=100=10 unitsAC=-6-62+-7-92=-122+-162=400=20 units
AB+BC=10+10 units=20 units=AC
Hence, the given points are collinear.

(iii)
Let A(−1, −1), B(2, 3) and C(8, 11) be the given points. Then
AB=2+12+3+12=32+42=25=5 unitsBC=8-22+11-32=62+82=100=10 unitsAC=8+12+11+12=92+122=225=15 units
AB+BC=5+10 units=15 units=AC
Hence, the given points are collinear.

(iv)
Let A(−2, 5), B(0, 1) and C(2, −3) be the given points. Then
AB=0+22+1-52=22+-42=20=25 unitsBC=2-02+-3-12=22+-42=20=25 unitsAC=2+22+-3-52=42+-82=80=45 units
AB+BC=25+25 units=45 units=AC
Hence, the given points are collinear.

Page No 308:

Question 19:

Prove that the points A(7, 10), B(−2, 5) and C(3, −4) are the vertices of an isosceles right triangle.

Answer:

The given points are A(7, 10), B(−2, 5) and C(3, −4).
AB = -2-72+5-102 = -92+-52 =81+25 =106BC = 3--22+-4-52 = 52+-92 =25+81 = 106AC = 3-72+-4-102 = -42+-142 = 16+196 =212
Since, AB and BC are equal, they form the vertices of an isosceles triangle.
Also, (AB)2+(BC)21062+ 1062=212 
and (AC)2 = 2122 = 212
Thus, (AB)2+(BC)2 = (AC)2
This show that ABC is right- angled at B.
Therefore, the points A(7, 10), B(−2, 5) and C(3, −4) are the vertices of an isosceles right-angled triangle.

Page No 308:

Question 20:

Show that the points A(3, 0), B(6, 4) and C(− 1, 3) are the vertices of an isosceles right triangle.

Answer:

The given points are A(3, 0), B(6, 4) and C(− 1, 3). Now
AB=3-62+0-42=-32+-42     =9+16=25=5BC=6+12+4-32=72+12     =49+1=50=52
AC=3+12+0-32=42+-32     =16+9=25=5
AB=AC and AB2+AC2=BC2
Therefore, A(3, 0), B(6, 4) and C(− 1, 3) are the vertices of an isosceles right triangle.

Page No 308:

Question 21:

Show that the points A(5, 2), B(2, − 2) and C(− 2, t) are the vertices of a right triangle with B=90, then find the value of t.

Answer:

B=90AC2=AB2+BC25+22+2-t2=5-22+2+22+2+22+-2-t272+t-22=32+42+42+t+22
49+t2-4t+4=9+16+16+t2+4t+48-4t=4t8t=8t=1
Hence, t = 1.

Page No 308:

Question 22:

Prove that the points A(2, 4), B(2, 6) and C2+3, 5 are the vertices of an equilateral triangle.

Answer:

The given points are A(2, 4), B(2, 6) and C2+3, 5. Now
AB=2-22+4-62=02+-22     =0+4=2BC=2-2-32+6-52=-32+12     =3+1=2
AC=2-2-32+4-52=-32+-12     =3+1=2
Hence, the points A(2, 4), B(2, 6) and C2+3, 5 are the vertices of an equilateral triangle.

Page No 308:

Question 23:

Show that the points (− 3, − 3), (3, 3) and -33, 33 are the vertices of an equilateral triangle.

Answer:

Let the given points be A(− 3, − 3), B(3, 3) and C-33, 33. Now
AB=-3-32+-3-32=-62+-62     =36+36=72=62BC=3+332+3-332      =9+27+183+9+27-183=72=62
AC=-3+332+-3-332=3-332+3+332     =9+27-183+9+27+183      =72=62
Hence, the given points are the vertices of an equilateral triangle.

Page No 308:

Question 24:

Show that the points A(−5, 6) B(3, 0) and C(9, 8) are the vertices of an isosceles right-angled triangle. Calculate its area.

Answer:

Let the given points be A(−5, 6) B(3, 0) and C(9, 8).
AB = 3--52+0-62 = 82+-62 =64+36 =100 =10 unitsBC = 9-32+8-02 = 62+82 =36+64 = 100 = 10 unitsAC = 9--52+8-62 = 142+22 = 196+4 =200 = 102 unitsTherefore, AB= BC = 10 units

Also, (AB)2+(BC)2102+ 102=200 
and (AC)2 = 1022=200
Thus, (AB)2+(BC)2 = (AC)2
This show that ABC is right- angled at B.
Therefore, the points A(−5, 6) B(3, 0) and C(9, 8) are the vertices of an isosceles right-angled triangle.
Also, area of a triangle = 12×base×height
If AB is the height and BC is the base,Area = 12×10×10 = 50 square units



Page No 309:

Question 25:

Show that the points O(0, 0) A(3, 3) and B(3, −3) are the vertices of an equilateral triangle. Find the area of this triangle.

Answer:

The given points are O(0, 0) A(3, 3) and B(3, − 3).
OA = 3-02+3-02 = 32+32 = 9+3 = 12 =23 unitsAB = 3-32+-3 -32 = 0+232 =43 = 12 = 23 unitsOB = 3-02+-3 -02 =32+32 = 9+3 =12 =23 unitsTherefore, OA = AB = OB = 23 units
Thus, t
he points O(0, 0) A(3, 3)and B(3, − 3) are the vertices of an equilateral triangle.
Also, the area of the triangle OAB = 34×side2
34×232  =34×12  = 33 square units

Page No 309:

Question 26:

Show that the following points are the vertices of a square.

(i) A(3, 2), B(0, 5), C(−3, 2) and D(0, −1)    
(ii) A(6, 2), B(2, 1), C(1, 5) and D(5, 6)   
(iii) P(0, −2), Q(3, 1), R(0, 4) and S(−3, 1)

Answer:

(i) The given points are A(3, 2), B(0, 5), C(−3, 2) and D(0, −1).
AB = 0-32+5-22 = -32+32 = 9+9 = 18 =32 unitsBC = -3-02+2-52 = -32+-32 = 9+9 = 18 =32 unitsCD = 0+32+-1-22 = 32+-32 = 9+9 = 18 =32 unitsDA =0-32+-1-22 = -32+-32 = 9+9 = 18 =32 unitsTherefore, AB = BC = CD = DA = 32 unitsAlso, AC = -3-32+2-22 = -62+02 = 36 = 6 unitsBD = 0-02+-1-52 = 02+-62 = 36 = 6 unitsThus, diagonal AC = diagonal BD
Therefore, the given points form a square.


(ii) The given points are A(6, 2), B(2, 1), C(1, 5) and D(5, 6).
AB = 2-62+1-22 = -42+-12 = 16+1 = 17 unitsBC = 1-22+5-12 = -12+-42 = 1+16 = 17 unitsCD = 5-12+6-52 = 42+12 = 16+1 = 17 unitsDA =5-62+6-22 = 12+42 = 1+16 = 17 unitsTherefore AB = BC = CD = DA = 17 unitsAlso AC = 1-62+5-22 = -52+32 = 25+9 = 34 unitsBD = 5-22+6-12 = 32+52 = 9+25 = 34 unitsThus, diagonal AC = diagonal BD
Therefore, the given points form a square.


(iii) The given points are P(0, −2), Q(3, 1), R(0, 4) and S(−3, 1).
PQ = 3-02+1+22 = 32+32 = 9+9 = 18 = 32 unitsQR = 0-32+4-12 = -32+32 = 9+9 = 18 =32 unitsRS = -3-02+1-42 = -32+-32 = 9+9 = 18 = 32unitsSP =-3-02+1+22 = -32+32 = 9+9 = 18  = 32unitsTherefore, PQ = QR = RS = SP = 32 unitsAlso, PR = 0-02+4+22 = 02+62 = 36 = 6 unitsQS= -3-32+1-12 = -62+02 = 36 = 6 unitsThus, diagonal PR = diagonal QS
Therefore, the given points form a square.

Page No 309:

Question 27:

Show that the points A(−3, 2), B(−5, −5), C(2, −3) and D(4, 4) are the vertices of a rhombus. Find the area of this rhombus.

Answer:

The given points are A(−3, 2), B(−5, −5), C(2, −3) and D(4, 4).
AB = -5+32+-5-22 = -22+-72 = 4+49 = 53 units.BC = 2+52+-3+52 = 72+22 = 49+4 = 53 units.CD = 4-22+4+32 = 22+72 = 4+49 = 53 units.DA= 4+32+4-22 = 72+22 = 49+4 = 53 units.Therefore, AB= BC= CD = DA = 53 unitsAC = 2+32+-3-22 = 52+-52 =25+25 = 50 =25×2= 52 unitsBD = 4+52+4+52 = 92+92 = 81+81 = 162 =81×2= 92 unitsThus, diagonal AC is not equal to diagonal BD.
Therefore, ABCD is a quadrilateral with equal sides and unequal diagonals.
Hence, ABCD is a rhombus.
Area of a rhombus = 12×product of diagonals                                    = 12×52×92                                    = 4522                                    = 45 square units

Page No 309:

Question 28:

Show that the points A(3, 0), B(4, 5), C(−1, 4) and D(−2, −1) are the vertices of a rhombus. Find its area.

Answer:

The given points are A(3, 0), B(4, 5), C(− 1, 4) and D(− 2, − 1).
AB=3-42+0-52=-12+-52     =1+25=26
BC=4+12+5-42=52+12     =25+1=26
CD=-1+22+4+12=12+52     =1+25=26AD=3+22+0+12=52+12     =25+1=26
AC=3+12+0-42=42+-42     =16+16=42BD=4+22+5+12=62+62     =36+36=62
AB=BC=CD=AD=62 and ACBD
Therefore, the given points are the vertices of a rhombus.
AreaABCD=12×AC×BD                          =12×42×62=24 sq. units
Hence, the area of the rhombus is 24 sq. units.

Page No 309:

Question 29:

Show that the points A(6, 1), B(8, 2), C(9, 4) and D(7, 3) are the vertices of a rhombus. Find its area.

Answer:

The given points are A(6, 1), B(8, 2), C(9, 4) and D(7, 3).
AB=6-82+1-22=-22+-12     =4+1=5
BC=8-92+2-42=-12+-22     =1+4=5
CD=9-72+4-32=22+12     =4+1=5AD=7-62+3-12=12+22     =1+4=5
AC=6-92+1-42=-32+-32     =9+9=32BD=8-72+2-32=12+-12     =1+1=2
AB=BC=CD=AD=5 and ACBD
Therefore, the given points are the vertices of a rhombus. Now
AreaABCD=12×AC×BD                          =12×32×2=3 sq. units
Hence, the area of the rhombus is 3 sq. units.

Page No 309:

Question 30:

Show that the points A(2, 1), B(5, 2), C(6, 4) and D(3, 3) are the angular points of a parallelogram. Is this figure a rectangle?

Answer:

The given points are A(2, 1), B(5, 2), C(6, 4) and D(3, 3).
AB = 5-22+2-12 = 32+12 = 9+1 =10 unitsBC = 6-52+4-22 = 12+22 = 1+4 =5 unitsCD = 3-62+3-42 =-32+-12 = 9+1 =10 unitsAD = 3-22+3-12 = 12+22 = 1+4 =5 unitsThus, AB = CD = 10 units and BC = AD =5 unitsSo, quadrilateral ABCD is a parallelogramAlso, AC = 6-22+4-12 = 42+32 = 16+9 = 25=5 unitsBD = 3-52+3-22 =-22+12 = 4+1 = 5 units
But diagonal AC is not equal to diagonal BD.
Hence, the given points do not form a rectangle.

