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RS Aggarwal 2015 Solutions for Class 10 Math Chapter 16 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 2015 Book of class 10 Math Chapter 16 are provided here for you for free. You will also love the ad-free experience on Meritnation’s RS Aggarwal 2015 Solutions. All RS Aggarwal 2015 Solutions for class 10 Math are prepared by experts and are 100% accurate.

Page No 636:

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, a − b) and Q(a −b, 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, a − b) and Q(a −b, 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 637:

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 637:

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 = 25⇒5-a2+3+12 = 25⇒5-a2+42 = 25⇒5-a2+16 = 25⇒5-a2 = 25-16⇒5-a2 = 9⇒5-a =± 9⇒5-a = ±3⇒5-a = 3  or   5-a=-3⇒a=2 or 8
Therefore, a = 2 or 8.

Page No 637:

Question 4:

Find the ordinate of a point whose abscissa is 10 and which is at a distance of 10 units from the points P(2, −3).

Answer:

Let Q be the other point whose coordinates are Q(10, y).
Given, PQ = 10 units
Therefore, (PQ)2 = 100
⇒10-22+y--32 = 100⇒82+y+32 = 100⇒64+ y+32= 100⇒ y+32= 100-64⇒ y+32= 36⇒ y+3=± 36⇒ y+3=± 6⇒ y+3=6   or   y+3=-6⇒ y=3   or  -9
Therefore, y = 3 or −9.

Page No 637:

Question 5:

If A(6, −1), B(1, 3) and C(k, 8) are three points such that AB = BC, find the value of k.

Answer:

The given points are A(6, −1), B(1, 3) and C(k, 8).
Also, it is given that AB = BC.
Therefore, (AB)2 = (BC)2
⇒1-62+3--12 =k-12+8-32 ⇒1-62+3+12 =k-12+8-32 ⇒-52+42 =k-12+52 ⇒25+16 =k-12+25⇒25+16 -25=k-12⇒k-12 =16 ⇒k-1 =±16⇒k-1 = ±4⇒k-1 = 4   or   k-1=-4⇒k =5  or -3
Therefore, the value of k will be either 5 or −3.

Page No 637:

Question 6:

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

Answer:

The given points are A(a, 2), B(8, −2) and C(2, −2).
Also, it is given that AB = AC.
Therefore, (AB)2 = (AC)2
⇒8-a2+-2-22 =2-a2+-2-22 ⇒8-a2+-42 =2-a2+-42⇒8-a2+42 =2-a2+-42⇒8-a2+16 =2-a2+16⇒8-a2 =2-a2⇒64+a2-16a=4+a2-4a⇒-12a=-60⇒a=-60-12=5
Therefore, the value of a is 5.

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Question 7:

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

Answer:

Let the given points be A(−2, 5) and B(−2, 9) and the required point be P(x,0). Then, AP = PB.
That is, (AP)2 = (PB)2
⇒x--22+0-52 = -2-x2+9-02⇒x+22+0-52 = -2-x2+92⇒x+22+-52 = -2-x2+92⇒x+22+25 = -2-x2+81⇒x2+4x+4+25 =x2+4x+4+81⇒x2+4x+29 =x2+4x+85⇒x2+4x-x2-4x =85-29⇒0 = 56

Disclaimer : As it is never possible that 0 will become equal to 56
                     Therefore, we can say that there is no point on the x-axis which is equidistance from A(-2,5) and B(-2,9)

Page No 637:

Question 8:

Find the point on the y-axis that is equidistant from the points (5, −2) and (−3, 2).

Answer:

Let the given points be A(5, −2) and B(−3, 2) and the required point be P(0, y). Then, AP = BP.
Also, (AP)2 = (BP)2
⇒0-52+y+22 = 0+32+y-22⇒-52+y+22 = 32+y-22⇒25 +y2+4+4y = 9+y2+4-4y⇒y2+4+4y +25 = y2+4-4y+9⇒y2+4y-y2+4y =4+9-4-25⇒8y = -16⇒y = -168=-2
So, the point is (0, −2).

Page No 637:

Question 9:

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+5-32⇒22+02 = x-22+22⇒4= x-22+4⇒x-22 = 0⇒x-2 = 0⇒x=2
Therefore, x = 2.

Page No 637:

Question 10:

If P(x, y) is a point equidistant from the points A(6, −1) and B(2, 3), show that x − y = 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-y2⇒x2-12x+36+y2+2y+1 = x2-4x+4+y2-6y+9⇒x2+y2-12x+2y+37 = x2+y2-4x-6y+13⇒x2+y2-12x+2y-x2-y2+4x+6y =13-37⇒-8x+8y = -24⇒-8x-y = -24⇒x-y =-24-8⇒x-y =3
Hence proved.

Page No 637:

Question 11:

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

Answer:

A(11,−8) is the given point. Let P(x,0) be the required point on the x-axis; then, AP = 10 units.
Also (AP)2 = (10)2 = 100
⇒x-112+0+82 = 100⇒x-112+82 = 100⇒x-112+64 = 100⇒x-112= 100-64⇒x-112= 36⇒x-11=±36⇒x-11=±6If x-11=6,x =6+11⇒x = 17 Or if  x-11=-6,x=11-6=5
Therefore, the coordinates of the point are (17, 0) or (5, 0). It is 10 units away from the point A(11, −8).

Page No 637:

Question 12:

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)2 
⇒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 637:

Question 13:

Using the distance formula, show that the points (1, −1), (5, 2) and (9, 5) are collinear.

Answer:

Let A(1, −1), B (5, 2) and C(9, 5) be the given points. Then:
AB =5-12+2+12 = 42+32 =16+9 =25 = 5 unitsBC =9-52+5-22 = 42+32 =16+9 =25 = 5 unitsAC = 9-12+5+12 = 82+62 =64+36 =100 = 10 unitsAB + BC = 5+5 = 10 units = AC

Hence, the given points are collinear.

Page No 637:

Question 14:

Show that the following points are collinear:

(i) A(6, 9) B(0, 1) and C(−6, −7)
(ii) A(−1, −1) B(2, 3) and C(8, 11)
(iii) P(1, 1) Q(−2, 7) and R(3, −3)
(iv) P(2, 0) Q(11, 6) and R(−4, −4)

Answer:

(i) A(6, 9) B(0, 1) and C(−6, −7)
AB =0-62+1-92 = -62+-82 =36+64 =100 = 10 unitsBC =-6-02+-7-12 = -62+-82 =36+64 =100 = 10 unitsAC = -6-62+-7-92 = -122+-162 =144+256 =400 = 20 units AB + BC = 10+10 = 20 units = AC

Hence, the points A, B and C are collinear.

(ii) A(−1, −1) B(2, 3) and C(8, 11)
AB =2+12+3+12 = 32+42 =9+16 =25 = 5 unitsBC =8-22+11-32 = 62+82 =36+64 =100 = 10 unitsAC = 8+12+11+12 = 92+122 =81+144 =225 = 15 unitsAB + BC = 5+10 = 15 units = AC
Hence, the points A, B and C are collinear.

(iii) P(1, 1) Q(−2, 7) and R(3, −3)
PQ =-2-12+7-12 = -32+62 =9+36 =45 =9×5= 35 unitsQR =3+22+-3-72 = 52+-102 =25+100 =125=25×5=55  unitsPR = 3-12+-3-12 = 22+-42 =4+16 =20 =4×5 = 25 unitsPQ + PR = 35+ 25 =55  units = QR

Hence, the points Q, P and R are collinear.

