RD Sharma XI 2018 Solutions for Class 11 Humanities Math Chapter 22 Brief Review Of Cartesian System Of Rectanglar Co Ordinates are provided here with simple step-by-step explanations. These solutions for Brief Review Of Cartesian System Of Rectanglar Co Ordinates are extremely popular among class 11 Humanities students for Math Brief Review Of Cartesian System Of Rectanglar Co Ordinates Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the RD Sharma XI 2018 Book of class 11 Humanities Math Chapter 22 are provided here for you for free. You will also love the ad-free experience on Meritnation’s RD Sharma XI 2018 Solutions. All RD Sharma XI 2018 Solutions for class 11 Humanities Math are prepared by experts and are 100% accurate.

Page No 22.12:

Question 1:

If the line segment joining the points P (x1, y1) and Q (x2, y2) subtends an angle α at the origin O, prove that
OP · OQ cos α = x1 x2 + y1, y2

Answer:


From the figure,
OP2=x12+y12
OQ2=x22+y22
PQ2=x2-x12+y2-y12
Using cosine formula in OPQ, we get:
PQ2=OP2+OQ2-2OP·OQcosα
x2-x12+y2-y12=x12+y12+x22+y22-2OP·OQcosα
x22+x12-2x1x2+y22+y12-2y1y2=x12+y12+x22+y22-2OP·OQcosα
-2x1x2-2y1y2=-2OP·OQcosα
OP·OQcosα=x1x2+y1y2



Page No 22.13:

Question 2:

The vertices of a triangle ABC are A (0, 0), B (2, −1) and C (9, 2). Find cos B.

Answer:

We know that cosB=a2+c2-b22ac, where a = BC, b = CA and c = AB are the lengths of the sides of ABC.
Thus,
a=BC=2-92+-1-22=49+9=58
b=AC=0-92+0-22=81+4=85
c=AB=2-02+-1-02=4+1=5

Using cosine formula in ABC, we get:
cosB=a2+c2-b22ac

cosB=58+5-85258×5=-11290

Page No 22.13:

Question 3:

Four points A (6, 3), B (−3, 5), C (4, −2) and D (x, 3x) are given in such a way that Δ DBCΔ ABC=12. Find x.

Answer:

We know that the area of a triangle with vertices x1, y1, x2, y2 and x3, y3 is given by:

Area=12x1y2-y3+x2y3-y1+x3y1-y2
Area of DBC=12-3-2-3x+43x-5+x5+2
Area of DBC=72x-1

Area of ABC=1265+2-3-2-3+43-5
=492

It is given that Δ DBCΔ ABC=12.
72x-1×249=12

x=118

Page No 22.13:

Question 4:

The points A (2, 0), B (9, 1), C (11, 6) and D (4, 4) are the vertices of a quadrilateral ABCD. Determine whether ABCD is a rhombus or not.

Answer:

The given points are A (2, 0), B (9, 1), C (11, 6) and D (4, 4).
Let us find the length of all the sides of the quadrilateral ABCD.

AB=2-92+0-12=50=52

BC=11-92+6-12=29

CD=4-112+4-62=49+4=53

AD=4-22+4-02=4+16=25

ABBCCDAD, quadrilateral ABCD is not a rhombus.

Page No 22.13:

Question 5:

Find the coordinates of the centre of the circle inscribed in a triangle whose vertices are (−36, 7), (20, 7) and (0, −8).

Answer:

The coordinates of the in-centre of a triangle whose vertices are Ax1,y1, Bx2,y2 and Cx3,y3 are ax1+bx2+cx3a+b+c, ay1+by2+cy3a+b+c, where a = BC, b = AC and c = AB.
Let A(−36, 7), B(20, 7) and C(0, −8) be the coordinates of the vertices of the given triangle.
Now,
a=BC=20-02+7+82=25
b=AC=0+362+-8-72=39
c=AB=20+362+7-72=56

Thus, the coordinates of the in-centre of the given triangle are:

25×-36+39×20+025+39+56, 25×7+39×7+56-825+39+56

= -120120, 0

= -1, 0

Hence, the coordinates of the centre of the circle inscribed in a triangle whose vertices are (−36, 7), (20, 7) and (0, −8) is -1, 0.

Page No 22.13:

Question 6:

The base of an equilateral triangle with side 2a lies along the y-axis, such that the mid-point of the base is at the origin. Find the vertices of the triangle.