Page No 309:

Question 31:

Show that A(1, 2), B(4, 3), C(6, 6) and D(3, 5) are the vertices of a parallelogram. Show that ABCD is not a rectangle.

Answer:

The given vertices are A(1, 2), B(4, 3), C(6, 6) and D(3, 5).
AB=1-42+2-32=-32+-12     =9+1=10
BC=4-62+3-62=-22+-32     =4+9=13
CD=6-32+6-52=32+12     =9+1=10AD=1-32+2-52=-22+-32     =4+9=13
AB=CD=10 units and BC=AD=13 units
Therefore, ABCD is a parallelogram. Now
AC=1-62+2-62=-52+-42     =25+16=41BD=4-32+3-52=12+-22     =1+4=5
Thus, the diagonals AC and BD are not equal and hence ABCD is not a rectangle.

Page No 309:

Question 32:

Show that the following points are the vertices of a rectangle.
(i) A(−4, −1), B(−2, −4) C(4, 0) and D(2, 3)   
(ii) A(2, −2), B(14, 10) C(11, 13) and D(−1, 1)
(iii) A(0, −4), B(6, 2) C(3, 5) and D(−3, −1)

Answer:

(i)
Given: A-4,-1,B-2,-4,C4,0 and D2,3In rectangle the opposite sides are equalUsing d=x2-x12+y2-y12AB=-4+22+-1+42AB=13BC=4+22+0+42BC=52CD=2-42+3-02CD=13AD=2+42+3+12AD=52AB=CD and BC=AD
Now, we will prove for rectangles because diagnols in a rectangle are also equalAC=4--42+0--12AC=4+42+12AC=65AndBD=2--22+3--42BD=42+72BD=65We can see that diagonals are also equal Therefore, given coordinates are of rectangle.

(ii) The given points are A(2, −2), B(14, 10) C(11, 13) and D(−1, 1).
AB = 14-22+10--22 = 122+122 = 144+144 =288 =122 unitsBC= 11-142+13-102 = -32+32 = 9+9 = 18 =32 unitsCD = -1-112+1-132 = -122+-122 = 144+144 =288 =122 unitsAD = -1-22+1--22 = -32+32 = 9+9 =18 =32 unitsThus, AB = CD = 122 units and BC = AD =32 unitsAlso, AC = 11-22+13--22 = 92+152 = 81+225 =306 =334 unitsBD = -1-142+1-102 = -152+-92 = 81+225 = 306=334 units
Also, diagonal AC = diagonal BD
Hence, the given points form a rectangle.

(iii) The given points are A(0, −4), B(6, 2) C(3, 5) and D(−3, −1).
AB = 6-02+2--42 = 62+62 = 36+36 =72 =62 unitsBC = 3-62+5-22 = -32+32 = 9+9 =18=32 unitsCD = -3-32+-1-52 = -62+-62 = 36+36 =72 =62 unitsAD = -3-02+-1--42 = -32+32 = 9+9 =18 =32 unitsThus,  AB = CD = 10 units and BC = AD =5 unitsAlso, AC = 3-02+5--42 = 32+92 = 9+81 =90=310 unitsBD = -3-62+-1-22 = -92+-32 = 81+9 = 90=310 units
Also, diagonal AC = diagonal BD.
Hence, the given points form a rectangle.

Page No 309:

Question 33:

Show that ΔABC with vertices A(–2, 0), B(0, 2) and C(2, 0) is similar to ΔDEF with vertices D(–4, 0), F(4, 0) and E(0, 4).        [CBSE 2017]

Answer:

In ΔABC, the coordinates of the vertices are A(–2, 0), B(0,2), C(2,0). 

AB=0+22+2-02=8=22CB=0-22+2-02=8=22AC=2+22+0-02=4

In ΔDEF, the coordinates of the vertices are D(–4, 0), E(4, 0), F(0, 4).

DF=4+42+0-02=8FE=0-42+4-02=42DE=0+42+4-02=42

Now, for ΔABC and ΔDEF to be similar, the corresponding sides should be proportional. 

So, ACDF=BCFE=ABDE48=2242=224212=12=12Since, the corresponding sides are proportional Therefore, given two triangles are similar.

Page No 309:

Question 34:

Show that ΔABC, where A(22, 0), B(2, 0), C(0, 2) and ΔPQR where P(–4, 0), Q(4, 0), R(0, 4) are similar triangles.           [CBSE 2017]

Answer:

Disclaimer: Instead of 22 it should be -2
 

In ΔABC, the coordinates of the vertices are A(–2, 0), B(2, 0), C(0, 2). 

AB=2+22+0-02=4CB=0-22+2-02=8=22AC=0+22+2-02=8=22

In ΔPQR, the coordinates of the vertices are P(–4, 0), Q(4, 0), R(0, 4).

PQ=4+42+0-02=8QR=0-42+4-02=42PR=0+42+4-02=42

Now, for ΔABC and ΔPQR to be similar, the corresponding sides should be proportional. 

So, ABPQ=BCQR=CAPR48=2242=2242=12
Thus, ΔABC is similar to ΔPQR. 



Page No 320:

Question 1:

(i) Find the coordinates of the point that divides the join of A(−1, 7) and B(4, −3) in the ratio 2 : 3.

(ii) Find the coordinates of the point which divides the join of A(−5, 11) and B(4, −7) in the ratio 7 : 2.

Answer:

(i) The end points of AB are A(−1, 7) and B(4, −3).
Therefore, (x1 = −1, y1 = 7) and (x2 = 4, y2 = −3)
Also, m = 2 and n = 3
Let the required point be P(x, y).
By section formula, we get:
x =mx2+nx1m+n, y = my2+ny1m+nx= 2×4 + 3×-12+3, y =2×-3+3×72+3x= 8-35, y = -6+215x = 55, y = 155Therefore, x = 1 and y = 3
Hence, the coordinates of the required point are (1, 3).

(ii) The end points of AB are A(−5, 11) and B(4, −7).
Therefore, (x1 = −5, y1 = 11) and (x2 = 4, y2 = −7).
Also, m = 7 and n = 2
Let the required point be P(x, y).
By section formula, we have:
x =mx2+nx1m+n, y = my2+ny1m+nx= 7×4 + 2×-57+2, y =7×-7+2×117+2x= 28-109, y = -49+229x = 189, y = -279Therefore, x = 2 and y = -3
Hence, the required point is P(2, −3).

Page No 320:

Question 2:

Find the coordinates of the points of trisection of the line segment joining the points A(7, –2) and B(1, –5).                [CBSE 2017]

Answer:


Consider the figure.
Here points P and Q trisect AB.
Therefore, P divides AB into 1 : 2 and Q divides AB into 2 : 1.
Using section formula, coordinates of P are;
P(x,y)=1×1+2×73,1×-5+2×-23P(x,y)=153,-93=5,-3
Similarly, coordinates of Q are;
Q(a,b)=2×1+1×73,2×-5+1×-23Q(a,b)=93,-123=3,-4
Therefore, coordinates of points P and Q are (5, -3) and (3, -4) respectively.

Page No 320:

Question 3:

If the coordinates of points A and B are (−2, −2) and (2, −4) respectively, find the coordinates of the point P
such that AP=37AB, where P lies on the line segment AB.     [CBSE 2015]

Answer:

The coordinates of the points A and B are (−2, −2) and (2, −4) respectively, where AP=37AB and P lies on the line segment AB. So
AP+BP=ABAP+BP=7AP3                 AP=37ABBP=7AP3-APAPBP=34
Let (x, y) be the coordinates of P which divides AB in the ratio 3 : 4 internally. Then
x=3×2+4×-23+4=6-87=-27y=3×-4+4×-23+4=-12-87=-207
Hence, the coordinates of point P are -27, -207.



Page No 321:

Question 4:

Point A lies on the line segment PQ joining P(6, −6) and Q(−4, −1) in such a way that PAPQ=25.
If the point A also lies  on the line 3x + k (y + 1) = 0, find the value of k.                               [CBSE 2015]

Answer:

Let the coordinates of A be (x, y). Here, PAPQ=25. So,
PA+AQ=PQPA+AQ=5PA2                 [PA=25PQ]AQ=5PA2-PAAQPA=32PAAQ=23
Let (x, y) be the coordinates of A, which divides PQ in the ratio 2 : 3 internally. Then using section formula, we get
x=2×-4+3×62+3=-8+185=105=2y=2×-1+3×-62+3=-2-185=-205=-4
Now, the point (2, −4) lies on the line 3x + k (y + 1) = 0, therefore
3×2+k-4+1=03k=6k=63=2
Hence, k = 2.

Page No 321:

Question 5:

Point P, Q, R and S divide the line segment joining the points A(1, 2) and B(6, 7) in five equal parts.
Find the coordinates of the points P, Q and R.

Answer:

Since, the points P, Q, R and S divide the line segment joining the points A(1, 2) and B(6, 7) in five equal parts, so
AP = PQ = QR = RS = SB
Here, point P divides AB in the ratio of 1 : 4 internally. So using section formula, we get
Coordinates of P=1×6+4×11+4, 1×7+4×21+4                           =6+45, 7+85=2, 3
The point Q divides AB in the ratio of 2 : 3 internally. So using section formula, we get
Coordinates of Q=2×6+3×12+3, 2×7+3×22+3                           =12+35, 14+65=3, 4
The point R divides AB in the ratio of 3 : 2 internally. So using section formula, we get
Coordinates of R=3×6+2×13+2, 3×7+2×23+2                           =18+25, 21+45=4, 5
Hence, the coordinates of the points P, Q and R are (2, 3), (3, 4) and (4, 5) respectively.

Page No 321:

Question 6:

Points P, Q and R, in that order, divide a line segment joining A(1, 6) and B(5, −2) in four equal parts. Find the coordinates of P, Q and R.

Answer:

The given points are A(1, 6) and B(5, −2).
Then, P(x, y) is a point that divides the line AB in the ratio 1:3.
By the section formula:
x =mx2+nx1m+n, y = my2+ny1m+nx= 1×5 + 3×11+3, y =1×-2+3×61+3x= 5+34, y = -2+184x = 84, y = 164 x = 2 and y = 4
Therefore, the coordinates of point P are (2, 4).
Let Q be the mid point of AB.
Then, Q(x, y):
x = x1+x22, y =y1+y22x = 1+52, y =6+-22x = 62, y = 42x = 3, y = 2
Therefore, the coordinates of Q are (3, 2).
Let R (x, y) be a point that divides AB in the ratio 3:1.
Then, by the section formula:
x =mx2+nx1m+n, y = my2+ny1m+nx= 3×5 + 1×13+1, y =3×-2+1×63+1x= 15+14, y = -6+64x = 164, y = 04 x = 4 and y = 0
Therefore, the coordinates of R are (4, 0).
Hence, the coordinates of point P, Q and R are (2, 4), (3, 2) and (4, 0) respectively.

Page No 321:

Question 7:

The line segment joining the points A(3, −4) and B(1, 2) is trisected at the points P(p, −2) and Q53,q. Find the values of p and q.