(iv) P(2, 0) Q(11, 6) and R(−4, −4)

PQ =11-22+6-02 = 92+62 =81+36 =117 =9×13= 313 unitsQR =-4-112+-4-62 = -152+-102 =225+100 =325=25×13=513  unitsPR = -4-22+-4-02 = -62+-42 =36+16 =52 =4×13 = 213 unitsPQ + PR = 313+ 213 =513  units = QR

Hence, the points Q, P and R are collinear.

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Question 15:

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

Answer:

A(3, 0), B (6, 4) and C(−1, 3) are the given points. Then:
AB =6-32+4-02 = 32+42 =9+16 =25 = 5 unitsBC =-1-62+3-42 = -72+-12 =49+1 =50 = 25×2 = 52 unitsAC = -1-32+3-02 = -42+32 =16+9 =25 = 5 units
Thus, AB = BC = 5 units
Also, (AB)2+(AC)2 = (5)2+(5)2 = 50
and (BC)2 = (52)2 = 25×2 = 50
Thus, (AB)2+(AC)2 = (BC)2
This show that ∆ABC is right- angled at A.
This proves that the the points A(3, 0), B(6, 4) and C(−1, 3) are the vertices of an isosceles right- angled triangle.

Page No 637:

Question 16:

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)2 = 1062+ 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 638:

Question 17:

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)2 = 102+ 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 638:

Question 18:

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, the 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 638:

Question 19:

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 638:

Question 20:

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

(i) A(0, −4), B(6, 2) C(3, 5) and D(−3, −1)
(ii) A(2, −2), B(14, 10) C(11, 13) and D(−1, 1)
(iii) A(−4, −1), B(−2, −4) C(−3, 2) and D(0, −1)

Answer:

(i) 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.

(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(−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.

Page No 638:

Question 21:

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

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

Answer:

(i) 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.

(ii) 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.

(iii) 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.

Page No 638:

Question 22:

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 649:

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+n⇒x= 2×4 + 3×-12+3, y =2×-3+3×72+3⇒x= 8-35, y = -6+215⇒x = 55, y = 155Therefore, x = 1 and y = 3
Hence, the coordinates of the required point are (1, 3).

Page No 649:

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+n⇒x= 7×4 + 2×-57+2, y =7×-7+2×117+2⇒x= 28-109, y = -49+229⇒x = 189, y = -279Therefore, x = 2 and y = -3
Hence, the required point is P(2, −3).

Page No 649:

Question 3:

Find the coordinates of the points of trisection of the line segment AB, whose end points are A(2, 1) and B(5, −8).

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= 1×5 + 2×21+2, y =1×-8+2×11+2x= 5+43, y = -8+23x = 93, y = -63So,  x = 3 and y = -2
Therefore, the coordinates of P are (3, −2).
Also, Q divides AB in the ratio 2:1.
So, the coordinates of Q are
x =mx2+nx1m+n, y = my2+ny1m+nx= 2×5 + 1×22+1, y =2×-8+1×12+1x= 10+23, y = -16+13x = 123, y = -153So, x = 4 and y = -5
Therefore, the coordinates of Q are (4, −5).
Hence, the points are (3, −2) and (4, −5).

Page No 649:

Question 4:

Find the coordinates of the points that divide the join of A(−4, 0) and B(0, 6) in three equal parts.

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+n⇒x= 1×0 + 2×-41+2, y =1×6+2×01+2⇒x= -83, y = 63⇒x = -83, y = 2Therefore, coordinates of P are -83, 2.
Also Q divides AB in the ratio 2:1
So the coordinates of Q are
x =mx2+nx1m+n, y = my2+ny1m+nx= 2×0 + 1×-42+1, y =2×6+1×02+1x= -43, y = 123x = -43, y = 4Therefore, coordinates of Q are -43, 4.
Hence, the points are -83, 2 and -43, 4.

Page No 649:

Question 5:

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+n⇒x= 1×1 + 2×31+2, y =1×2+2×-41+2⇒x= 1+63, y = 2-83⇒x= 73, y = -63⇒x = 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+n⇒x= 2×1 + 1×32+1, y =2×2+1×-42+1⇒x= 2+33, y = 4-43⇒x = 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 649:

Question 6:

The line segment joining the points A(4, −5) and B(4, 5) is divided by the point P such that APAB=25. Find the coordinates of P.

Answer:

The given points are A(4, −5) and B(4, 5).
P divides AB, such that APAB = 25.
Here, AB = AP+PBTherefore, APAP+PB = 25⇒5AP = 2AP + 2PB⇒5AP-2AP = 2PB⇒3AP = 2PBTherefore, APPB = 23So, m = 2 and n = 3.

x =mx2+nx1m+n, y = my2+ny1m+n⇒x= 2×4 + 3×42+3, y =2×5+3×-52+3⇒x= 8+125, y = 10-155⇒x = 205, y = -55x = 4 and y = -1Hence, the coordinates of P are (4, -1).

Page No 649:

Question 7:

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+y22⇒x = 3+-52, y =0+42⇒x = -22, y = 42⇒x = -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+y22⇒x = -11+82, y =-8-22⇒x =- 32, y = -102⇒x = -32, y = -5
Therefore, -32, -5 are the coordinates of midpoint of PQ.

Page No 649:

Question 8:

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+y22⇒x = 6+-22, y =-5+112⇒x = 6-22, y = -5+112⇒x = 42, y = 62⇒x = 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 649:

Question 9:

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+y22⇒1 = 2a+-22, 2a+1 =4+3b2⇒2=2a-2,  4a+2 = 4+3b⇒2a = 2+2  , 4a-3b = 4-2⇒a=42, 4a-3b = 2⇒a = 2, 4a-3b = 2Putting the value of a in the equation  4a+3b = 2, we get:42-3b = 2⇒-3b = 2-8 = -6⇒b = 63 = 2Therefore, a=2 and b=2.

Page No 649:

Question 10:

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+n⇒x= 1×5 + 3×11+3, y =1×-2+3×61+3⇒x= 5+34, y = -2+184⇒x = 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+y22⇒x = 1+52, y =6+-22⇒x = 62, y = 42⇒x = 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+n⇒x= 3×5 + 1×13+1, y =3×-2+1×63+1⇒x= 15+14, y = -6+64⇒x = 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 650:

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+y22⇒x = -2+62, y =9+32⇒x = 42, y = 122⇒x = 2, y = 6
Therefore, the coordinates of point C are (2, 6).

Page No 650:

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+42⇒4 = a+1 , -6 = b+4⇒a = 4-1, b=-6-4⇒a = 3, b= -10
Therefore, the coordinates of point A are (3, -10).

Page No 650:

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+1⇒2k+2 = -6k+8 ,  5k+5 = 9k+2⇒2k+6k = 8-2 ,  5-2 =9k-5k⇒8k = 6, 4k = 3⇒k = 68, k = 34⇒k = 34 in each case.
Therefore, the point P(2, 5) divides AB in the ratio 3 : 4.

Page No 650:

Question 14:

Find the ratio in which the point P(−6, a) divides the join of A(−3, −1) and B(−8, 9), Also, find the value of a.