Answer:

Let ABC be an equilateral triangle, where BC = 2a. Let A(x, 0) be the third vertex of ABC.

In equilateral triangle ABC,
AB = BC = AC
AB2= BC2= AC2

a2+x2=2a2       BC=2ax2=3a2x=±3a

So, the vertices of the triangle are 0,-a, 0,a and 3a,0 or 0,-a, 0,a and -3a,0.

Page No 22.13:

Question 7:

Find the distance between P (x1, y1) and Q (x2, y2) when (i) PQ is parallel to the y-axis (ii) PQ is parallel to the x-axis.

Answer:

The given points are Px1,y1 and Qx2,y2.

Distance between P and Q is:

PQ=x1-x22+y1-y22

(i) When PQ is parallel to the y-axis:

In this case, x1=x2.

PQ=x1-x12+y1-y22=y1-y2

(ii) When PQ is parallel to the x-axis:

In this case, y1=y2.

PQ=x1-x22+y1-y12=x1-x2

Page No 22.13:

Question 8:

Find a point on the x-axis, which is equidistant from the points (7, 6) and (3, 4).

Answer:

Let C(x, 0) be a point on the x-axis, which is equidistant from the points A(7, 6) and B(3, 4).

AC = BC

AC2=BC2

7-x2+6-02=3-x2+4-0249+x2-14x+36=9+x2-6x+1685-14x=25-6x60=8x152=x

Thus, the point on the x-axis, which is equidistant from the points (7, 6) and (3, 4) is 152, 0.



Page No 22.18:

Question 1:

Find the locus of a point equidistant from the point (2, 4) and the y-axis.

Answer:

Let P(h, k) be the point which is equidistant from the point (2, 4) and the y-axis.
The distance of point P(h, k) from the y-axis is h.

h=h-22+k-42h2-4h+4+k2-8k+16=h2k2-4h-8k+20=0

Hence, the locus of (h, k) is y2-4x-8y+20=0.

Page No 22.18:

Question 2:

Find the equation of the locus of a point which moves such that the ratio of its distances from (2, 0) and (1, 3) is 5 : 4.

Answer:

Let A(2, 0) and B(1, 3) be the given points. Let P (h, k) be a point such that PA:PB = 5:4

PAPB=54h-22+k-02h-12+k-32=54

Squaring both sides, we get:

 16h2-4h+4+k2=25h2-2h+1+k2-6k+99h2+9k2+64h-50h-150k-64+250=09h2+9k2+14h-150k+186=0

Hence, the locus of (h, k) is 9x2+9y2+14x-150y+186=0.

Page No 22.18:

Question 3:

A point moves so that the difference of its distances from (ae, 0) and (−ae, 0) is 2a. Prove that the equation to its locus is
x2a2-y2b2=1, where b2 = a2 (e2 − 1).

Answer:

Let A-ae,0 and Bae,0 be the given points. Let P(h, k) be a point such that PA-PB=2a.

h+ae2+k-02-h-ae2+k-02=2ah+ae2+k2=2a+h-ae2+k2

Squaring both sides, we get:

h2+a2e2+2aeh+k2=4a2+h2+a2e2-2aeh+k2+4ah-ae2+k2aeh=2a2-aeh+2ah-ae2+k2eh-a=h-ae2+k2    a0

Squaring both sides again, we get:

e2h2+a2-2aeh=h2+a2e2-2aeh+k2    e2h2+a2=h2+a2e2+k2 a2e2-1=h2e2-1-k2h2a2-k2a2e2-1=1

Hence, the locus of (h, k) is x2a2-y2b2=1, where b2=a2e2-1.

Page No 22.18:

Question 4:

Find the locus of a point such that the sum of its distances from (0, 2) and (0, −2) is 6.

Answer:

Let P(h, k) be a point. Let the given points be A0,2 and B0,-2.
According to the given condition,

AP + BP = 6

h-02+k-22+h-02+k+22=6

h2+k-22=6-h2+k+22

Squaring both sides, we get:

h2+k-22=36+h2+k+22-12h2+k+22

h2+k2+4-4k=36+h2+k2+4+4k-12h2+k+22

3h2+k+22=9+2k

9h2+k2+4+4k=81+4k2+36k        (Squaring both sides)

9h2+9k2+36+36k=81+4k2+36k

9h2+5k2-45=0

Hence, the locus of (h, k) is 9x2+5y2-45=0.