Answer:

Let P and Q be the points of trisection of AB.
Then, P divides AB in the ratio 1:2.
So, the coordinates of P are
x =mx2+nx1m+n, y = my2+ny1m+nx= 1×1 + 2×31+2, y =1×2+2×-41+2x= 1+63, y = 2-83x= 73, y = -63x = 73, y = -2
Hence, the coordinates of P are (73, −2).
But (p, −2) are the coordinates of P.
So, p=73
Also, Q divides the line AB in the ratio 2:1.
So, the coordinates of Q are
x =mx2+nx1m+n, y = my2+ny1m+nx= 2×1 + 1×32+1, y =2×2+1×-42+1x= 2+33, y = 4-43x = 53, y = 0Hence, coordinates of Q are 53, 0.
But the given coordinates of Q are 53, q.
So, q = 0
Thus, p=73 and q=053

Page No 321:

Question 8:

Find the coordinates of the midpoints of the line segment joining:

(i) A(3, 0) and B(−5, 4)
(ii) P(−11, −8) and Q(8, −2)

Answer:

(i) The given points are A(3, 0) and B(−5, 4).
Let (x, y) be the mid point of AB. Then:
x = x1+x22, y =y1+y22x = 3+-52, y =0+42x = -22, y = 42x = -1, y = 2
Therefore, (−1, 2) are the coordinates of mid point of AB.

(ii) The given points are P(−11, −8) and Q(8, −2).
Let (x, y) be the mid point of PQ. Then:
x = x1+x22, y =y1+y22x = -11+82, y =-8-22x =- 32, y = -102x = -32, y = -5
Therefore, -32, -5 are the coordinates of midpoint of PQ.

Page No 321:

Question 9:

If (2, p) is the midpoint of the line segment joining the points A(6, −5) and B(−2, 11), find the value of p.

Answer:

The given points are A(6, −5) and B(−2, 11).
Let (x, y) be the mid point of AB. Then:
x = x1+x22, y =y1+y22x = 6+-22, y =-5+112x = 6-22, y = -5+112x = 42, y = 62x = 2, y = 3
So, the midpoint of AB is (2, 3).
But it is given that the midpoint of AB is (2, p).
Therefore, the value of p = 3.

Page No 321:

Question 10:

The midpoint of the line segment joining A(2a, 4) and B(−2, 3b) is C(1, 2a + 1). Find the values of a and b.

Answer:

The points are A(2a, 4) and B(−2, 3b).
Let C(1, 2a + 1) be the mid point of AB. Then:
x = x1+x22, y =y1+y221 = 2a+-22, 2a+1 =4+3b22=2a-2,  4a+2 = 4+3b2a = 2+2  , 4a-3b = 4-2a=42, 4a-3b = 2a = 2, 4a-3b = 2Putting the value of a in the equation  4a+3b = 2, we get:42-3b = 2-3b = 2-8 = -6b = 63 = 2Therefore, a=2 and b=2.

Page No 321:

Question 11:

The line segment joining A(−2, 9) and B(6, 3) is a diameter of a circle with centre C. Find the coordinates of C.

Answer:

The given points are A(−2, 9) and B(6, 3).
Then, C(x, y) is the midpoint of AB.
x = x1+x22, y =y1+y22x = -2+62, y =9+32x = 42, y = 122x = 2, y = 6
Therefore, the coordinates of point C are (2, 6).

Page No 321:

Question 12:

Find the coordinates of a point A, where AB is the diameter of a circle with centre C(2, −3) and the other end of the diameter is B(1, 4).

Answer:

C(2, −3) is the centre of the given circle. Let A(a, b) and B(1, 4) be the two end-points of the given diameter AB. Then, the coordinates of C are
x = a+12, y =b+42It is given that x = 2 and y = -3.2 =a+12, -3 = b+424 = a+1 , -6 = b+4a = 4-1, b=-6-4a = 3, b= -10
Therefore, the coordinates of point A are (3, -10).

Page No 321:

Question 13:

In what ratio does the point P(2, 5) divide the join of A(8, 2) and B(−6, 9)?

Answer:

Let the point P(2, 5) divide AB in the ratio k : 1.
Then, by section formula, the coordinates of P are
x=-6k+8k+1, y=9k+2k+1It is given that the coordinates of P are P(2, 5).2 = -6k+8k+1, 5 =9k+2k+12k+2 = -6k+8 ,  5k+5 = 9k+22k+6k = 8-2 ,  5-2 =9k-5k8k = 6, 4k = 3k = 68, k = 34k = 34 in each case.
Therefore, the point P(2, 5) divides AB in the ratio 3 : 4.

Page No 321:

Question 14:

Find the ratio in which the point P34, 512 divides the line segment joining the points
A12, 32 and 2,-5.                                        [CBSE 2015]

Answer:

Let k : 1 be the ratio in which the point P34, 512 divides the line segment joining the points A12, 32 and 2,-5. Then
34,512=k2+12k+1,k-5+32k+1k2+12k+1=34   and k-5+32k+1=5128k+2=3k+3   and -60k+18=5k+5k=15   and k=15
Hence, the required ratio is 1 : 5.

Page No 321:

Question 15:

Find the ratio in which the point P(m, 6) divides the join of A(−4, 3) and B(2, 8). Also, find the value of m.

Answer:

Let the point P(m, 6) divide the line AB in the ratio k : 1.
Then, by the section formula:
x = mx2+nx1m+n, y =my2+ny1m+nThe coordinates of P are (m, 6).m =2k-4k+1 , 6 = 8k+3k+1m(k+1) = 2k-4 ,  6k+6 = 8k+3m(k+1) = 2k-4 ,  6 -3= 8k - 6km(k+1) = 2k-4, 2k =3m(k+1) = 2k-4 , k = 32Therefore, the point P divides the line AB in the ratio 3 : 2.Now, putting the value of k in the equation m(k+1) = 2k-4, we get:m32+1 = 232-4m3+22 = 3-45m2 =-15m = -2m = -25 Therefore, the value of m = -25So, the coordinates of P are (-25, 6).

Page No 321:

Question 16:

Find the ratio in which the point (−3, k) divides the join of A(−5, −4) and B(−2, 3). Also, find the value of k.

Answer:

Let the point P(−3, k) divide the line AB in the ratio s : 1.
Then, by the section formula:
x = mx2+nx1m+n, y =my2+ny1m+nThe coordinates of P are (-3, k).-3 = -2s-5s+1, k = 3s-4s+1-3s-3 = -2s-5,  ks+1 = 3s-4-3s+2s =-5+3,  ks+1 = 3s-4-s = -2,  ks+1 = 3s-4s =2,  ks+1 = 3s-4Therefore, the point P divides the line AB in the ratio 2 : 1.Now, putting the value of s in the equation ks+1 = 3s-4, we get:k2+1 = 32-43k = 6-43k = 2 k = 23Therefore, the value of k = 23That is, the coordinates of P are (-3, 23).

Page No 321:

Question 17:

In what ratio is the line segment joining A(2, −3) and B(5, 6) divided by the x-axis? Also, find the coordinates of the point of division.

Answer:

Let AB be divided by the x-axis in the ratio k : 1 at the point P.
Then, by section formula the coordinates of P are
P=5k+2k+1, 6k-3k+1
But P lies on the x-axis; so, its ordinate is 0.
Therefore, 6k-3k+1=06k-3 =06k = 3k =36 k = 12
Therefore, the required ratio is 12 : 1, which is same as 1 : 2.
Thus, the x-axis divides the line AB in the ratio 1 : 2 at the point P.
Applying k = 12, we get the coordinates of point :
  P5k+2k+1, 0 = P5×12+212+1, 0 = P5+421+22, 0 = P93, 0  = P3, 0 
Hence, the point of intersection of AB and the x-axis is P(3, 0).



Page No 322:

Question 18:

In what ratio is the line segment joining the points A(−2, −3) and B(3, 7) divided by the y-axis? Also, find the coordinates of the points of division.

Answer:

Let AB be divided by the x-axis in the ratio k : 1 at the point P.
Then, by section formula the coordinates of P are
P3k-2k+1,7k-3k+1
But P lies on the y-axis; so, its abscissa is 0.
Therefore, 3k-2k+1=03k-2 =03k = 2k =23 k = 23
Therefore, the required ratio is 23 : 1, which is same as 2 : 3.
Thus, the x-axis divides the line AB in the ratio 2:3 at the point P.
Applying k=23, we get the coordinates of point P:
P0,7k-3k+1 = P0,7×23-323+1 = P0,14-932+33 = P0,55  = P0, 1
Hence, the point of intersection of AB and the x-axis is P(0, 1).

Page No 322:

Question 19:

In what ratio does the line xy − 2 = 0 divide the line segment joining the points A(3, −1) and B(8, 9)?

Answer:

Let the line xy − 2 = 0 divide the line segment joining the points A(3, −1) and B(8, 9) in the ratio k : 1 at P.
Then, the coordinates of P are
P8k+3k+1, 9k-1k+1
Since, P lies on the line xy − 2 = 0, we have:

8k+3k+1-9k-1k+1-2=08k+3-9k+1-2k-2=08k-9k-2k+3+1-2=0-3k+2=0-3k=-2k=23
So, the required ratio is 23 : 1, which is equal to 2 : 3.

Page No 322:

Question 20:

Find the lengths of the median of ∆ABC whose vertices are A(0, −1), B(2, 1) and C(0, 3).

Answer:

The vertices of ∆ABC are A(0, −1), B(2, 1) and C(0, 3).
Let AD, BE and CF be the medians of ∆ABC.

Let D be the midpoint of BC. So, the coordinates of D are
D2+02,1+32 i.e. D22,42 i.e. D1,2
Let E be the midpoint of AC. So, the coordinates of E are
E0+02,-1+32 i.e. E02,22 i.e. E0,1
Let F be the midpoint of AB. So, the coordinates of F are
F0+22,-1+12  i.e. F22,02  i.e. F1,0
AD = 1-02+2--12 =12+32 = 1+9 = 10 units.BE = 0-22+1-12 =-22+02 = 4+0 = 4 = 2 units.CF = 1-02+0-32 =12+-32 = 1+9 = 10 units.
Therefore, the lengths of the medians: AD10 units, BE = 2 units and CF = 10 units

Page No 322:

Question 21:

Find the centroid of ∆ABC whose vertices are A(−1, 0), B(5, −2) and C(8, 2).

Answer:

Here, (x1 = −1, y1 = 0), (x2 = 5, y2 = −2) and (x3 = 8, y3 = 2).
Let G(x, y) be the centroid of the ∆ABC. Then,
x = 13x1+x2+x3 = 13-1+5+8 = 1312 = 4y = 13y1+y2+y3 = 130-2+2 = 130 = 0
Hence, the centroid of ∆ABC is G(4, 0).

Page No 322:

Question 22:

If G(−2, 1) is the centroid of a ∆ABC and two of its vertices are A(1, −6) and B(−5, 2), find the third vertex of the triangle.

Answer:

Two vertices of ∆ABC are A(1, −6) and B(−5, 2). Let the third vertex be C(a, b).
Then the coordinates of its centroid are
C1-5+a3, -6+2+b3C-4+a3, -4+b3
But it is given that G(−2, 1) is the centroid. Therefore,
-2 =-4+a3, 1=-4+b3-6 = -4+a , 3 =-4+b-6+4 = a, 3+4 = ba =-2, b= 7
Therefore, the third vertex of ∆ABC is C(−2, 7).

Page No 322:

Question 23:

Find the third vertex of ∆ABC if two of its vertices are B(−3, 1) and C(0, −2) and its centroid is at the origin.