Answer:

Let the point P(−6, a) divide the line AB in the ratio k : 1.
Then, by the section formula:
x = -8k-3k+1, y =9k-1k+1The coordinates of P are P(-6, a).-6 = -8k-3k+1, a = 9k-1k+1⇒-6k-6 = -8k-3 , ak+1 = 9k-1⇒-6k+8k = -3+6 ,  ak+1 = 9k-1⇒2k = 3,  ak+1 = 9k-1⇒k = 32,  ak+1 = 9k-1Therefore, the point P divides the line AB in the ratio 3 : 2.Now, putting the value of k in the equation ak+1 = 9k-1, we get:a32+1 = 932-1⇒a3+22 = 27-22⇒5a2 = 252⇒5a = 25⇒a = 255 = 5Therefore, the value of a = 5That is, the coordinates of P are (-6, 5).Point P divides the line AB in the ratio 3 : 2.

Page No 650:

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+1⇒m(k+1) = 2k-4 ,  6k+6 = 8k+3⇒m(k+1) = 2k-4 ,  6 -3= 8k - 6k⇒m(k+1) = 2k-4, 2k =3⇒m(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 = 232-4⇒m3+22 = 3-4⇒5m2 =-1⇒5m = -2⇒m = -25 Therefore, the value of m = -25So, the coordinates of P are (-25, 6).

Page No 650:

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-4⇒s =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 = 32-4⇒3k = 6-4⇒3k = 2 ⇒k = 23Therefore, the value of k = 23That is, the coordinates of P are (-3, 23).

Page No 650:

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=0⇒6k-3 =0⇒6k = 3⇒k =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 650:

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=0⇒3k-2 =0⇒3k = 2⇒k =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 650:

Question 19:

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

Answer:

Let the line x − y − 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 x − y − 2 = 0, we have:

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

Page No 650:

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: AD = 10 units, BE = 2 units and CF = 10 units

Page No 650:

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 650:

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 = b⇒a =-2, b= 7
Therefore, the third vertex of ∆ABC is C(−2, 7).

Page No 650:

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+b3⇒0 = -3+a , 0 =-1+b⇒3 = a, 1= b⇒a =3, b= 1
Therefore, the third vertex of ∆ABC is A(3, 1).

Page No 650:

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 650:

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=2⇒a+2 = 6 , b+1 = 4⇒a = 6-2 , b = 4-1⇒a = 4 and b = 3

Page No 650:

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 = 4⇒3+a =6 and 6+b =8⇒a = 6-3 and b= 8-6⇒a= 3 and b= 2
Therefore, the fourth vertex is D(3, 2).



Page No 655:

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 655:

Question 2:

(i) Find the area of a quadrilateral ABCD whose vertices are
A(−4, −2), B(−3, −5), C(3, −2) and D(2, 3)

(ii) Find the area of the quadrilateral ABCD whose vertices are
A(0, 0), B(6, 0), C(4, 3) and D(0, 3)

(iii) Find the area of the quadrilateral ABCD whose vertices are
A(1, 0), B(5, 3), C(2, 7) and D(−2, 4).

Answer:

(i) The vertices are A(−4, −2), B(−3, −5), C(3, −2) and D(2, 3).
Join AC. Then,
Area of quadrilateral ABCD = Area of triangle ABC + Area of triangle ACD

Area of triangle ABC=12x1y2-y3+x2y3-y1+x3y1-y2= 12-4-5--2+-3-2--2+3-2--5=12-4-3-30+33=1212+9=1221= 10.5 sq. units
Now,
Area of triangle ACD=12x1y2-y3+x2y3-y1+x3y1-y2= 12-4-2-3+33--2+2-2--2=12-4-5+35+20=1220+15=1235= 17.5 sq. units
Area of quadrilateral ABCD = Area of triangle ABC + Area of triangle ACD
Area of quadrilateral ABCD = 10.5 sq. units + 17.5 sq. units = 28 sq. units

(ii) The vertices are A(0, 0), B(6, 0), C(4, 3) and D(0, 3).
Join AC. Then,
Area of quadrilateral ABCD = Area of triangle ABC + Area of triangle ACD

Area of triangle ABC=12x1y2-y3+x2y3-y1+x3y1-y2= 1200-3+63-0+40-0=120+18+0=1218= 9 sq. units
Now,
Area of triangle ACD=12x1y2-y3+x2y3-y1+x3y1-y2= 1203-3+43-0+00-3=120+12+0=1212= 6 sq. units
Area of quadrilateral ABCD = Area of triangle ABC + Area of triangle ACD
Area of quadrilateral ABCD = 9 sq. units + 6 sq. units = 15 sq. units

(iii) The vertices are A(1, 0), B(5, 3), C(2, 7) and D(−2, 4).
Join AC. Then,
Area of quadrilateral ABCD = Area of triangle ABC + Area of triangle ACD

Area of triangle ABC=12x1y2-y3+x2y3-y1+x3y1-y2= 1213-7+57-0+20-3=121-4+57+2-3=12-4+35-6=1225= 12.5 sq. units
Now,
Area of triangle ACD=12x1y2-y3+x2y3-y1+x3y1-y2= 1217-4+24-0-20-7=1213+24-2-7=123+8+14=1225Area of triangle ACD = 12.5 sq. units
Area of quadrilateral ABCD = Area of triangle ABC + Area of triangle ACD
Area of quadrilateral ABCD = 12.5 sq. units + 12.5 sq. units = 25 sq. units

Page No 655:

Question 3:

Show that the following points are collinear:

(i) A(0, 1), B(1, 2) and C(−2, −1)
(ii) A(−5, 1), B(5, 5) and C(10, 7)
(iii) P(a, b + c), Q(b, c + a) and R(c, a + b)

Answer:

(i) A(0, 1), B(1, 2) and C(−2, −1) are the given points. Then:
(x1 = 0, y1 = 1), (x2 = 1, y2 = 2) and (x3 =−2, y3 = −1)
∴x1y2-y3+x2y3-y1+x3y1-y2=02--1+1-1-1+-21-2=03+1-2-2-1=0-2+2=0
Hence, the given points are collinear.

(ii) A(−5, 1), B(5, 5) and C(10, 7) are the given points. Then:
(x1 = −5, y1 = 1), (x2 = 5, y2 = 5) and (x3 = 10, y3 = 7)
∴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) P(a, b + c), Q(b, c + a) and R(c, a + b) are the given points. Then:
(x1 = a, y1 = b + c), (x2 = b, y2 = c + a) and (x3 = c, y3 = a + b)
∴x1y2-y3+x2y3-y1+x3y1-y2=ac+a-a-b+ba+b-b-c+cb+c-c-a=a(c-b)+b(a-c)+c(b-a)=ac-ab+ba-bc+cb-ca=0
Hence, the given points are collinear.

Page No 655:

Question 4:

Find the values of p for which the given points are collinear:

(i) A(−1, 3) B(2, p) and C(5, −1)
(ii) A(3, 2) B(4, p) and C(5, 3)
(iii) A(−3, 9) B(2, p) and C(4, −5)

Answer:

(i) A(−1, 3), B(2, p) and C(5, −1) are the given points. Then:
(x1 = −1, y1 = 3), (x2 = 2, y2 = p) and (x3 = 5, y3 = −1)
It is given that points A, B and C are collinear.
⇒x1y2-y3+x2y3-y1+x3y1-y2=0⇒-1p--1+2-1-3+53-p=0⇒-1p+1+2-4+53-p=0⇒-p-1-8+15-5p=0⇒6-6p=0⇒6p = 6⇒p =66 =1
Therefore, the value of p is 1.

