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Page No 724:

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 724:

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 724:

Question 3:

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

Answer:

Given AB = 5 units
Therefore, (AB)2 = 25 units
5-a2+3--12 = 255-a2+3+12 = 255-a2+42 = 255-a2+16 = 255-a2 = 25-165-a2 = 95-a =± 95-a = ±35-a = 3  or   5-a=-3a=2 or 8
Therefore, a = 2 or 8.

Page No 724:

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 724:

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 724:

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 724:

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 725:

Question 8:

Find the point on the x-axis which is equidistant from the points (2, −5) and (−2, 9).

Answer:

Let (x, 0) be the point on the x-axis. Then as per the question, we have
x-22+0+52=x+22+0-92x-22+52=x+22+-92x-22+52=x+22+-92                          Squaring both sidesx2-4x+4+25=x2+4x+4+81
8x=25-81x=-568=-7
Hence, the point on the x-axis is (− 7, 0).

Page No 725:

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 725:

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 725:

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 725:

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 725:

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 725:

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 725:

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 725:

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 725:

Question 17:

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

Answer:

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.

Page No 725:

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 725:

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 725:

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 725:

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 725:

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 725:

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 725:

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 726:

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 726:

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 726:

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 726:

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 726:

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 726:

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 726:

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 726:

Question 32:

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

(i) A(−4, −1), B(−2, −4) C(−3, 2) and D(0, −1)    
(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) The given points are A(−4, −1), B(−2, −4) C(4, 0) and D(2, 3).
AB = -2--42+-4--12 = 22+-32 = 4+9 =13 unitsBC = 4--22+0--42 = 62+42 = 36+16 =52 =213 unitsCD = 2-42+3-02 = -22+32 = 4+9 =13 unitsAD = 2--42+3--12 = 62+42 =36+16 =52 =213 unitsThus, AB = CD = 13 units and BC = AD =213 unitsAlso, AC = 4--42+0--12 = 82+12 = 64+1 =65 unitsBD = 2--22+3--42 = 42+72 = 16+49 = 65 units
Also, diagonal AC = diagonal BD
Hence, the given points form a 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 737:

Question 1:

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

Answer:

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).

Page No 737:

Question 2:

Find the coordinates of the points that divide the join of A(−5, 11) and B(4, −7) in the ratio 7 : 2.

Answer:

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 737:

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 738:

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 738:

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 738:

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 738:

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 738:

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 738:

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 738:

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 738:

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 738:

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 738:

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 738:

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 738:

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 738:

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 738:

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 739:

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 739:

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 739:

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

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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).

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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).

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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).

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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.

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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

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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).

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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.

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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.

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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.

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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.

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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.

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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.



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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.

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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.



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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

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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.

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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.

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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.

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Question 5:

Find the area of quadrilateral ABCD whose vertices are A(−5, 7), B(−4, −5), C(−1, −6) and D(4, 5).

Answer:

By joining A and C, we get two triangles ABC and ACD.
Let Ax1, y1=A-5, 7, Bx2, y2=B-4, -5, Cx3, y3=C-1, -6 and Dx4, y4=D4, 5. Then
Area of ABC=12x1y2-y3+x2y3-y1+x3y1-y2                       =12-5-5+6-4-6-7-17+5                       =12-5+52-12=352 sq. units
Area of ACD=12x1y3-y4+x3y4-y1+x4y1-y3                       =12-5-6-5-15-7+47+6                       =1255+2+52=1092 sq. units
So, the area of the quadrilateral ABCD is 352+1092=72 sq. units.

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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.

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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.

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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.

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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.

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Question 10:

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.

Answer:

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.



Page No 754:

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.

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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.

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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.

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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.

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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.

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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.

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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

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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.

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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.

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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.

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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.

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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.

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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.



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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.

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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.

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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..

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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.

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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.

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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.

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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.

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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.

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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.

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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

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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.

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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.

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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

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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

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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.

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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.

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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 760:

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).

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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).

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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).

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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).

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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).

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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).

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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 761:

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).

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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).

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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).

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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).

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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).

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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).

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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).

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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).

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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).

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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).



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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).

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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).

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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.

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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.

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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.

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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

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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.

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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

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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).

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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.

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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.

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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.

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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



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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

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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

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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.

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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



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