Answer:

Two vertices of ∆ABC are B(−3,1) and C(0, −2). Let the third vertex be A(a, b).
Then, the coordinates of its centroid are
  -3+0+a3, 1-2+b3 i.e. -3+a3, -1+b3
But it is given that the centroid is at the origin, that is G(0, 0). Therefore,
0 =-3+a3, 0=-1+b30 = -3+a , 0 =-1+b3 = a, 1= ba =3, b= 1
Therefore, the third vertex of ∆ABC is A(3, 1).

Page No 322:

Question 24:

Show that the points A(3, 1), B(0, −2), C(1, 1) and D(4, 4) are the vertices of a parallelogram ABCD.

Answer:

The points are A(3, 1), B(0, −2), C(1, 1) and D(4, 4).
Join AC and BD, intersecting at O.   
      

We know that the diagonals of a parallelogram bisect each other.
Midpoint of AC = 3+12,1+12 = 42,22 = 2,1Midpoint of BD = 0+42,-2+42 = 42,22 = 2,1
Thus, the diagonals AC and BD have the same midpoint.
Therefore, ABCD is a parallelogram.

Page No 322:

Question 25:

If the points P(a, −11), Q(5, b), R(2, 15) and S(1, 1) are the vertices of a parallelogram PQRS, find the values of a and b.

Answer:

The points are P(a, −11), Q(5, b), R(2, 15) and S(1, 1).

Join PR and QS, intersecting at O.
We know that the diagonals of a parallelogram bisect each other.
Therefore, O is the midpoint of PR as well as QS.
Midpoint of PR = a+22,-11+152 = a+22,42 = a+22,2Midpoint of QS = 5+12,b+12 = 62,b+12 = 3,b+12 Therefore, a+22=3, b+12=2a+2 = 6 , b+1 = 4a = 6-2 , b = 4-1a = 4 and b = 3

Page No 322:

Question 26:

If three consecutive vertices of a parallelogram ABCD are A(1, −2), B(3, 6) and C(5, 10), find its fourth vertex D.

Answer:

Let A(1, −2), B(3, 6) and C(5, 10) be the three vertices of a parallelogram ABCD and the fourth vertex be D(a, b).
Join AC and BD intersecting at O.

We know that the diagonals of a parallelogram bisect each other.
Therefore, O is the midpoint of AC as well as BD.
Midpoint of AC = 1+52,-2+102 = 62,82 = 3,4Midpoint of BD = 3+a2,6+b2 Therefore, 3+a2 = 3 and 6+b2 = 43+a =6 and 6+b =8a = 6-3 and b= 8-6a= 3 and b= 2
Therefore, the fourth vertex is D(3, 2).

Page No 322:

Question 27:

In what ratio does y-axis divide the line segment joining the points (−4, 7) and (3, −7)?     [CBSE 2012]

Answer:

Let y-axis divides the line segment joining the points (−4, 7) and (3, −7) in the ratio k : 1. Then
0=3k-4k+13k=4k=43
Hence, the required ratio is 4 : 3.

Page No 322:

Question 28:

If the point P12, y lies on the line segment joining the points A(3, −5) and B(−7, 9) then find the
ratio in which P divides AB. Also, find the value of y.

Answer:

Let the point P12, y divides the line segment joining the points A(3, −5) and B(−7, 9) in the ratio k : 1. Then
12, y=k-7+3k+1, k9-5k+1-7k+3k+1=12 and 9k-5k+1=yk+1=-14k+6k=13
Now, substituting k=13 in 9k-5k+1=y, we get
93-513+1=yy=9-151+3=-32
Hence, required ratio is 1 : 3 and y=-32.

Page No 322:

Question 29:

Find the ratio in which the line segment joining the points A(3, −3) and B(−2, 7) is divided by x-axis.
Also, find the point of division.                                [CBSE 2015]

Answer:

The line segment joining the points A(3, − 3) and B(− 2, 7) is divided by x-axis. Let the required ratio be k : 1. So,
0=k7-3k+1k=37
Now
Point of division=k-2+3k+1, k7-3k+1                         =37×-2+337+1, 37×7-337+1             k=37                         =-6+213+7, 21-213+7                          =32, 0
Hence, the required ratio is 3 : 7 and the point of division is 32, 0.

Page No 322:

Question 30:

The base QR of an equilateral triangle PQR lies on x-axis. The coordinates of the point Q are (−4, 0)
and origin is the midpoint of the base. Find the coordinates of the points P and R.

Answer:

Let (x, 0) be the coordinates of R. Then
0=-4+x2x=4
Thus, the coordinates of R are (4, 0).
Here, PQ = QR = PR and the coordinates of P lies on y-axis. Let the coordinates of P be (0, y). Then
PQ=QRPQ2=QR20+42+y-02=82y2=64-16=48y=±43
Hence, the required coordinates are R4, 0 and P0, 43 or P0, -43.

Page No 322:

Question 31:

The base BC of an equilateral triangle ABC lies on y-axis. The coordinates of point C are (0, −3).
The origin is the midpoint of the base. Find the coordinates of the points A and B. Also, find
the coordinates of another point D such that ABCD is a rhombus.

Answer:

Let (0, y) be the coordinates of B. Then
0=-3+y2y=3
Thus, the coordinates of B are (0, 3).
Here, AB = BC = AC and by symmetry the coordinates of A lies on x-axis. Let the coordinates of A be (x, 0). Then
AB=BCAB2=BC2x-02+0-32=62x2=36-9=27x=±33
If the coordinates of point A are 33, 0, then the coordinates of D are -33, 0.
If the coordinates of point A are -33, 0, then the coordinates of D are 33, 0.
Hence, the required coordinates are A33, 0, B0, 3 and D-33, 0 or A-33, 0, B0, 3 and D33, 0.

Page No 322:

Question 32:

Find the ratio in which the point P( −1, y), lying on the line segment joining points A(−3, 10) and B(6, −8) divides it. Also, find the value of y. Also, find the value of y.          [CBSE 2013]

Answer:

Let k be the ratio in which P( −1, y) divides the line segment joining the points A(−3, 10) and B(6, −8). Then
-1, y=k6-3k+1, k-8+10k+1k6-3k+1=-1 and y=k-8+10k+1k=27
Substituting k=27 in  y=k-8+10k+1, we get
y=-8×27+1027+1=-16+709=6
Hence, the required ratio is 2 : 7 and y = 6.



Page No 323:

Question 33:

ABCD is a rectangle formed by the points A-1,-1, B-1, 4, C5, 4 and D5, -1. If P, Q, R and S be the mid points of
AB, BC, CD and DA respectively, show that PQRS is a rhombus.

Answer:

Here, the points P, Q, R and S are the mid points of AB, BC, CD and DA respectively. Then
Coordinates of P=-1-12, -1+42=-1, 32Coordinates of Q=-1+52, 4+42=2, 4Coordinates of R=5+52, 4-12=5, 32Coordinates of S=-1+52, -1-12=2, -1
Now
PQ=2+12+4-322=9+254=612QR=5-22+32-42=9+254=612RS=5-22+32+12=9+254=612SP=2+12+-1-322=9+254=612PR=5+12+32-322=36=6QS=2-22+-1-42=25=5
Thus, PQ = QR = RS = SP and PRQS therefore PQRS is a rhombus.

Page No 323:

Question 34:

The midpoint P of the line segment joining the points A(−10, 4) and B(−2, 0) lies on the line segment joining the points C(−9, −4) and D(−4, y). Find the ratio in which P divides CD. Also find the value of y.

Answer:

The midpoint of AB is -10-22, 4+02=P-6, 2.
Let k be the ratio in which P divides CD. So
-6, 2=k-4-9k+1, ky-4k+1k-4-9k+1=-6 and ky-4k+1=2k=32
Now, substituting k=32 in ky-4k+1=2, we get
y×32-432+1=23y-85=2y=10+83=6
Hence, the required ratio is 3 : 2 and y = 6.

Page No 323:

Question 35:

A line intersects the y-axis and x-axis  at the points P and Q respectively. If (2, –5) is the midpoint of PQ then find the coordinates of P and Q.     [CBSE 2017]

Answer:

Suppose the line intersects the y-axis at P(0, y) and the x-axis at Q(x, 0). 
It is given that (2, –5) is the mid-point of PQ.

Using mid-point formula, we have
x+02,0+y2=2, -5x2, y2=2, -5x2=2 and y2=-5x=4, y=-10
Thus, the coordinates of P and Q are (0, −10) and (4, 0), respectively.

Page No 323:

Question 36:

In what ratio does the point 2411, y divide the line segment joining the points P(2, –2) and Q(3, 7)? Also, find the value of y.        [CBSE 2017]

Answer:

Let the point P2411, y divides the line PQ in the ratio k : 1.
Then, by the section formula:
x = mx2+nx1m+n, y =my2+ny1m+nThe coordinates of R are 2411, y.2411 =3k+2k+1 , y = 7k-2k+124(k+1)=33k+22 ,  y(k+1)=7k-224k+24=33k+22  ,   yk+y=7k-22=9kk=29
Now consider the equation yk+y =7k-2 and put k=29. 29y+y =149-2 119y =-49y =-411Therefore, the point R divides the line PQ in the ratio 2 : 9.And, the coordinates of R are 2411, -411.
  

Page No 323:

Question 37:

The midpoints of the sides BC, CA and AB of a ΔABC are D(3, 4), E(8, 9) and F(6, 7) respectively. Find the coordinates of the vertices of the triangle.            [CBSE 2017]

Answer:


Let the coordinates of A, B, C be (x1,y1), (x2,y2) and (x3,y3) respectively.
Because D is the mid-point of BC, using mid-point formula, we have
x2+x32=3           and          y2+y32=4x2+x3=6        and          y2+y3=8               .....i
Similarly, E is the mid point of AC. Using mid-point formula, we have;
x1+x32=8          and          y1+y32=9x1+x3=16     and          y1+y3=18               .....ii
Again, F is the mid point of AB. Using mid point formula, we have
x1+x22=6          and          y1+y22=7x1+x2=12          and          y1+y2=14               .....iii
Adding (i), (ii) and (iii), we get
2x1+x2+x3=34          and           2y1+y2+y3=40x1+x2+x3=17          and             y1+y2+y3=20               .....iv
On solving equation (iv)using equations (i), (ii) and (iii), we get
x1=11x2=1x3=5
Similarly,
y1=12y2=2y3=6
Hence, the points are: A(11,12), B(1,2) and C(5,6).

Page No 323:

Question 38:

If two adjacent vertices of a parallelogram are (3, 2) and (–1, 0) and the diagonals intersect at (2, –5) then find the coordinates of the other two vertices.   [CBSE 2017]

Answer:

Let ABCD be the parallelogram with two adjacent vertices A(3, 2) and B(−1, 0). Suppose O(2, −5) be the point of intersection of the diagonals AC and BD.
Let C(x1y1) and D(x2y2) be the coordinates of the other vertices of the parallelogram.

We know that the diagonals of the parallelogram bisect each other. Therefore, O is the mid-point of AC and BD.
Using the mid-point formula, we have
x1+32,y1+22=2,-5x1+32=2 and y1+22=-5x1+3=4 and y1+2=-10x1=4-3=1 and y1=-10-2=-12
So, the coordinates of C are (1, −12).
Also,
x2+-12,y2+02=2,-5x2-12=2 and y22=-5x2-1=4 and y2=-10x2=4+1=5 and y2=-10
So, the coordinates of D are (5, −10).