(ii) A(3, 2) B(4, p) and C(5, 3) are the given points. Then:
(x1 = 3, y1 = 2), (x2 = 4, y2 = p) and (x3 = 5, y3 = 3)
It is given that points A, B and C are collinear.
⇒x1y2-y3+x2y3-y1+x3y1-y2=0⇒3p-3+43-2+52-p=0⇒3p-3+41+52-p=0⇒3p-9+4+10-5p=0⇒5-2p=0⇒2p = 5⇒p =52 
Therefore, the value of p is 52.

(iii) A(−3, 9) B(2, p) and C(4, −5) are the given points. Then:
(x1 = −3, y1 = 9), (x2 = 2, y2 = p) and (x3 = 4, y3 = −5)
It is given that points A, B and C are collinear.
⇒x1y2-y3+x2y3-y1+x3y1-y2=0⇒-3p--5+2-5-9+49-p=0⇒-3p+5+2-14+49-p=0⇒-3p-15-28+36-4p=0⇒-7-7p=0⇒7p = -7⇒p =-77= -1
Therefore, the value of p is −1.

Page No 655:

Question 5:

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=0⇒9-21+6x=0⇒6x-12=0⇒6x = 12⇒x =126 =2
Therefore, when x= 2, the given points are collinear.

Page No 655:

Question 6:

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=0⇒1y-16+316-4+-34-y=0⇒1y-16+312-34-y=0⇒y-16+36-12+3y=0⇒8+4y=0⇒4y = -8⇒y =-84 =-2
When y = −2, the given points are collinear.

Page No 655:

Question 7:

If the points A(x, y), B(−5, 7) and C(−4, 5) are collinear, then show that 2x + y + 3 = 0.

Answer:

The given points are A(x, y), B(−5, 7) and C(−4, 5).
So, (x1 = x, y1 = y), (x2 = −5, y2 = 7) and (x3 = −4, y3 = 5).
If the given points are collinear, then:
x1y2-y3+x2y3-y1+x3y1-y2 =0⇒x7-5+-55-y+-4y-7=0⇒x2-55-y-4y-7=0⇒2x-25+5y-4y+28=0⇒2x+y+3 = 0

Page No 655:

Question 8:

A(−4, −2), B(−3, −5) C(3, −2) and D(2, k) are the vertices of a quadrilateral ABCD. Find the value of k, if the area of the quadrilateral is 28 sq units.

Answer:

Let A(−4, −2), B(−3, −5), C(3, −2) and D(2, k) be the vertices of quadrilateral ABCD.
Join AC. Then, area of quadrilateral ABCD = Area of triangle ABC + Area of triangle ACD.

Area of the triangle ABC=12x1y2-y3+x2y3-y1+x3y1-y2= 12-4-5--2+-3-2--2+3-2--5=12-4-3-30+33=1212+9=1221Area of the triangle ABC = 10.5 sq. units.
Therefore, area of triangle ACD = Area of the quadrilateral ABCD-Area of the triangle ABCArea of triangle ACD = 28 sq units - 10.5 sq units = 17.5 sq. units.
To find the value of k, let us consider the area of triangle ACD.
Area of the triangle ACD=12x1y2-y3+x2y3-y1+x3y1-y217.5= 12-4-2-k+3k--2+2-2--217.5=12-4-2-k+3k+2+2017.5=128+4k+3k+617.5=127k+1435 = 7k+147k = 35-14k=217 = 3
Therefore, the value of k = 3.

Page No 655:

Question 9:

For ∆ABC, whose vertices are A(4, −6), B(3, −2) and C(5, 2), verify that a median of the triangle divides it into two triangles of equal areas.

Answer:

To prove : Area of triangle ABD = Area of triangle ACD
Given ∆ABC, whose vertices are A(4, −6), B(3, −2) and C(5, 2).
Join AD. Let AD be the median that divides the triangle into two triangles.

So, D is the midpoint of BC.
Therefore, the coordinates of D are
D3+52, -2+22 = D82, 02 =D4,0
Now,
Area of triangle ABD = 12x1y2-y3+x2y3-y1+x3y1-y2=124-2-0+30--6+4-6--2=124-2+36+4-4=12-8+18-16=12-6=3 sq. units.                               Area cannot be -ve
and
Area of triangle ACD = 12x1y2-y3+x2y3-y1+x3y1-y2=1242-0+50--6+4-6-2=1242+56+4-8=128+30-32=126=3 sq. units.
Therefore, area of triangle ABD = Area of triangle ACD
Hence, the median of ∆ABC divides it into two triangles of equal areas.

Page No 655:

Question 10:

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

Answer:

Consider the points A(2, 1), B(x, y) and C(7, 5) .
Here, (x1 = 2, y1 = 1), (x2 = x, y2 = y) and (x3 = 7, y3 = 5).
It is given that the points are collinear. So,
x1y2-y3+x2y3-y1+x3y1-y2 = 0⇒2y-5+x5-1+71-y=0⇒2y-10+5x-x+7-7y=0⇒4x-5y-3 = 0



Page No 656:

Question 11:

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 = 0⇒ab-1+01-0+10-b=0⇒ab-a-b = oDividing the equation by ab:⇒1-1b-1a = 0⇒1-1a+1b = 0⇒1a+1b=1 
Therefore, the given points are collinear if 1a+1b = 1.

Page No 656:

Question 1:

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 656:

Question 2:

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=5⇒2×3-3×a=5
              ⇒6-3a=5⇒3a=1⇒a=13
Hence, the value of a is 13.

Page No 656:

Question 3:

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+3-32⇒x-22+22=22+02⇒x-22=22-22⇒x-22=0⇒x-2=0⇒x=2
Hence, the value of x=2.

Page No 656:

Question 4:

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=PB⇒PA2=PB2⇒x-72+y-12=x-32+y-52⇒x2+y2-14x-2y+50=x2+y2-6x-10y+34⇒8x-8y=16⇒x-y=2

Page No 656:

Question 5:

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=0⇒a+b+c=0

Page No 656:

Question 6:

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 656:

Question 7:

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 given⇒7k+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 656:

Question 8:

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=0⇒2k+3+4-3-3+63-k=0⇒2k+6-24+18-6k=0⇒-4k=0⇒k=0



Page No 658:

Question 1:

In which quadrant does the point (−3, 5) lie?

(a) I
(b) II
(c) III
(d) IV

Answer:

(b) II
The point (−3, 5) lies in quadrant II, since the signs of the coordinates are (−, +).

Page No 658:

Question 2:

In which quadrant does the point (2, −4) lie?

(a) I
(b) II
(c) III
(d) IV

Answer:

(d) IV
The point (2, −4) lies in quadrant IV since the signs of the coordinates are (+, −).

Page No 658:

Question 3:

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 658:

Question 4:

P is a point on x-axis at a distance of 3 units from y-axis, to its right. The coordinates of P are

(a) (3, 0)
(b) (0, 3)
(c) (3, 3)
(d) (−3, 3)

Answer:

(a) (3, 0)
It is given that point P is on the x-axis at a distance of 3 units from the y-axis, to its right.Therefore, the coordinates of P are 3,0.

Page No 658:

Question 5:

A is a point on the y-axis at a distance of 4 units from the x-axis and lying below the x-axis. The coordinates of A are

(a) (4, 0)
(b) (0, 4)
(c) (−4, 0)
(d) (0, −4)

Answer:

(d) (0, −4)
It is given that point A is on the y-axis at a distance of 4 units from the x-axis and lies below the x-axis.Therefore, the coordinates of A are 0,-4.