Page No 22.18:

Question 5:

Find the locus of a point which is equidistant from (1, 3) and the x-axis.

Answer:

Let P(h, k) be a point that is equidistant from A(1, 3) and the x-axis.

Now, the distance of the point P(h, k) from the x-axis is k.
AP = k

 AP2=k2

h-12+k-32=k2h2-2h+1+k2-6k+9=k2h2-2h-6k+10=0

Hence, the locus of (h, k) is x2-2x-6y+10=0.

Page No 22.18:

Question 6:

Find the locus of a point which moves such that its distance from the origin is three times its distance from the x-axis.

Answer:

Let P(h, k) be a point. Let O(0, 0) be the origin.
So, the distance of point P(h, k) from the x-axis is k.

OP=3k

OP2=3k2

h-02+k-02=3k2h2+k2=9k2h2=8k2

Hence, the locus of (h, k) is x2=8y2 .

Page No 22.18:

Question 7:

A (5, 3), B (3, −2) are two fixed points; find the equation to the locus of a point P which moves so that the area of the triangle PAB is 9 units.

Answer:

Let P(h, k) be a point. Let the given points be A(5, 3) and B(3, -2).

 Area of ABP=12x1y2-y3+x2y3-y1+x3y1-y29=125-2-k+3k-3+h3+25h-2k-19=185h-2k-19=18 or 5h-2k-19=-185h-2k-37=0 or 5h-2k-1=0

Hence, the locus of (h, k) is 5x-2y-37=0 or 5x-2y-1=0.

Page No 22.18:

Question 8:

Find the locus of a point such that the line segments with end points (2, 0) and (−2, 0) subtend a right angle at that point.

Answer:

Let the given points be A2,0 and B-2,0. Let P(h, k) be a point such that APB=90.

Thus, APB is a right angled triangle.

AB2=AP2+BP2

2+22+0=h-22+k2+h+22+k216=h2+4-4h+k2+h2+4+4h+k2h2+k2=4

Hence, the locus of (h, k) is x2+ y2 = 4.

Page No 22.18:

Question 9:

If A (−1, 1) and B (2, 3) are two fixed points, find the locus of a point P, so that the area of ∆PAB = 8 sq. units.

Answer:

Let the coordinates of P be (h, k).
Let the given points be A-1,1 and B2,3.

Area of PAB=12x1y2-y3+x2y3-y1+x3y1-y28×2=-13-k+2k-1+h1-316=-3+k+2k-2-2h16=2h-3k+52h-3k+5=16 or 2h-3k+5=-162h-3k-11=0 or 2h-3k+21=0

Hence, the locus of (h, k) is 2x-3y-11=0 or 2x-3y+21=0 .

Page No 22.18:

Question 10:

A rod of length l slides between two perpendicular lines. Find the locus of the point on the rod which divides it in the ratio 1 : 2.

Answer:

Let the two perpendicular lines be the coordinate axes. Let AB be a rod of length l and the coordinates of A and B be (a, 0) and (0, b) respectively.
As the rod AB slides, the values of a and b change. Let P(h, k) be a point on AB.



Here, BP:AP = 1:2 .

h=a+03, k=0+2b3a=3h, b=3k2          ... (1)

The length of the given rod is l.

AB=la2+b2=la2+b2=l2

Using equation (1), we get:

9h2+3k22=l2h2+k24=l29

Hence, the locus of (h, k) is x2+y24=l29 .

Page No 22.18:

Question 11:

Find the locus of the mid-point of the portion of the line x cos α + y sin α = p which is intercepted between the axes.

Answer:

The given line is xcosα+ysinα=p.
We need to find the intersection of the above line with the coordinate axes.
Let us put x = 0, and y = 0, respectively.
Thus,
at x = 0, 0+ysinα=py=pcosecα

at y = 0, xcosα+0=px=psecα

So, the points on the axes are Apsecα,0 and B0, pcosecα.

Let P(h, k) be the mid-point of the line AB.

h=psecα+02 and k=0+pcosecα2cosα=p2h and sinα=p2k

We know that sin2α+cos2α=1.

p2h2+p2k2=11h2+1k2=4p2

Hence, the locus of (h, k) is 1x2+1y2=4p2.

Page No 22.18:

Question 12:

If O is the origin and Q is a variable point on y2 = x, find the locus of the mid-point of OQ.