Page No 336:

Question 1:

Find the area of ∆ABC whose vertices are:

(i) A(1, 2), B(−2, 3) and C(−3, −4)
(ii) A(−5, 7), B(−4, −5) and C(4, 5)
(iii) A(3, 8), B(−4, 2) and C(5, −1)
(iv) A(10, −6), B(2, 5) and C(−1, 3)

Answer:

(i) A(1, 2), B(−2, 3) and C(−3, −4) are the vertices of ∆ABC. Then,
(x1 = 1, y1 = 2), (x2 = −2, y2 = 3) and (x3 = −3, y3 = -4)
Area of triangle ABC=12x1y2-y3+x2y3-y1+x3y1-y2= 1213--4+-2-4-2+-32-3=1213+4-2-6-3-1=127+12+3=1222= 11 sq. units

(ii) A(−5, 7), B(−4, −5) and C(4, 5) are the vertices of ∆ABC. Then,
(x1 = −5, y1 = 7), (x2 = −4, y2 = −5) and (x3 = 4, y3 = 5)
Area of triangle ABC=12x1y2-y3+x2y3-y1+x3y1-y2= 12-5-5-5+-45-7+47--5=12-5-10-4-2+412=1250+8+48=12106= 53 sq. units

(iii) A(3, 8), B(−4, 2) and C(5, −1) are the vertices of ∆ABC. Then,
(x1 = 3, y1 = 8), (x2 = −4, y2 = 2) and (x3 = 5, y3 = −1)
Area of triangle ABC=12x1y2-y3+x2y3-y1+x3y1-y2= 1232--1+-4-1-8+58-2=1232+1-4-9+56=129+36+30=1275= 37.5 sq. units

(iv) A(10, −6), B(2, 5) and C(−1, −3) are the vertices of ∆ABC. Then,
(x1 = 10, y1 = −6), (x2 = 2, y2 = 5) and (x3 = −1, y3 = 3)
Area of triangle ABC=12x1y2-y3+x2y3-y1+x3y1-y2= 12105-3+23--6+-1-6-5=12102+29-1-11=1220+18+11=1249= 24.5 sq. units

Page No 336:

Question 2:

Find the area of quadrilateral ABCD whose vertices are A(3, −1), B(9, −5), C(14, 0) and D(9, 19).                [CBSE 2012]

Answer:

By joining A and C, we get two triangles ABC and ACD.
Let Ax1, y1=A3, -1, Bx2, y2=B9, -5, Cx3, y3=C14, 0 and Dx4, y4=D9, 19. Then
Area of ABC=12x1y2-y3+x2y3-y1+x3y1-y2                       =123-5-0+90+1+14-1+5                       =12-15+9+56=25 sq. units
Area of ACD=12x1y3-y4+x3y4-y1+x4y1-y3                       =1230-19+1419+1+9-1-0                       =12-57+280-9=107 sq. units
So, the area of the quadrilateral is 25 + 107 = 132 sq. units.

Page No 336:

Question 3:

Find the area of quadrilateral PQRS whose vertices are P(−5, −3), Q(−4, −6), R(2, −3) and S(1, 2).
[CBSE 2015]

Answer:

By joining P and R, we get two triangles PQR and PRS.
Let Px1, y1=P-5, -3, Qx2, y2=Q-4, -6, Rx3, y3=R2, -3 and Sx4, y4=S1, 2. Then
Area of PQR=12x1y2-y3+x2y3-y1+x3y1-y2                       =12-5-6+3-4-3+3+2-3+6                       =1215-0+6=212 sq. units
Area of PRS=12x1y3-y4+x3y4-y1+x4y1-y3                       =12-5-3-2+22+3+1-3+3                       =1225+10+0=352 sq. units
So, the area of the quadrilateral PQRS is 212+352=28 sq. units sq. units.

Page No 336:

Question 4:

Find the area of quadrilateral ABCD whose vertices are A(−3, −1), B(−2, −4), C(4, −1) and D(3, 4).
[CBSE 2013C]

Answer:

By joining A and C, we get two triangles ABC and ACD.
Let Ax1, y1=A-3, -1, Bx2, y2=B-2, -4, Cx3, y3=C4, -1 and Dx4, y4=D3, 4. Then
Area of ABC=12x1y2-y3+x2y3-y1+x3y1-y2                       =12-3-4+1-2-1+1+4-1+4                       =129-0+12=212 sq. units
Area of ACD=12x1y3-y4+x3y4-y1+x4y1-y3                       =12-3-1-4+44+1+3-1+1                       =1215+20+0=352 sq. units
So, the area of the quadrilateral ABCD is 212+352=28 sq. units sq. units.

Page No 336:

Question 5:

If A(–7, 5), B(–6, –7), C(–3, –8) and D(2, 3) are the vertices of a quadrilateral ABCD then find the area of the quadrilateral.

Answer:


Consider the figure. 
Construction: Produce AC by joining points A to C to form two triangles, ABC and ADC
In ABC,
x1=-7, x2=-6 and x3=-3; y1=5, y2=-7 and y3=-8
We know that,
ar(ABC)=12(x1)(y2-y3)+(x2)(y3-y1)+(x3)(y1-y2)ar(ABC)=12(-7)(-7+8)+(-6)(-8-5)+(-3)(5+7)ar(ABC)=12(-7)(1)+(-6)(-13)+(-3)(12)ar(ABC)=12(-7)+78+(-36)ar(ABC)=1235ar(ABC)=352 sq. units
Similarly, in ADC,
x1=-7, x2=2, x3=-3, y1=5, y2=3 and y3=-8
ar(ADC)=12(-7)(3+8)+(2)(-8-5)+(-3)(5-3)ar(ADC)=12(-7)(11)+(2)(-13)+(-3)(2)ar(ADC)=12(-77)-26-6ar(ADC)=12-109ar(ADC)=1092 sq. units
Now, ar(quad. ABCD) = ar(ABC)+ar(ADC)

ar(quad. ABCD) = 352+1092=1442=72
Therefore, area of quadrilateral ABCD is 72 sq. units

Disclaimer: The answer thus calculated does not match with the answer given in the book.

Page No 336:

Question 6:

Find area of the triangle formed by joining the midpoints of the sides of the triangle
whose vertices are A(2, 1), B(4, 3) and C(2, 5).

Answer:

The vertices of the triangle are A(2, 1), B(4, 3) and C(2, 5).
Coordinates of midpoint of AB=Px1, y1=2+42,1+32=3, 2Coordinates of midpoint of BC=Qx2, y2=4+22,3+52=3, 4Coordinates of midpoint of AC=Rx3, y3=2+22,1+52=2, 3
Now
Area of PQR=12x1y2-y3+x2y3-y1+x3y1-y2                       =1234-3+33-2+22-4                       =123+3-4=1 sq. unit
Hence, the area of the required triangle is 1 sq. unit.



Page No 337:

Question 7:

A(7, −3), B(5, 3), C(3, −1) are the vertices of a ABC and AD is its median. Prove that the median
AD divides ABC into two triangles of equal areas.

Answer:

The vertices of the triangle are A(7, −3), B(5, 3), C(3, −1).
Coordinates of D=5+32,3-12=4, 1
For the area of the triangle ADC, let Ax1,y1=A7,-3, Dx2,y2=D4,1 and Cx3,y3=C3,-1. Then
Area of ADC=12x1y2-y3+x2y3-y1+x3y1-y2                       =1271+1+4-1+3+3-3-1                       =1214+8-12=5 sq. unit
Now, for the area of triangle ABD, let Ax1, y1=A7,-3, Bx2, y2=B5, 3 and Dx3, y3=D4, 1. Then
Area of ABD=12x1y2-y3+x2y3-y1+x3y1-y2                       =1273-1+51+3+4-3-3                       =1214+20-24=5 sq. unit
Thus, AreaADC=AreaABD=5 sq. units.
Hence, AD divides ABC into two triangles of equal areas.

Page No 337:

Question 8:

Find the area of ABC with A(1, −4) and midpoints of sides through A being (2, −1) and (0, −1).      [CBSE 2015]

Answer:

Let x2, y2 and x3, y3 be the coordinates of B and C respectively. Since, the coordinates of A are (1, −4), therefore
1+x22=2x2=3-4+y22=-1y2=21+x32=0x3=-1-4+y32=-1y3=2
Let Ax1, y1=A1, -4, Bx2, y2=B3, 2 and Cx3, y3=C-1, 2. Now
AreaABC=12x1y2-y3+x2y3-y1+x3y1-y2                   =1212-2+32+4-1-4-2                   =120+18+6                   =12 sq. units
Hence, the area of the triangle ABC is 12 sq. units.

Page No 337:

Question 9:

A(6, 1), B(8, 2) and C(9, 4) are the vertices of a parallelogram ABCD. If E is the midpoint of DC, find the area of ADE.                                                                                                                                           [CBSE 2015]

Answer:

Let (x, y) be the coordinates of D and x', y' be the coordinates of E. Since, the diagonals of a parallelogram bisect
each other at the same point, therefore
x+82=6+92x=7y+22=1+42y=3
Thus, the coordinates of D are (7, 3).
E is the midpoint of DC, therefore
x'=7+92x'=8y'=3+42y'=72 
Thus, the coordinates of E are 8, 72.
Let Ax1, y1=A6, 1, Ex2, y2=E8, 72 and Dx3, y3=D7, 3. Now
AreaABC=12x1y2-y3+x2y3-y1+x3y1-y2                   =12672-3+83-1+71-72                   =1232                   =34 sq. unit
Hence, the area of the triangle ADE is 34 sq. units.

Page No 337:

Question 10:

(i) If the vertices of ABC be A(1, −3), B(4, p) and C(−9, 7) and its area is 15 square units, find the values of p.                             [CBSE 2012]
(ii) The area of a triangle is 5 sq units. Two of its vertices are (2, 1) and (3, –2). If the third vertex is 72, y, find the value of y.    [CBSE 2017]

Answer:

(i) Let Ax1, y1=A1, -3, Bx2, y2=B4, p and Cx3, y3=C-9, 7. Now
AreaABC=12x1y2-y3+x2y3-y1+x3y1-y215=121p-7+47+3-9-3-p15=1210p+6010p+60=30
Therefore
10p+60=-30 or 3010p=-90 or -30p=-9 or -3
Hence, p=-9 or p=-3.

(ii)
Let Ax1, y1=A2, 1, Bx2, y2=B3,-2 and Cx3, y3=C72,y.
Now
AreaABC=12x1y2-y3+x2y3-y1+x3y1-y25=122-2-y+3y-1+721+210=-4-2y+3y-3+21210=y+7210=y+72    or -10=y+72 y=132 or y = -272
Hence, 132 or -272.
 

Page No 337:

Question 11:

Find the value of k so that the area of the triangle with vertices A(k + 1, 1), B(4, −3) and C(7, −k) is 6 square units.

Answer:

Let Ax1, y1=Ak+1, 1, Bx2, y2=B4, -3 and Cx3, y3=C7, -k. Now
AreaABC=12x1y2-y3+x2y3-y1+x3y1-y26=12k+1-3+k+4-k-1+71+36=12k2-2k-3-4k-4+28k2-6k+9=0
k-32=0k=3
Hence, k = 3.

Page No 337:

Question 12:

For what value of k (k > 0) is the area of the triangle with vertices (−2, 5), (k, −4) and
(2k + 1, 10) equal to 53 square units?

Answer:

Let Ax1=-2, y1=5, Bx2=k, y2=-4 and Cx3=2k+1, y3=10 be the vertices of
the triangle. So
AreaABC=12x1y2-y3+x2y3-y1+x3y1-y253=12-2-4-10+k10-5+2k+15+453=1228+5k+92k+128+5k+18k+9=106

37+23k=10623k=106-37=69k=6923=3
Hence, k = 3.