Page No 658:

Question 6:

The distance of the point P(4, −3) from the origin is

(a) 1 unit
(b) 3 units
(c) 5 units
(d) 7 units

Answer:

(c) 5 units
The given point is P​(4, −3) and the origin is (0, 0).
Then, x1=4, y1=-3 and x2=0, y2=0
OP=0-42+0--32    =-42+32    =16+9    =25    =5 units

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Question 7:

The distance between the points A(2, −3) and B(2, 2) is

(a) 4 units
(b) 5 units
(c) 3 units
(d) 2 units

Answer:

(b) 5 units
The given points are A(2, −3) and B(2, 2).
Then, x1=2, y1=-3 and x2=2, y2=2
Therefore,
AB=x2-x12+y2- y12    =2-22+2+32    =02+52    =0+25    =25    =5 units

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Question 8:

What point on the 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:

(c) (3, 0)
Let P(x,0) be equidistant from the points A(7,6) and B(−3,4).
Then,
PA=PB⇒PA2=PB2⇒x-72+0-62=x+32+0-42⇒x2-14x+49+36=x2+6x+9+16⇒x2-14x+85=x2+6x+25⇒20x=60⇒x=3 units
Hence,
x=3, y=0⇒3,0

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Question 9:

The distance between the points A(0, 5) and B(−5, 0) is

(a) 5 units
(b) 25 units
(c) 10 units
(d) 52 units

Answer:

(d) 52 units
The given points are A(0, 5) and B(−5, 0).
Then, x1=0, y1=5 and x2=-5, y2=0
Therefore,
AB=x2-x12+y2- y12    =-5-02+0-52    =-52+-52    =25+25    =50    =52 units

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Question 10:

A point P divides the join of A(5, −2) and B(9, 6) in the ratio 3 : 1. The coordinates of P are

(a) (4, 7)
(b) (8, 4)
(c) (12, 8)
(d) 112,5

Answer:

(b) (8, 4)
The given points are A(5, −2) and B(9, 6)
Then, x1=5, y1=-2 and x2=9, y2=6
Also, m ​= 3 and n = 1
Let the required point be P(x, y).
By section formula, we have
x=mx2+nx1m+n,  y=my2+ny1m+n⇒x=3×9+1×53+1, y=3×6+1×-23+1⇒x=324, y=164⇒x=8, y=4
Hence, the coordinates of P are (8, 4).



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Question 11:

In what ratio does the point P(1, 2) divide the join of A(−2, 1) and B(7, 4)?

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

Answer:

(a) 1 : 2
Let the required ratio be k : 1.
Then, by section formula, the coordinates of P are
P7k-2k+1,4k+1k+1
Therefore,
7k-2k+1=1 and 4k+1k+1=2   Since P1, 2 is given
⇒7k-2=k+1 and 4k+1=2k+2
⇒6k=3 and 2k=1
⇒k=12 in each case
So, the required ratio is 12 : 1, which is same as 1 : 2.

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Question 12:

The point which divides the line segment joining the points A(7, −6) and B(3, 4) in the ratio 1 : 2 lies in

(a) I quadrant
(b) II quadrant
(c) III quadrant
(d) IV quadrant

Answer:

(d) IV quadrant
The given points are A7, -6 and B3, 4.
Then, x1=7, y1=-6 and x2=3, y2=4
Also, m = 1 and n = 2
Let the required point be Px, y.
By section formula, we have
x=1×3+2×71+2, y=1×4+2×-61+2 
⇒x=173, y=-83
Hence, the coordinates of P are 173,-83, which lies in the IV quadrant.

Page No 659:

Question 13:

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 P are
P5k+2k+1,6k-3k+1
But P lies 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 14:

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 P 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 15:

The midpoint of the line segment joining the points A(−2, 8) and B(−6, −4) is

(a) (4, 2)
(b) (−4, 2)
(c) (2, 6)
(d) (−4, −6)

Answer:

(b) (−4, 2)
The given points are A(−2, 8) and B(−6, −4).
Then, x1=-2, y1=8 and x2=-6, y2=-4
Let M(x,y) be the midpoint of AB. Then,
x=-2+-62  =-82  =-4
and
y=8+-42  =42  =2
Hence, the required point is M-4, 2.

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Question 16:

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=1⇒b=-1

Page No 659:

Question 17:

If Pa3,4 is the midpoint of the line segment joining A(−6, 5) and B(−2, 3), then a = ?

(a) −4
(b) −12
(c) 12
(d) −6

Answer:

(b) −12
The given points are A(−6, 5) and B(−2, 3).
Then, x1=-6, y1=5 and x2=-2, y2=3
Therefore,
x=-6+-22  =-82  =-4
and
y=5+32  =82  =4
But the given midpoint is Pa3, 4 .
Therefore,
-4=a3⇒a=-12

Page No 659:

Question 18:

If the distance between the points A(2, −2) and B(−1, x) is 5, then

(a) x = −3 or x = 4
(b) x = 3 or x = −4
(c) x = −6 or x = 2
(d) x = 6 or x = −2

Answer:

(c) x = −6 or x = 2
The given points are A(2, −2) and B(−1, x) and AB = 5.
Then, x1=2, y1=-2 and x2=-1, y2=x
Therefore,
AB = 5⇒x2-x12+y2-y12=5⇒-1-22+x+22=5⇒-32+x+22=25⇒9+x2+4x+4=25⇒x2+4x+13=25⇒x2+4x-12=0⇒x+6x-2=0⇒x=-6 or x=2

Page No 659:

Question 19:

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=0⇒6k+4+7k-2-4k+4=0⇒9k=2⇒k=29
Hence, the required ratio is 29 : 1, which is same as 2 : 9.

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Question 20:

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, 4⇒x1=6, y1=5 and x2=1, y2=4i.e. D72, 92

Page No 659:

Question 21:

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 22:

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 660:

Question 23:

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=5⇒x2-x12+y2-y12=5⇒1-42+0-p2=5⇒-32+-p2=25⇒9+p2=25⇒p2=16⇒p=±16⇒p=±4

Page No 660:

Question 24:

The three vertices of a parallelogram ABCD are A(−2, 3) B(6, 7) and C(8, 3). The fourth vertex D is

(a) (1, 0)
(b) (0, 1)
(c) (−1, 0)
(d) (0, −1)

Answer:

(d) (0, −1)
Let the fourth vertex be D(a, b).
Join AC and BD.
Let AC and BD intersect at the point O.
We know that the diagonals of a parallelogram bisect each other.
Midpoint of AC is -2+82,3+32⇒3, 3Midpoint of BD is 6+a2,7+b2
Therefore,
6+a2=3 and 7+b2=3⇒6+a=6 and 7+b=6
⇒a=0 and b=-1
Hence, the fourth vertex is D(0,−1).

Page No 660:

Question 25:

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 26:

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+PR2⇒342+342=68QR2⇒217 2=68
Thus, PQ2+PR2=QR2
Therefore, ∆PQR is right-angled.

Page No 660:

Question 27:

The points (a, a), (−a, a) and (-3a,3a) form the vertices of

(a) an equilateral triangle
(b) a scalene triangle
(c) an isosceles triangle
(d) a right triangle

Answer:

(a) an equilateral triangle
Let A(a, a), B(−a, −a) and C-3a, 3a be the given points. Then,
AB=-a-a2+-a-a2    =-2a2+-2a2    =4a2+4a2    =8a2    =22a unitsBC=-3a+a2+3a+a2    =a1-32+1+32    =a21+3    =a8    =22a unitsAC=-3a-a2+3a-a2    =a3+12+3-12    =a23+1    =a8    =22a units
AB = BC = AC
Hence, the points form the vertices of an equilateral triangle.