Answer:

Let the coordinates of Q be (a, b), which lies on the parabola y2=x.

b2=a                ... (1)

Let P(h, k) be the mid-point of OQ.

Now,

h=0+a2 and k=0+b2a=2h and b=2k

Putting a = 2h and b = 2k in equation (1), we get:

2k2=2h2k2=h

Hence, the locus of the mid-point of OQ is 2y2=x.



Page No 22.21:

Question 1:

What does the equation (xa)2 + (yb)2 = r2 become when the axes are transferred to parallel axes through the point (ac, b)?

Answer:

Substituting x=X+a-c, y=Y+b in the given equation, we get:
X+a-c-a2+Y+b-b2=r2X-c2+Y2=r2X2+Y2-2cX=r2-c2

Hence, the transformed equation is X2+Y2-2cX=r2-c2.

Page No 22.21:

Question 2:

What does the equation (ab) (x2 + y2) −2abx = 0 become if the origin is shifted to the point aba-b,0 without rotation?

Answer:

Substituting x=X+aba-b, y=Y+0 in the given equation, we get:

a-bX+aba-b2+Y2-2ab×X+aba-b=0a-bX2+a2b2a-b2+2abXa-b+Y2-2abX-2a2b2a-b=0a-bX2+Y2+a2b2a-b+2abX-2abX-2a2b2a-b=0a-bX2+Y2-a2b2a-b=0a-b2X2+Y2=a2b2

Hence, the transformed equation is a-b2X2+Y2=a2b2.

Page No 22.21:

Question 3:

Find what the following equations become when the origin is shifted to the point (1, 1).
(i) x2 + xy − 3xy + 2 = 0
(ii) x2y2 − 2x +2y = 0
(iii) xy xy + 1 = 0
(iv) xyy2x + y = 0

Answer:

(i) Substituting x=X+1, y=Y+1 in the given equation, we get:

X+12+X+1Y+1-3X+1-Y+1+2=0X2+2X+1+XY+X+Y+1-3X-3-Y-1+2=0X2+XY=0

Hence, the transformed equation is x2+xy=0.

(ii) Substituting x=X+1, y=Y+1 in the given equation, we get:

X+12-Y+12-2X+1+2Y+1=0X2+2X+1-Y2-2Y-1-2X-2+2Y+2=0X2-Y2=0

Hence, the transformed equation is x2-y2=0.

(iii) Substituting x=X+1, y=Y+1 in the given equation, we get:

X+1Y+1-X+1-Y+1+1=0XY+X+Y+1-X-1-Y-1+1XY=0

Hence, the transformed equation is xy = 0.

(iv) Substituting x=X+1, y=Y+1 in the given equation, we get:

X+1Y+1-Y+12-X+1+Y+1=0XY+X+Y+1-Y2-1-2Y-X-1+Y+1=0XY-Y2=0

Hence, the transformed equation is xy-y2=0.

Page No 22.21:

Question 4:

To what point should the origin be shifted so that the equation x2 + xy − 3x − y + 2 = 0 does not contain any first degree term and constant term?

Answer:

Let the origin be shifted to (h, k). Then, x = X + h and y = Y + k.

Substituting x = X + h and y = Y + k in the equation x2 + xy − 3x − y + 2 = 0, we get:

X+h2+X+hY+k-3X+h-Y+k+2=0X2+2hX+h2+XY+kX+hY+hk-3X-3h-Y-k+2=0X2+XY+X2h+k-3+Yh-1+h2+hk-3h-k+2=0

For this equation to be free from the first-degree terms and constant term, we must have

2h+k-3=0, h-1=0, h2+hk-3h-k+2=0 h=1, k=1, h2+hk-3k-h+2=0

Also, h =1 and k = 1 satisfy the equation h2+hk-3k-h+2=0.

Hence, the origin should be shifted to the point (1, 1).

Page No 22.21:

Question 5:

Verify that the area of the triangle with vertices (2, 3), (5, 7) and (− 3 − 1) remains invariant under the translation of axes when the origin is shifted to the point (−1, 3).

Answer:

Let A(2, 3), B(5, 7) and C(− 3 − 1) represent the vertices of the triangle.

 Area of ABC=12x1y2-y3+x2y3-y1+x3y1-y2                                =1227+1+5-1-3-33-7                                =1216-20+12                                =4

Since the origin is shifted to the point (−1, 3), the vertices of the ABC will be

A'2+1,3-3, B'5+1, 7-3, and C'-3+1, -1-3or A'3,0, B'6, 4, and C'-2, -4

Now, area of A'B'C' :

                                                   12x1y2-y3+x2y3-y1+x3y1-y2                                            =1234+4+6-4-0-20-4                                            =4

Hence, area of the triangle remains invariant.