Page No 337:

Question 13:

Show that the following points are collinear:

(i) A(2, −2), B(−3, 8) and C(−1, 4)
(ii) A(−5, 1), B(5, 5) and C(10, 7)
(iii) A(5, 1), B(1, −1) and C(11, 4)
(iv) A(8, 1), B(3, −4) and C(2, −5)

Answer:

(i)
Let A(x1 = 2, y1 = −2), B(x2 = −3, y2 = 8) and C(x3 = −1, y3 = 4) be the given points. Now
x1y2-y3+x2y3-y1+x3y1-y2=28-4+-34+2+-1-2-8=8-18+10=0
Hence, the given points are collinear.

(ii)
Let A(x1 = −5, y1 = 1), B(x2 = 5, y2 = 5) and C(x3 = 10, y3 = 7) be the given points. Now
x1y2-y3+x2y3-y1+x3y1-y2=-55-7+57-1+101-5=-5-2+56+10-4=10+30-40=0
Hence, the given points are collinear.

(iii)
Let A(x1 = 5, y1 = 1), B(x2 = 1, y2 = −1) and C(x3 = 11, y3 = 4) be the given points. Now
x1y2-y3+x2y3-y1+x3y1-y2=5-1-4+14-1+111+1=-25+3+22=0
Hence, the given points are collinear.

(iv)
Let A(x1 = 8, y1 = 1), B(x2 = 3, y2 = −4) and C(x3 = 2, y3 = −5) be the given points. Now
x1y2-y3+x2y3-y1+x3y1-y2=8-4+5+3-5-1+21+4=8-18+10=0
Hence, the given points are collinear.

Page No 337:

Question 14:

Find the value of x for which the points Ax, 2, B-3, -4 and C7, -5 are collinear.      [CBSE 2015]

Answer:

Let Ax1, y1=Ax, 2, Bx2, y2=B-3, -4 and Cx3, y3=C7, -5. So, the condition for three collinear points is
x1y2-y3+x2y3-y1+x3y1-y2=0x-4+5-3-5-2+72+4=0x+21+42=0x=-63
Hence, x = − 63.

Page No 337:

Question 15:

For what value of x are the points A(−3, 12), B(7, 6) and C(x, 9) collinear?

Answer:

 A(−3, 12), B(7, 6) and C(x, 9) are the given points. Then:
(x1 = −3, y1 = 12), (x2 = 7, y2 = 6) and (x3 = x, y3 = 9)
It is given that the points A, B and C are collinear. Therefore,
x1y2-y3+x2y3-y1+x3y1-y2=0-36-9+79-12+x12-6=0-3-3+7-3+x6=09-21+6x=06x-12=06x = 12x =126 =2
Therefore, when x= 2, the given points are collinear.

Page No 337:

Question 16:

For what value of y, are the points P(1, 4), Q(3, y) and R(−3, 16) collinear?

Answer:

P(1, 4), Q(3, y) and R(−3, 16) are the given points. Then:
(x1 = 1, y1 = 4), (x2 = 3, y2 = y) and (x3 = −3, y3 = 16)
It is given that the points P, Q and R are collinear.
Therefore,
x1y2-y3+x2y3-y1+x3y1-y2=01y-16+316-4+-34-y=01y-16+312-34-y=0y-16+36-12+3y=08+4y=04y = -8y =-84 =-2
When y = −2, the given points are collinear.

Page No 337:

Question 17:

Find the value of y for which the points A(−3, 9), B(2, y) and C(4, −5) are collinear.

Answer:

Let A(x1 = −3, y1 = 9), B(x2 = 2, y2 = y) and C(x3 = 4, y3 = −5) be the given points.
The given points are collinear if
x1y2-y3+x2y3-y1+x3y1-y2=0-3y+5+2-5-9+49-y=0-3y-15-28+36-4y=07y=36-43
y=-1

Page No 337:

Question 18:

For what values of k are the points A(8, 1), B(3, −2k) and C(k, −5) collinear?                  [CBSE 2015]

Answer:

Let A(x1 = 8, y1 = 1), B(x2 = 3, y2 = −2k) and C(x3 = ky3 = −5) be the given points.
The given points are collinear if
x1y2-y3+x2y3-y1+x3y1-y2=08-2k+5+3-5-1+k1+2k=0-16k+40-18+k+2k2=02k2-15k+22=0
2k2-11k-4k+22=0k2k-11-22k-11=0k-22k-11=0k=2 or k=112
Hence, k=2 or k=112.

Page No 337:

Question 19:

Find a relation between x and y, if the points A(2, 1), B(x, y) and C(7, 5) are collinear.                 [CBSE 2009C]

Answer:

Let A(x1 = 2, y1 = 1), B(x2 = xy2 = y) and C(x3 = 7, y3 = 5) be the given points.
The given points are collinear if
x1y2-y3+x2y3-y1+x3y1-y2=02y-5+x5-1+71-y=02y-10+4x+7-7y=04x-5y-3=0
Hence, the required relation is 4x − 5y − 3 = 0.

Page No 337:

Question 20:

Find a relation between x and y, if the points A(x, y), B(5, 7) and C(4, 5) are collinear.      [CBSE 2015]

Answer:

Let A(x1 = xy1 = y), B(x2 = −5, y2 = 7) and C(x3 = −4, y3 = 5) be the given points.
The given points are collinear if
x1y2-y3+x2y3-y1+x3y1-y2=0x7-5+-55-y+-4y-7=07x-5x-25+5y-4y+28=02x+y+3=0
Hence, the required relation is 2x + y + 3 = 0.

Page No 337:

Question 21:

Prove that the points A(a, 0), B(0, b) and C(1, 1) are collinear if 1a+1b=1.

Answer:

Consider the points A(a, 0), B(0, b) and C(1, 1).
Here, (x1 = a, y1 = 0), (x2 = 0, y2 = b) and (x3 = 1, y3 = 1).
It is given that the points are collinear. So,
x1y2-y3+x2y3-y1+x3y1-y2 = 0ab-1+01-0+10-b=0ab-a-b = oDividing the equation by ab:1-1b-1a = 01-1a+1b = 01a+1b=1 
Therefore, the given points are collinear if 1a+1b = 1.



Page No 338:

Question 22:

If the points P(−3, 9), Q(a, b) and R(4, 5) are collinear and a + b = 1, find the values of a and b.     [CBSE 2014]

Answer:

Let A(x1 = −3, y1 = 9), B(x2 = ay2 = b) and C(x3 = 4, y3 = 5) be the given points.
The given points are collinear if
x1y2-y3+x2y3-y1+x3y1-y2=0-3b+5+a-5-9+49-b=0-3b-15-14a+36-4b=02a+b=3
Now, solving a + b = 1 and 2a + b = 3, we get a = 2 and b = −1.
Hence, a = 2 and b = −1.

Page No 338:

Question 23:

Find the area of ABC with vertices A(0, −1), B(2, 1) and C(0, 3). Also, find the area of the triangle formed
by joining the midpoints of its sides. Show that the ratio of the areas of two triangles is 4 : 1.           [CBSE 2014]

Answer:

Let A(x1 = 0, y1 = −1), B(x2 = 2, y2 = 1) and C(x3 = 0, y3 = 3) be the given points. Then
AreaABC=12x1y2-y3+x2y3-y1+x3y1-y2                     =1201-3+23+1+0-1-1                     =12×8=4 sq. units
So, the area of the triangle ABC is 4 sq. units.
Let D(a1, b1), E(a2, b2) and F(a3, b3) be the midpoints of AB, BC and AC respectively. Then
a1=0+22=1              b1=-1+12=0a2=2+02=1              b2=1+32=2a3=0+02=0              b3=-1+32=1
Thus, the coordinates of D, E and F are D(a1 = 1, b1 = 0), E(a2 = 1, b2 = 2) and F(a3 = 0, b3 = 1). Now
AreaDEF=12a1b2-b3+a2b3-b1+a3b1-b2                     =1212-1+11-0+00-2                     =121+1+0=1 sq. unit
So, the area of the triangle DEF is 1 sq. unit.
Hence, ABC : DEF=4 : 1.

Page No 338:

Question 24:

If a b c, prove that (a, a2), (b, b2), (0, 0) will not be collinear.         [CBSE 2017]

Answer:

Let A(aa2), B(bb2) and C(0, 0) be the coordinates of the given points.
We know that the area of triangle having vertices x1, y1, x2, y2 and x3, y3 is 12x1y2-y3+x2y3-y1+x3y1-y2 square units.
So,
Area of ∆ABC
=12ab2-0+b0-a2+0a2-b2=12ab2-a2b=12abb-a0             ab0
Since the area of the triangle formed by the points (aa2), (bb2) and (0, 0) is not zero, so the given points are not collinear.



Page No 340:

Question 1:

Points A(−1, y) and B(5, 7) lie on a circle with centre O(2, −3y). Find the values of y.     [CBSE 2014]

Answer:

The given points are A(−1, y), B(5, 7) and O(2, −3y).
Here, AO and BO are the radii of the circle. So
AO=BOAO2=BO22+12+-3y-y2=2-52+-3y-729+4y2=-32+3y+729+16y2=9+9y2+49+42y
7y2-42y-49=0y2-6y-7=0y2-7y+y-7=0yy-7+1y-7=0
y-7y+1=0y=-1 or y=7
Hence, y = 7 or y = −1.

Page No 340:

Question 2:

If the point A(0, 2) is equidistant from the points B(3, p) and C(p, 5), find p.                [CBSE 2014]

Answer:

The given points are A(0, 2), B(3, p) and C(p, 5).
AB=ACAB2=AC23-02+p-22=p-02+5-229+p2-4p+4=p2+94p=4p=1
Hence, p = 1.

Page No 340:

Question 3:

ABCD is a rectangle whose three vertices are B(4, 0), C(4, 3) and D(0, 3). Find the length of one of its diagonal.
                                                                                                                                                     [CBSE 2014]

Answer:

The given vertices are B(4, 0), C(4, 3) and D(0, 3). Here, BD is one of the diagonals. So
BD=4-02+0-32    =42+-32    =16+9    =25    =5
Hence, the length of the diagonal is 5 units..

Page No 340:

Question 4:

If the point P(k − 1, 2) is equidistant from the points A(3, k) and B(k, 5), find the values of k.             [CBSE 2014]

Answer:

The given points are P(k − 1, 2), A(3, k) and B(k, 5).
 AP=BP  AP2=BP2k-1-32+2-k2=k-1-k2+2-52k-42+2-k2=-12+-32
k2-8y+16+4+k2-4k=1+9k2-6y+5=0k-1k-5=0k=1 or k=5
Hence, k = 1 or k = 5.

Page No 340:

Question 5:

Find the ratio in which the point P(x, 2) divides the join of A(12, 5) and B(4, −3).          [CBSE 2014]

Answer:

Let k be the ratio in which the point P(x, 2) divides the line joining the points A(x1 = 12, y1 = 5) and B(x2 = 4, y2 = −3). Then
x=k×4+12k+1          and                2=k×-3+5k+1
Now
2=k×-3+5k+12k+2=-3k+5k=35
Hence, the required ratio is 3 : 5.