Page No 660:

Question 28:

Three points A(1, −2) B(3, 4) and C(4, 7) form

(a) a straight line
(b) an equilateral triangle
(c) a right-angled triangle
(d) None of these

Answer:

(a) a straight line
The given points are A(1, −2), B(3, 4) and C(4, 7).
Here, x1=1, y1=-2, x2=3, y2=4 and x3=4, y3=7.
Therefore,
Area of ∆ABC=12x1y2-y3+x2y3-y1+x3y1-y2
 
                  =1214-7+37+2+4-2-4=12-3+27-24=0
Thus, the given points are collinear, i.e. they form a straight line.

Page No 660:

Question 29:

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=0⇒2k-7+57-3+63-k=0⇒2k-14+20+18-6k=0⇒-4k=-24⇒k=6

Page No 660:

Question 30:

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=0⇒10-b+0b-2+a2-0=0⇒-b+2a=0⇒2a=b

Page No 660:

Question 31:

The area of ∆ABC with vertices A(a, b + c), B(b, c + a) and C(c, a + b) is

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

Answer:

(d) 0
The given points are Aa,b+c, Bb,c+a and Cc,a+b.
Here, x1=a, y1=b+c, x2=b, y2=c+a and x3=c, y3=a+b
Therefore,
Area of ∆ABC=12x1y2-y3+x2y3-y1+x3y1-y2

                  =12ac+a-a+b+ba+b-b+c+cb+c-c+a=12ac-b+ba-c+cb-a=12ac-ab+ab-bc+bc-ac=0

Page No 660:

Question 32:

The point which lies on the perpendicular bisector of the line segment joining the points A(−2, −5) and B(2, 5) is

(a) (0, 0)
(b) (0, 2)
(c) (2, 0)
(d) (−2, 0)

Answer:

(a) (0, 0)
The point which lies on the perpendicular bisector of AB is the midpoint of AB.
The given points are A-2,-5 and B2,5.
Here, x1=-2, y1=-5 and x2=2, y2=5
Midpoint of AB=-2+22,-5+52i.e. 0,0
Hence, the required point is (0,0).

Page No 660:

Question 33:

In the given figure, A(0, 2y) and B(2x, 0) are the end-points of line segment AB. The coordinates of point C, which is equidistant from the three vertices of ∆AOB, are


(a) (x , y)
(b) (y , x)
(c) x2,y2
(d) y2,x2

Answer:

(a) (x , y)
The midpoint of AB is equidistant from the three vertices of ∆AOB.
Therefore,
Midpoint of AB=0+2x2,2y+02   A0,2y and B2x,0⇒x1=0, y1=2y and x2=2x, y2=0i.e. x, y
Hence, the coordinates of C are (x, y).



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Question 34:

The points A(9, 0), B(9, 6), C(−9, 6) and D(−9, 0) are the vertices of a

(a) square
(b) rectangle
(c) rhombus
(d) trapezium

Answer:

(b) rectangle
A9, 0, B9, 6, C-9, 6 and D-9, 0 are the given vertices.
Then,
AB2=9-92+6-02    =02+62    =0+36    =36 unitsBC2=-9-92+6-62    =-182+02    =324+0    =324 unitsCD2=-9+92+0-62     =02+-62     =0+36     =36 unitsDA2=-9-92+0-02     =-182+02     =324+0     =324 units
Therefore, we have:
AB2=CD2 and BC2=DA2
Now, the diagonals are:
AC2=-9-92+6-02    =-182+62    =324+36    =360 unitsBD2=-9-92+0-62     =-182+-62     =324+36     =360 units
Therefore,
AC2=BD2
Hence, ABCD is a rectangle.

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Question 35:

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 661:

Question 36:

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 661:

Question 37:

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 the coordinates of P and Q, respectively, are

(a) (0, −5) and (2, 0)
(b) (0, −10) and (4, 0)
(c) (0, 10) and (−4, 0)
(d) (0, 4) and (−10, 0)

Answer:

(b) (0, −10) and (4, 0)
Let the line PQ intersect y axis at P(0, a) and x axis at Q(b, 0).
Midpoint of PQ=0+b2,a+02
But the midpoint of PQ is (2, −5).
Therefore,
0+b2=2 and a+02=-5⇒b=4 and a=-10.
Therefore,
P0, a⇒P0, -10Qb, 0⇒Q4, 0
Hence, the coordinates of P and Q are (0, −10) and (4, 0).

Page No 661:

Question 38:

A circle drawn with origin as the centre passes through the point A132,0. Which of the following points does not lie inside the circle?

(a) -34,1
(b) 2,73
(c) 5,-13
(d) -6,52

Answer:

(d) -6,52
Let the given point be A132, 0 and the origin (0, 0). Then,
the radius of the circle,
   OA=0-132+0-02
⇒OA=-1322+02⇒OA=1694⇒OA=132 units
Now, the distance of the point -34, 1 from the origin is
  0+34+0-12=342+-12=916+1=2516=54 units, which lies inside the circle.
Now, the distance of the point 2, 73 from the origin is
  0-22+0-732=-22+-732=4+499=859=853 units, which lies inside the circle.
Now, the distance of the point 5, -12 from the origin is
  0-52+0+122=-52+122=25+14=1014=1012 units, which lies inside the circle.
Now, the distance of the point -6, 52 from the origin is
  0+62+0-522=62+-522=36+254=1694=132 units=radius OA
Hence, it does not lie inside the circle.

Page No 661:

Question 39:

The coordinates of one end point of a diameter AB of a circle are A(4, −1) and the coordinates of the centre of the circle are C(1, −3). Then, the coordinates of B are


(a) (2, −5)
(b) (−2, 5)
(c) (−2, −5)
(d) (2, 5)

Answer:

(c) (−2, −5)
Let the coordinates of B be (a, b). Then,
Midpoint of AB=4+a2,-1+b2
But the midpoint C(1, −3) is given.
Therefore,
4+a2=1 and -1+b2=-3⇒4+a=2 and -1+b=-6⇒a=-2 and b=-5.
Hence, the coordinates of B are (−2, −5).