Page No 22.21:

Question 6:

Find what the following equations become when the origin is shifted to the point (1, 1).
(i) x2 + xy − 3y2y + 2 = 0
(ii) xyy2x + y = 0
(iii) xyxy + 1 = 0
(iv) x2y2 − 2x + 2y = 0

Answer:

(i) The given equation is x2 + xy − 3y2y + 2 = 0.

Substituting x=X+1, y=Y+1 in the given equation, we get:

X+12+X+1Y+1-3Y+12-Y+1+2=0X2+1+2X+XY+X+Y+1-3Y2-3-6Y-Y-1+2=0X2+XY-3Y2+3X-6Y=0

Hence, the transformed equation is x2+xy-3y2+3x-6y=0.

(ii) The given equation is xyy2x + y = 0.

Substituting x=X+1, y=Y+1 in the given equation, we get:

X+1Y+1-Y+12-X+1+Y+1=0XY+X+Y+1-Y2-2Y-1-X-1+Y+1=0XY-Y2=0

Hence, the transformed equation is xy-y2=0.

(iii) The given equation is xyxy + 1 = 0.

Substituting x=X+1, y=Y+1 in the given equation, we get:

X+1Y+1-X+1-Y+1+1=0XY+X+Y+1-X-1-Y-1+1=0XY=0

Hence, the transformed equation is xy=0.

(iv) The given equation is x2y2 − 2x + 2y = 0.

Substituting x=X+1, y=Y+1 in the given equation, we get:

X+12-Y+12-2X+1+2Y+1=0X2+2X+1-Y2-2Y-1-2X-2+2Y+2=0X2-Y2=0

Hence, the transformed equation is x2-y2=0.

Page No 22.21:

Question 7:

Find the point to which the origin should be shifted after a translation of axes so that the following equations will have no first degree terms:
(i) y2 + x2 − 4x − 8y + 3 = 0
(ii) x2 + y2 − 5x + 2y − 5 = 0
(iii) x2 − 12x + 4 = 0

Answer:

Let the origin be shifted to (h, k). Then, x = X + h and y = Y + k.

(i) Substituting x = X + h and y = Y + k in the equation y2 + x2 − 4x − 8y + 3 = 0, we get:

Y+k2+X+h2-4X+h-8Y+k+3=0Y2+2kY+k2+X2+2hX+h2-4X-4h-8Y-8k+3=0X2+Y2+X2h-4+Y2k-8+k2+h2-4h-8k+3=0

For this equation to be free from the terms containing X and Y, we must have

2h-4=0,2k-8=0 h=2, k=4

Hence, the origin should be shifted to the point (2, 4).

(ii) Substituting x = X + h and y = Y + k in the equation x2 + y2 − 5x + 2y − 5 = 0, we get:

X+h2+Y+k2-5X+h+2Y+k-5=0X2+2hX+h2+Y2+2kY+k2-5X-5h+2Y+2k-5=0X2+Y2+X2h-5+Y2k+2+k2+h2-5h+2k-5=0

For this equation to be free from the terms containing X and Y, we must have

2h-5=0,2k+2=0 h=52, k=-1

Hence, the origin should be shifted to the point 52, -1.

(iii) Substituting x = X + h and y = Y + k in the equation x2 − 12x + 4 = 0, we get:

X+h2-12X+h+4=0X2+2hX+h2-12X-12h+4=0X2+X2h-12+h2-12h+4=0

For this equation to be free from the terms containing X and Y, we must have

2h-12  h=6

Hence, the origin should be shifted to the point 6, k, kR.

Page No 22.21:

Question 8:

Verify that the area of the triangle with vertices (4, 6), (7, 10) and (1, −2) remains invariant under the translation of axes when the origin is shifted to the point (−2, 1).

Answer:

Let the vertices of the given triangle be A(4, 6), B(7, 10) and C(1,− 2).