Page No 340:

Question 6:

Prove that the diagonals of a rectangle ABCD with vertices A(2, −1), B(5, −1), C(5, 6) and D(2, 6) are
equal and bisect each other.                                         [CBSE 2014]

Answer:

The vertices of the rectangle ABCD are A(2, −1), B(5, −1), C(5, 6) and D(2, 6). Now
Coordinates of midpoint of AC=2+52, -1+62=72, 52Coordinates of midpoint of BD=5+22, -1+62=72, 52
Since, the midpoints of AC and BD coincide, therefore the diagonals of rectangle ABCD bisect each other.

Page No 340:

Question 7:

Find the lengths of the medians AD and BE of ABC whose vertices are A(7, −3), B(5, 3) and C(3, −1).       [CBSE 2014]

Answer:

The given vertices are A(7, −3), B(5, 3) and C(3, −1).
Since D and E are the midpoints of BC and AC respectively, therefore
Coordinates of D=5+32, 3-12=4, 1Coordinates of E=7+32, -3-12=5, -2
Now
AD=7-42+-3-12=9+16=5BE=5-52+3+22=0+25=5
Hence, AD = BE = 5 units.

Page No 340:

Question 8:

If the point C(k, 4) divides the join of A(2, 6) and B(5, 1) in the ratio 2 : 3 then find the value of k.   [CBSE 2013C]

Answer:

Here, the point C(k, 4) divides the join of A(2, 6) and B(5, 1) in the ratio 2 : 3. So
k=2×5+3×22+3  =10+65   =165
Hence, k=165.

Page No 340:

Question 9:

Find the point on x-axis which is equidistant from points A(−1, 0) and B(5, 0).               [CBSE 2013C]

Answer:

Let P(x, 0) be the point on x-axis. Then
AP=BPAP2=BP2x+12+0-02=x-52+0-02x2+2x+1=x2-10x+2512x=24x=2
Hence, x = 2.

Page No 340:

Question 10:

Find the distance between the points -85,2 and 25,2.

Answer:

The given points are A-85,2 and B25,2.
Then, x1=-85, y1=2 and x2=25, y2=2
Therefore,
AB=x2-x12+y2- y12    =25--852+2-22    =22+02    =4+0    =4    =2 units



Page No 341:

Question 11:

Find the value of a, so that the point (3, a) lies on the line 2x − 3y = 5.

Answer:

The point  3, a lies on the line 2x-3y=5.If point 3, a lies on the line 2x-3y=5 , then2x-3y=52×3-3×a=5
              6-3a=53a=1a=13
Hence, the value of a is 13.

Page No 341:

Question 12:

If the points A(4, 3) and B(x, 5) lie on the circle with centre O(2, 3), find the value of x.

Answer:

The given points  A4, 3 and Bx, 5 lie on the circle with centre O2,3.Then, OA = OB x-22+5-32=4-22+3-32x-22+22=22+02x-22=22-22x-22=0x-2=0x=2
Hence, the value of x=2.

Page No 341:

Question 13:

If P(x, y) is equidistant from the points A(7, 1) and B(3, 5), find the relation between x and y.

Answer:

Let the point Px, y be equidistant from the points A(7, 1) and B(3, 5).
Then,
PA=PBPA2=PB2x-72+y-12=x-32+y-52x2+y2-14x-2y+50=x2+y2-6x-10y+348x-8y=16x-y=2

Page No 341:

Question 14:

If the centroid of ∆ABC, which has vertices A(a, b), B(b, c) and C(c, a), is the origin, find the value of (a + b + c).

Answer:

The given points are A(a, b), B(b, c) and C(c, a).
Here,
 x1=a, y1=b, x2=b, y2=c and x3=c, y3=a
Let the centroid be (x, y).
Then,
x=13x1+x2+x3  =13a+b+c  =a+b+c3y=13 y1+ y2+ y3  =13b+c+a  =a+b+c3
But it is given that the centroid of the triangle is the origin.
Then, we have:
a+b+c3=0a+b+c=0

Page No 341:

Question 15:

Find the centroid of ∆ABC whose vertices are A(2, 2), B(−4, −4) and C(5, −8).

Answer:

The given points are A(2, 2), B(−4, −4) and C(5, −8).
Here, x1=2, y1=2, x2=-4, y2=-4 and x3=5, y3=-8
Let G(x, y) be the centroid of ABC. Then,
x=13x1+x2+x3  =132-4+5  =1
y=13y1+y2+y3  =132-4-8  =-103
Hence, the centroid of  ABC is G1,-103.

Page No 341:

Question 16:

In what ratio does the point C(4, 5) divide the join of A(2, 3) and B(7, 8)?

Answer:

Let the required ratio be k : 1.
Then, by section formula, the coordinates of C are
C7k+2k+1,8k+3k+1
Therefore,
7k+2k+1=4  and  8k+3k+1=5                 C4, 5 is given7k+2=4k+4  and  8k+3=5k+5 3k=2  and  3k=2
k=23in each case
So, the required ratio is 23:1, which is same as 2:3.

Page No 341:

Question 17:

If the points A(2, 3) B(4, k) and C(6, −3) are collinear, find the value of k.

Answer:

The given points are A2, 3, B4, k and C6, -3.
Here, x1=2, y1=3, x2=4, y2=k and x3=6, y3=-3
It is given that the points A, B and C​ are collinear. Then,
x1y2-y3+x2y3-y1+x3y1-y2=02k+3+4-3-3+63-k=02k+6-24+18-6k=0-4k=0k=0



Page No 343:

Question 1:

The distance of the point P(−6, 8) from the origin is                              [CBSE 2013C]
(a) 8                       (b) 27                         (c) 6                               (d) 10

Answer:

The distance of a point (x, y) from the origin O(0, 0) is x2+y2.
Let P(x = −6, y = 8) be the given point. Then
OP=x2+y2     =-62+82     =36+64     =100=10
Hence, the correct answer is option (d).

Page No 343:

Question 2:

The distance of the point (−3, 4) from x-axis is                            [CBSE 2012]
(a) 3                              (b) −3                                 (c) 4                                      (d) 5

Answer:

The distance of a point (x, y) from x-axis is y.
Here, the point is (−3, 4). So, its distance from x-axis is 4=4.
Hence, the correct answer is option (c).

Page No 343:

Question 3:

The point on x-axis which is equidistant from points A(−1, 0) and B(5, 0) is                  [CBSE 2013]
(a) (0, 2)                                (b) (2, 0)                                      (c) (3, 0)                                              (d) (0, 3)

Answer:

Let P(x, 0) the point on x-axis, then
AP=BPAP2=BP2x+12+0-02=x-52+0-02x2+2x+1=x2-10x+2512x=24x=2
Thus, the required point is (2, 0).
Hence, the correct answer is option (b).

Page No 343:

Question 4:

If R(5, 6) is the midpoint of the line segment AB joining the points A(6, 5) and B(4, y) then y equals
(a) 5                             (b) 7                                (c) 12                                (d) 6

Answer:

Since R(5, 6) is the midpoint of the line segment AB joining the points A(6, 5) and B(4, y), therefore
5+y2=65+y=12y=12-5=7
Hence, the correct option is (b).

Page No 343:

Question 5:

If the point C(k, 4) divides the join of the points A(2, 6) and B(5, 1) in the ratio 2 : 3 then the value of k is                                                                                                                                                [CBSE 2013C]
(a) 16                                 (b) 285                                     (c) 165                                    (d) 85

Answer:

The point C(k, 4) divides the join of the points A(2, 6) and B(5, 1) in the ratio 2 : 3. So
k=2×5+3×22+3=10+65=165
Hence, the correct answer is option (c).



Page No 344:

Question 6:

The perimeter of the triangle with vertices (0, 4), (0, 0) and (3, 0) is                        [CBSE 2014]
(a) 7+5                                 (b) 5                                     (c) 10                                    (d) 12

Answer:

Let A(0, 4), B(0, 0) and C(3, 0) be the given vertices. So
AB=0-02+4-02=16=4BC=0-32+0-02=9=3AC=0-32+4-02=9+16=5
Therefore
AB + BC + AC = 4 + 3 + 5 = 12
Hence, the correct answer is option (d).

Page No 344:

Question 7:

If A(1, 3), B(−1, 2) and C(2, 5) and D(x, 4) are the vertices of a ||gm ABCD then the value of x is       [CBSE 2012]
(a) 3                                 (b) 4                                     (c) 0                                    (d) 32

Answer:

The diagonals of a parallelogram bisect each other. The vertices of the ||gm ABCD are A(1, 3), B(−1, 2) and C(2, 5) and D(x, 4).
Here, AC and BD are the diagonals. So
1+22=-1+x2x-1=3x=1+3=4
Hence, the correct answer is option (b).

Page No 344:

Question 8:

If the points A(x, 2), B(−3, −4) and C(7, −5) are collinear then the value of x is                        [CBSE 2014]
(a) −63                                 (b) 63                                     (c) 60                                    (d) −60

Answer:

Let A(x1 = x, y1 = 2), B(x2 = −3, y2 = −4) and C(x3 = 7, y3 = −5) be collinear points. Then
x1y2-y3+x2y3-y1+x3y1-y2=0x-4+5+-3-5-2+72+4=0x+21+42=0x=-63
Hence, the correct answer is option (a).

Page No 344:

Question 9:

The area of a triangle with vertices A(5, 0), B(8, 0) and C(8, 4) in square units is                   [CBSE 2012]
(a) 20                                 (b) 12                                     (c) 6                                    (d) 16

Answer:

Let A(x1 = 5, y1 = 0), B(x2 = 8, y2 = 0) and C(x3 = 8, y3 = 4) be the vertices of the triangle. Then
AreaABC=12x1y2-y3+x2y3-y1+x3y1-y2                     =1250-4+84-0+80-0                     =12-20+32+0                     =6 sq. units
Hence, the correct answer is option (c).

Page No 344:

Question 10:

The area of ABO with vertices A(a, 0), O(0, 0) and B(0, b) in square units is
(a) ab                                  (b) 12ab                                 (c) 12a2b2                                     (d) 12b2

Answer:

Let A(x1 = a, y1 = 0), O(x2 = 0, y2 = 0) and B(x3 = 0, y3 = b) be the given vertices. So
AreaABO=12x1y2-y3+x2y3-y1+x3y1-y2                    =12a0-b+0b-0+00-0                    =12-ab=                    =12ab
Hence, the correct answer is (b).

Page No 344:

Question 11:

If Pa2, 4 is the midpoint of the line segment joining the points A(−6, 5) and B(−2, 3) then the value of a is  [CBSE 2011]
(a) −8                                  (b) 3                                 (c) −4                                     (d) 4

Answer:

The point Pa2, 4 is the midpoint of the line segment joining the points A(−6, 5) and B(−2, 3). So
a2=-6-22a2=-4a=-8
Hence, the correct answer is option (a).

Page No 344:

Question 12:

ABCD is a rectangle whose three vertices are B(4, 0), C(4, 3) and D(0, 3). The length of one of its diagonals is  [CBSE 2014]
(a) 5                                  (b) 4                                 (c) 3                                     (d) 25

Answer:

Here, AC and BD are two diagonals of the rectangle ABCD. So
BD=4-02+0-32    =42+-32    =16+9    =25    =5 units
Hence, the correct answer is option (a).

Page No 344:

Question 13:

The coordinates of the point P dividing the line segment joining the points A(1, 3) and B(4, 6) in the ratio 2 : 1 is    [CBSE 2012]
(a) (2, 4)                                  (b) (3, 5)                                 (c) (4, 2)                                     (d) (5, 3)

Answer:

Here, the point P divides the line segment joining the points A(1, 3) and B(4, 6) in the ratio 2 : 1. Then
Coordinates of P=2×4+1×12+1, 2×6+1×32+1                           =8+13, 12+33                           =93, 153                           =3, 5
Hence, the correct answer is option (b).