Page No 661:

Question 40:

Match the following columns:

Column I Column II
(a) The coordinates of the point
that divides the join of
A(−1, 7) and B(4, −3) in the
ratio 2 : 3 are .......		 (p) (5, 6)
(b) Two vertices of ∆ABC are
A(6, 4) and B(−2, 2) and its
centroid is G(3, 4). The
coordinates of C are .......		 (q) 10
(c) If the points A(4, 3) and B(x, 5)
lie on a circle with the centre
O(2, 3), then x = ....... .		 (r) (1, 3)
(d) If A(0, −1), B(2, 1) and C(0, 3)
are the vertices of a ∆ABC, then
the length of median AD is		 (s) 2

Answer:

(a) ⇒ (r)
(b) ⇒ (p)
(c) ⇒ (s)
(d) ⇒ (q)

(a) The given points are  A-1, 7 and B4, -3.
Then, 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 have:
x=2×4+3×-12+3, y=2×-3+3×72+3⇒x=55, y=155⇒x=1, y=3
Therefore, the coordinates of P are (1, 3).
Hence, a⇒r

(b) The two vertices of ∆ABC are A6, 4 and B-2, 2.
Let the third vertex be C(a, b).
Then, the coordinates of its centroid are
G6-2+a3,4+2+b3i.e. G4+a3,6+b3
But it is given that the centroid is G(3, 4).
Therefore,
4+a3=3 and 6+b3=4⇒4+a=9 and 6+b=12
⇒a=5 and b=6
Therefore, the coordinates of C are (5, 6).
Hence, b⇒p

(c) The points A4, 3 and Bx, 5 lie on the circle with centre O(2, 3).
Then, we have:
    x-22+5-32=4-22+3-32
⇒x-22+22=22+02⇒x-22=22-22⇒x-22=0⇒x-2=0⇒x=2
Therefore, the value of x = 2
Hence, c⇒s

(d) D is the midpoint of BC.
So, the coordinates of D are
D2+02,1+32    B2, 1 and C1, 4⇒x1=2, y1=1 and x2=1, y2=4i.e. D1, 2
Now, the length of the median
AD=1-02+2+12    =12+32    =1+9    =10 units
Hence, d⇒q



Page No 662:

Question 41:

Assertion (A)
If the points A(8, 1), B(k, −4) and C(2, −5) are collinear, then k = 4.

Reason (R)
Three points A(x1, y1), B(x2, y2) and C(x3, y3) are collinear only when x1(y2 − y3) + x2(y3 − y1) + x3(y1 − y2) = 0.

(a) Both Assertion (A) and Reason (R) are true and Reason (R) is a correct explanation of Assertion (A).
(b) Both Assertion (A) and Reason (R) are true but Reason (R) is not a correct explanation of Assertion (A).
(c) Assertion (A) is true and Reason (R) is false.
(d) Assertion (A) is false and Reason (R) is true.

Answer:

The given points are A8, 1, Bk, -4 and C2, -5
Here, x1=8, y1=1, x2=k, y2=-4 and x3=2, y3=-5
Points A,B and C are collinear.
⇒x1y2-y3+x2y3-y1+x3y1-y2=0⇒8-4+5+k-5-1+21+4=0⇒81+k-6+25=0⇒-6k=-18⇒k=3
Hence, Assertion(A) is false.
Reason(R):
The given statement is true.
Hence, the correct answer is (d).

Page No 662:

Question 42:

Assertion (A)
The point P(−1, 6) divides the join of A(−3, 10) and B(6, −8) in the ratio 2 : 7.

Reason (R)
If the point C(x, y) divides the join of A(x1, y1) and B(x2, y2) in the ratio m : n, then x=mx2+nx1m+n and y=my2+ny1m+n.

(a) Both Assertion (A) and Reason (R) are true and Reason (R) is a correct explanation of Assertion (A).
(b) Both Assertion (A) Reason (R) are true but Reason (R) is not a correct explanation of Assertion (A).
(c) Assertion (A) is true and Reason (R) is false.
(d) Assertion (A) is false and Reason (R) is true.

Answer:

The correct answer is (a)
Let the required ratio be k : 1.
Then, by section formula, the coordinates of C are
P6k-3k+1,-8k+10k+1
Therefore,
6k-3k+1=-1 and -8k+10k+1=6    Since P-1, 6 is given⇒6k-3=-k-1 and -8k+10=6k+6

⇒7k=2 and -14k=-4
⇒k=27 in each case.
Therefore, the required ratio is 27 : 1, which is same as 2 : 7.
Hence, Assertion(A) is true.
Reason(R):
The given statement is true.
 Both Assertion (A) and Reason (R) are true and Reason (R) is a correct explanation of Assertion (A).



Page No 667:

Question 1:

The ratio in which the point P(6, −6) divides the line segment joining A(1, 4) and B(9, −12) is

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

Answer:

(d) 5 : 3
Let the required ratio be k : 1.
Then, by section formula, the coordinates of C are
P9k+1k+1,-12k+4k+1
Therefore,
9k+1k+1=6 and 9k+1k+1=-6  Since P6,-6 is given⇒9k+1=6k+6 and 9k+1=-6k-6⇒3k=5 and -6k=-10⇒k=53 in each case.
Therefore, the required ratio is 53 : 1, which is same as 5 : 3.

Page No 667:

Question 2:

In what ratio is the segment joining the points A(4, 6) and B(−7, −1) divided by x-axis?

(a) 3 : 1
(b) 6 : 1
(c) 1 : 2
(d) 2 : 3

Answer:

(b) 6 : 1
Let AB be divided by x-axis in the ratio k : 1 at the point P.
Then, by section formula, the coordinates of P are
P-7k+4k+1, -k+6k+1
But P lies on the x-axis. So, its ordinate is 0.
Therefore,
-k+6k+1=0⇒-k+6=0⇒k=6
Hence, the required ratio is 6 : 1.

Page No 667:

Question 3:

The two vertices of a triangle are A(6, 3) and B(−1, 7) and its centroid is G(1, 5). The third vertex C of ∆ABC is

(a) (2, 5)
(b) (2, −5)
(c) (−2, 5)
(d) (−2, −5)

Answer:

(c) (−2, 5)
The two vertices of ∆ABC are A6, 3 and B-1, 7.
Let the third vertex be C(a, b).
Then, the coordinates of its centroid are
G6-1+a3,3+7+b3i.e. G5+a3,10+b3
But it is given that the centroid is G(1,5).
Therefore,
5+a3=1 and 10+b3=5⇒5+a=3 and 10+b=15⇒a=-2 and b=5
Hence, the third vertex of ∆ABC is C-2, 5.



Page No 668:

Question 4:

The area of ∆ABC with vertices A(1, −1), B(−4, 6) and C(−3, −5) is

(a) 52 sq units
(b) 27 sq units
(c) 48 sq units
(d) 24 sq units

Answer:

(d) 24 sq units
The given points are A1, -1, B-4, 6 and C-3, -5.
Here, x1=1, y1=-1, x2=-4, y2=6 and x3=-3, y3=-5
Therefore,
Area of ∆ABC=12x1y2-y3+x2y3-y1+x3y1-y2
                   =1216+5+-4-5+1+-3-1-6=1211+16+21=12×48=24 sq units

Page No 668:

Question 5:

Find the length of the line segment AB whose end points are A(2, −5) and B(−4, 7).

Answer:

The given points are A2, -5 and B-4, 7.
Then, x1=2, y1=-5 and x2=-4, y2=7
Therefore,
AB=x2-x12+y2- y12    =-4-22+7+52    =-62+122    =36+144    =180    =65 units

Page No 668:

Question 6:

Find the point on x-axis that is equidistant from the points A(−3, 2) and B(5, −2).

Answer:

Let the point P(x,0) be equidistant from the points A-3, 2 and B5, -2.
Then,
PA=PB⇒PA2=PB2⇒x+32+0-22=x-52+0+22⇒x2+6x+13=x2-10x+29⇒16x=16⇒x=1
Therefore,
Px, 0≡P1, 0

Page No 668:

Question 7:

Find the coordinates of point P that divides the join of A(4, −3) and B(9, 7) in the ratio 3 : 2.

Answer:

The given points are A4, -3 and B9, 7.
Then, x1=4, y1=-3 and x2=9, y2=7
Also, m = 3 and n = 2
Let the required point be P(x, y).
By section formula, we have:
x=3×9+2×43+2, y=3×7+2×-33+2⇒x=355, y=155⇒x=7, y=3
Hence, the coordinates of P are (7, 3).

Page No 668:

Question 8:

If A(−2, 4), B(0, 0) and C(4, 2) are the vertices of a ∆ABC, find the length of the median through A.

Answer:

Let D be the midpoint of BC.
So, the coordinates of D are
D0+42, 0+22    B0, 0 and C4, 2 i.e, x1=0, y1=0 and x2=4, y2=2i.e.  D2, 1
Length of the median:

AD=2+22+1-42    =42+-32    =16+9    =25    =5 units

Page No 668:

Question 9:

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
Points A,B and C are collinear. Then,
x1y2-y3+x2y3-y1+x3y1-y2=0⇒2k+3+4-3-3+63-k=0⇒2k+6-24+18-6k=0⇒-4k=0⇒k=0

Page No 668:

Question 10:

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 are A(4, 3) and B(x, 5), which lie on the circle with centre O(2, 3).
Then, we have:
OB = OA
⇒x-22+5-32=4-22+3-32⇒x-22+22=22+02⇒x-22=22-22⇒x-22=0⇒x-2=0⇒x=2

Page No 668:

Question 11:

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

Answer:

Let the line x − y − 2=0 divide the line segment joining A(3, −1) and B(8, 9) in the ratio k : 1 at the point P.
Then, by the section formula, the coordinates of P are
P8k+3k+1,9k-1k+1
Since, P lies on the line x − y − 2=0, we have:
8k+3k+1-9k-1k+1-2=0⇒8k+3-9k-1-2k+2=0⇒3k=2⇒k=23
So, the required ratio is 23 : 1  ,which is same as 2 : 3.

Page No 668:

Question 12:

Find the area of ∆ABC with vertices A(1, −3), B(4, −3) and C(−9, 7).

Answer:

The given points are A1, -3, B4, -3 and C-9, 7.
Here, x1=1, y1=-3, x2=4, y2=-3 and x3=-9, y3=7
Therefore,
Area of ∆ABC=12x1y2-y3+x2y3-y1+x3y1-y2
                  
                  =121-3-7+47+3-9-3+3=12-10+40+0=12×30=15 sq units

Page No 668:

Question 13:

Show that A(−3, −1), B(−4, −1), C(3, 3) and D(4, 3) are the vertices of a rhombus.

Answer:

Disclaimer:- It should be parallelogram in place of rhombus in the question.
The given vertices are A(−3, −1), B(−4, −1), C(3, 3) and D​(4, 3).
Then,
AB=-4+32+-1+12    =-12+02    =1+0    =1 unitBC=3+42+3+12    =72+42    =49+16    =65 unitsCD=4-32+3-32    =12+02    =1+0    =1 unitDA=-3-42+-1-32    =-72+-42    =49+16    =65 units
Here, AB = CD
and  BC = DA
Therefore, ABCD is a parallelogram.

Page No 668:

Question 14:

Find the points on y-axis, which are at a distance of 13 units each from the point (−5, 7).

Answer:

Let the point on y-axis be B(0, b) and the given point be A(−5, 7).The distance between A and B is 13 units.
Now,
     AB=13⇒0+52+b-72=13⇒25+b2+49-14b=169⇒b2-14b-95=0⇒b-19b+5=0⇒b=19 or -5
Hence, the points are (0, 19) and (0, −5).

Page No 668:

Question 15:

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 the fourth vertex be D(a, b).
Join AC and BD.
Let AC and BD intersect at the point O.
We know that the diagonals of a parallelogram bisect each other.
Midpoint of AC is 1+52, -2+102⇒3, 4Midpoint of BD is 3+a2, 6+b2
Therefore,

3+a2=3 and 6+b2=4⇒3+a=6 and 6+b=8

⇒a=3 and b=2

Hence, the fourth vertex is D(3, 2).

Page No 668:

Question 16:

If the points A(x, y), B(−5, 7) and C(−4, 5) are collinear, show that 2x + y + 3 = 0.

Answer:

The given points are Ax, y, B-5, 7 and C-4, 5.
Here, x1=x, y1=y, x2=-5, y2=7 and x3=-4, y3=5
A,B and C​ are collinear. Then,
x1y2-y3+x2y3-y1+x3y1-y2=0⇒x7-5-55-y-4y-7=0⇒7x-5x-25+5y-4y+28=0⇒2x+y+3=0

Page No 668:

Question 17:

Find the area of quadrilateral ABCD whose vertices are A(−4, −2), B(−3, −5), C(3, −2) and D(2, 3).

Answer:

Join A and C . Then,

Area of quadilateral ABCD=area of ∆ABC+area of ∆ACD
Area of∆ABC=12-4-5+2-3-2+2+3-2+5=12-4-3-30+33=1212-0+9=212 sq units

Area of ∆ACD=12-4-2-3+33+2+2-2+2=12-4-5+35+20=1220+15+0=352 sq units
Therefore, area of quadilateral ABCD =212+352
                                                              =28 sq units

Page No 668:

Question 18:

Find the area of the triangle formed by joining the midpoints of the sides of the triangle whose vertices are A(2, 2), B(4, 4) and C(2, 6).

Answer:

Let A(2, 2), B(4, 4) and C(2, 6) be the vertices of a triangle ABC.
Let D,E and F be the midpoints of AB, BC and CA, respectively.
Then, the coordinates of D, E and F​ are:
D2+42, 2+42, E4+22, 4+62 and F2+22, 2+62
i.e. D3, 3, E3, 5 and F2, 4Then, x1= 3, y1= 3,x2= 3, y2= 5 and x3= 2, y3= 4
Area of ∆DEF=12x1y2-y3+x2y3-y1+x3y1-y2

                    =1235-4+34-3+23-5=1231+31+2-2=123+3-4=12×2=1 sq unit

Page No 668:

Question 19:

Prove that the points A(−3, 0), B(1, −3) and C(4, 1) are the vertices of an isosceles right-angled triangle. Find the area of this triangle.

Answer:

Let A-3, 0, B1, -3 and C4, 1 be the given vertices. Then,
AB=1--32+-3-02    =42+-32    =16+9    =25    =5 unitsBC=4-12+1--32    =32+42    =9+16    =25    =5 unitsAC=4--32+1-02    =72+12    =49+1    =50    =52 units
Thus, AB = BC = 5 units.
Therefore, ∆ABC is an isosceles triangle.
Also,
AB2+BC2=52+52
                = 50 units.
and AC2=522
          = 50 units
Thus, AB2+BC2=AC2
This shows that ∆ABC is right-angled at B.
So, ∆ABC is an isosceles right-angled triangle.
In ∆ABC, we have:
base BC = 5 units and height AB= 5 units
Area of ∆ABC=12×base×height
                   =12×5×5=12.5 sq units

Page No 668:

Question 20:

If the point P(−1, 2) divides the line segment joining A(2, 5) and B in the ratio 3 : 4, find the coordinates of B.

Answer:

Let the coordinates of B be x1,  y1.
Then, P divides the line joining A(2, 5) and Bx1,  y1 in the ratio 3 : 4.
So, by section formula, the coordinates of P are
P3x1+83+4,3y1+203+4i.e. P3x1+87,3y1+207

But, the coordinates of P are (−1, 2).

3x1+87=-1 and 3y1+207=2

⇒3x1+8=−7  and  3y1+20=14

⇒x1=-15 and 3y1=-6⇒x1=-5 and y1=-2

Hence, the coordinates of B are (−5, −2).



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