Page No 344:

Question 14:

If the coordinates of one end of a diameter of a circle are (2, 3) and the coordinates of its centre are (−2, 5), then the coordinates of the other end of the diameter are                     [CBSE 2012]
(a) (−6, 7)                                  (b) (6, −7)                                 (c) (4, 2)                                     (d) (5, 3)

Answer:

Let (x, y) be the coordinates of the other end of the diameter. Then
-2=2+x2x=-65=3+y2y=7
Hence, the correct answer is option (a).

Page No 344:

Question 15:

In the given figure P(5, −3) and Q(3, y) are the points of trisection of the line segment
joining A(7, −2) and B(1, −5). Then, y equals                                 [CBSE 2012]
                                             
(a) 2                                  (b) 4                                 (c) −4                                     (d) -52

Answer:

Here, AQ : BQ = 2 : 1. Then
y=2×-5+1×-22+1  =-10-23  =-4
Hence, the correct answer is option (c).



Page No 345:

Question 16:

The midpoint of segment AB is P(0, 4). If the coordinates of B are (−2, 3), then the coordinates of A are                [CBSE 2011]
(a) (2, 5)                             (b) (−2, −5)                               (c) (2, 9)                                  (d) (−2, 11)

Answer:

Let (x, y) be the coordinates of A. Then
0=-2+x2x=24=3+y2y=8-3=5
Thus, the coordinates of A are (2, 5).
Hence, the correct answer is option (a).

Page No 345:

Question 17:

The point P which divides the line segment joining the points A(2, −5) and B(5, 2) in the ratio 2 : 3 lies in the quadrant                                                [CBSE 2011]
(a) I                           (b) II                             (c) III                           (d) IV

Answer:

Let (x, y) be the coordinates of P. Then
x=2×5+3×22+3=10+65=165y=2×2+3×-52+3=4-155=-115
Thus, the coordinates of point P are 165, -115 and so it lies in the fourth quadrant.
Hence, the correct answer is option (d).

Page No 345:

Question 18:

If A(−6, 7) and B(−1, −5) are two given points then the distance 2AB is                             [CBSE 2011]
(a) 13                           (b) 26                             (c) 169                           (d) 238

Answer:

The given points are A(−6, 7) and B(−1, −5). So
AB=-6+12+7+52    =-52+122    =25+144    =169    =13
Thus, 2AB = 26.
Hence, the correct answer is option (b).

Page No 345:

Question 19:

Which point on x-axis is equidistant from the points A(7, 6) and B(−3, 4)?              
(a) (0, 4)                           (b) (−4, 0)                             (c) (3, 0)                           (d) (0, 3)

Answer:

Let P(x, 0) be the point on x-axis. Then as per the question
AP=BPAP2=BP2x-72+0-62=x+32+0-42x2-14x+49+36=x2+6x+9+1660=20xx=6020=3
Thus, the required point is (3, 0).
Hence, the correct answer is option (c).

Page No 345:

Question 20:

The distance of the point P(3, 4) from the x-axis is

(a) 3 units
(b) 4 units
(c) 7 units
(d) 1 unit

Answer:

(b) 4 units
The y-coordinate is the distance of the point from the x-axis.
Here, the y-coordinate is 4.

Page No 345:

Question 21:

In what ratio does the x-axis divide the join of A(2, −3) and B(5, 6)?

(a) 2 : 3
(b) 3 : 5
(c) 1 : 2
(d) 2 : 1

Answer:

(c) 1 : 2
Let AB be divided by the x axis in the ratio k : 1 at the point P.
Then, by section formula, the coordinates of are
P5k+2k+1,6k-3k+1
Butlies on the x axis: so, its ordinate is 0.
6k-3k+1=0
6k-3=0
6k=3
k=12
Hence, the required ratio is 12 : 1, which is same as 1 : 2.

Page No 345:

Question 22:

In what ratio does the y-axis divide the join of P(−4, 2) and Q(8, 3)?

(a) 3 : 1
(b) 1 : 3
(c) 2 : 1
(d) 1 : 2

Answer:

(d) 1 : 2
Let AB be divided by the y axis in the ratio k : 1 at the point P.
Then, by section formula, the coordinates of are
P8k-4k+1,3k+2k+1
But, P lies on the y axis; so, its abscissa is 0.
8k-4k+1=0
8k-4=0
8k=4
k=12
Hence, the required ratio is 12 : 1, which is same as 1 : 2.

Page No 345:

Question 23:

If P(−1, 1) is the midpoint of the line segment joining A(−3, b) and B(1, b + 4), then b = ?

(a) 1
(b) −1
(c) 2
(d) 0

Answer:

(b) −1
The given points are A(−3, b) and B(1, b+4).
Then, x1=-3, y1=b and x2=1, y2=b+4
Therefore,
x=-3+12  =-22  =-1
and
y=b+b+42  =2b+42  =b+2
But the midpoint is P-1, 1.
Therefore,
b+2=1b=-1

Page No 345:

Question 24:

The line 2x + y −4 = 0 divides the line segment joining A(2, −2) and B(3, 7) in the ratio

(a) 2 : 5
(b) 2 : 9
(c) 2 : 7
(d) 2 : 3

Answer:

(b) 2 : 9
Let the line​ 2x+y-4=0 divide the line segment in the ratio k : 1 at the point P.
Then, by section formula, the coordinates of P are
P3k+2k+1,7k-2k+1
Since P lies on the line 2x+y-4=0 , we have:
23k+2k+1+7k-2k+1-4=06k+4+7k-2-4k+4=09k=2k=29
Hence, the required ratio is 29 : 1, which is same as 2 : 9.

Page No 345:

Question 25:

If A(4, 2), B(6, 5) and C(1, 4) be the vertices of ∆ABC and AD is median, then the coordinates of D are

(a) 52,3
(b) 5,72
(c) 72,92
(d) None of these

Answer:

(c) 72,92
D is the midpoint of BC.
So, the coordinates of D are
D6+12,5+42    B6, 5 and C1, 4x1=6, y1=5 and x2=1, y2=4i.e. D72, 92

Page No 345:

Question 26:

If A(−1, 0), B(5, −2) and C(8, 2) are the vertices of a ∆ABC, then its centroid is

(a) (12, 0)
(b) (6, 0)
(c) (0, 6)
(d) (4, 0)

Answer:

(d) (4, 0)
The given points are A-1, 0, B5, -2 and C8, 2.
Here, x1=-1, y1=0, x2=5, y2=-2 and x3=8, y3=2
Let G(x, y) be the centroid of ABC. Then,
x=13x1+x2+x3  =13-1+5+8  =4
and
y=13y1+y2+y3  =130-2+2  =0
Hence, the centroid of ABC is G(4, 0).

Page No 345:

Question 27:

Two vertices of ∆ABC are A(−1, 4) and B(5, 2) and its centroid is G(0, −3). Then, the coordinates of C are

(a) (4, 3)
(b) (4, 15)
(c) (−4, −15)
(d) (−15, −4)

Answer:

(c) (−4, −15)
Two vertices of ABC are A-1,4 and B5,2.
Let the third vertex be C(a, b).
Then, the coordinates of its centroid are
G-1+5+a3,4+2+b3i.e. G4+a3,6+b3
But it is given that the centroid is G0,-3.
Therefore,
4+a3=0 and 6+b3=-3
4+a=0 and 6+b=-9
a=-4 and b=-15
Hence, the third vertex of ABC is C-4, -15.

Page No 345:

Question 28:

The points A(−4, 0), B(4, 0) and C(0, 3) are the vertices of a triangle, which is

(a) isosceles
(b) equilateral
(c) scalene
(d) right-angled

Answer:

(a) isosceles
Let A(−4, 0), B(4, 0) and C(0, 3) be the given points. Then,
AB=4+42+0-02    =82+02    =64+0    =64    =8 unitsBC=0-42+3-02    =-42+32    =16+9    =25    =5 unitsAC=0+42+3-02    =42+32    =16+9    =25    =5 units
BC = AC = 5 units
Therefore, ABC is isosceles.



Page No 346:

Question 29:

The points P(0, 6), Q(−5, 3) and R(3, 1) are the vertices of a triangle, which is

(a) equilateral
(b) isosceles
(c) scalene
(d) right-angled

Answer:

(d) right-angled

Let P(0, 6), Q(−5, 3) and R(3, 1) be the given points. Then,
PQ=-5-02+3-62    =-52+-32    =25+9    =34  unitsQR=3+52+1-32    =82+-22    =64+4    =68    =217 unitsPR=3-02+1-62    =32+-52    =9+25    =34 unitsPQ2+PR2342+342=68QR2217 2=68
Thus, PQ2+PR2=QR2
Therefore, ∆PQR is right-angled.

Page No 346:

Question 30:

If the points A(2, 3), B(5, k) and C(6, 7) are collinear, then

(a) k = 4
(b) k = 6
(c) k=-32
(d) k=114

Answer:

(b) k = 6
The given points are A(2, 3), B(5, k) and C(6, 7).
Here, x1=2, y1=3, x2=5, y2=k and x3=6, y3=7.
Points A,B and C are collinear. Then,
x1y2-y3+x2y3-y1+x3y1-y2=02k-7+57-3+63-k=02k-14+20+18-6k=0-4k=-24k=6

Page No 346:

Question 31:

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

(a) a = b
(b) a = 2b
(c) 2a = b
(d) a + b = 0

Answer:

(c) 2a = b
The given points are A(1, 2), O(0, 0) and C(a, b).
Here, x1=1, y1=2, x2=0, y2=0 and x3=a, y3=b.
Points A, O and C are collinear.
x1y2-y3+x2y3-y1+x3y1-y2=010-b+0b-2+a2-0=0-b+2a=02a=b

Page No 346:

Question 32:

The area of ∆ABC with vertices A(3, 0), B(7, 0) and C(8, 4) is

(a) 14 sq units
(b) 28 sq units
(c) 8 sq units
(d) 6 sq units

Answer:

(c) 8 sq units
The given points are A3, 0, B7, 0 and C8, 4.
Here, x1=3, y1=0, x2=7, y2=0 and x3=8, y3=4
Therefore,
Area of ABC=12x1y2-y3+x2y3-y1+x3y1-y2

                  =1230-4+74-0+80-0=12-12+28+0=12×16=8 sq units

Page No 346:

Question 33:

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

(a) 5 units
(b) 3 units
(c) 34 units
(d) 4 units

Answer:

(c) 34 units
A0,3, O0,0 and B5,0 are the three vertices of a rectangle; let C be the fourth vertex.
Then, the length of the diagonal,
AB=5-02+0-32    =52+-32    =25+9    =34 units
Since, the diagonals of rectangle are equal .
Hence, the length of its diagonals is 34 units.

Page No 346:

Question 34:

If the distance between the points A(4, p) and B(1, 0) is 5, then

(a) p = 4 only
(b) p = −4 only
(c) p = ±4 only
(d) p = 0

Answer:

(c) p = ±4 only
The given points are A(4, p) and B(1, 0) and AB = 5.
Then, x1=4, y1=p and x2=1, y2=0
Therefore,
AB=5x2-x12+y2-y12=51-42+0-p2=5-32+-p2=259+p2=25p2=16p=±16p=±4



View NCERT Solutions for all chapters of Class 10