 Area of triangle ABC=12x1y2-y3+x2y3-y1+x3y1-y2 Area of triangle ABC=12410+2+7-2-6+16-10 Area of triangle ABC=1248-56-4=6

As the origin is shifted to the point (−2, 1), the vertices of the triangle ABC will be

A'4+2,6-1, B'7+2, 10-1 and C'1+2, -2-1or A'6, 5, B'9, 9 and C'3, -3

Now, area of triangle A'B'C':

12x1y2-y3+x2y3-y1+x3y1-y2=1269+3+9-3-5+35-9=6

So, in both the cases, the area of the triangle is 6 sq. units.

Hence, area of the triangle remains invariant.

Page No 22.21:

Question 1:

The vertices of a triangle are O (0, 0), A (a, 0) and B (0, b). Write the coordinates of its circumcentre.

Answer:

The coordinates of circumcentre of a triangle are the intersection of perpendicular bisectors of any two sides of the triangle.

Thus, the coordinates of circumcentre of triangle OAB are a2, b2, as shown in the figure.

Page No 22.21:

Question 2:

In Q.No. 1, write the distance between the circumcentre and orthocentre of ∆OAB.

Answer:

The coordinates of circumcentre of a triangle are the point of intersection of perpendicular bisectors of any two sides of the triangle.

Thus, the coordinates of the circumcentre of triangle OAB is a2, b2 ,as shown in the figure.
We know that the orthocentre of a triangle is the intersection of any two altitudes of the triangle.
So, the orthocentre of triangle OAB is the origin O(0, 0).

Distance between the circumcentre and orthocentre of ∆OAB = OC

OC=a2-02+b2-02=a2+b22

Page No 22.21:

Question 3:

Write the coordinates of the orthocentre of the triangle formed by points (8, 0), (4, 6) and (0, 0).

Answer:

The intersection point of three altitudes of a triangle is called orthocentre.

In the figure, two altitudes ON and BM of OAB are shown.

Slope of AB = 6-04-8=-32

Slope of ON  =23   Product of slopes=-1
Equation of ON:

y-0=23x-0y=23x              ... (1)

Equation of BM:

x = 4                ... (2)

On solving equations (1) and (2), we get 4, 83 as the coordinates of the orthocentre.

Page No 22.21:

Question 4:

Three vertices of a parallelogram, taken in order, are (−1, −6), (2, −5) and (7, 2). Write the coordinates of its fourth vertex.

Answer:

Let A-1,-6, B2,-5 and C7, 2 be the vertices of the parallelogram ABCD.
Let the coordinates of D be (x, y).

Since, diagonals of a parallelogram bisect each other,

-1+72=2+x2 and -6+22=-5+y2x=4 and y=1

Hence, the coordinates of the fourth vertex D are (4, 1).



Page No 22.22:

Question 5:

If the points (a, 0), (at12, 2at1) and (at22, 2at2) are collinear, write the value of t1 t2.

Answer:

For the points (a, 0), (at12, 2at1) and (at22, 2at2) to be collinear, the following condition has to be met:

a01at122at11at222at21=0a2at1-2at2-0+12a2t12t2-2a2t1t22=02a2t1-t2+2a2t1t2t1-t2=02a2t1-t21+t1t2=0

t1-t2=0 or 1+t1t2=0      a01+t1t2=0      t1t2t1t2=-1

Page No 22.22:

Question 6:

If the coordinates of the sides AB and AC of  ∆ABC are (3, 5) and (−3, −3), respectively, then write the length of side BC.

Answer:

Disclaimer: In the question it should have been the coordinates of the mid points of AB and AC are (3, 5) and (-3, -3)

Given: the coordinates of the midpoints of AB and AC are (3,5) and (-3, -3).
Let, D and E be the midpoints of AB and AC, respectively.

DE=3--32+5--32          =62+82          =100          =10 units

Now, as D and E are midpoints of AB and AC respectively,
by the mid-points theorem,

BC=2×DE      =2×10 units       =20 units

Page No 22.22:

Question 7:

Write the coordinates of the circumcentre of a triangle whose centroid and orthocentre are at (3, 3) and (−3, 5), respectively.

Answer:

Let the coordinates of the circumcentre of the triangle be C(x, y).
Let the points O(-3, 5) and G(3, 3) represent the coordinates of the orthocentre and centroid, respectively.

We know that the centroid of a triangle divides the line joining the orthocentre and circumcentre in the ratio 2:1.

3=-3×1+2x3 and 3=5×1+2y3x=6, y=2

Hence, the coordinates of the circumcentre is (6, 2).

Page No 22.22:

Question 8:

Write the coordinates of the in-centre of the triangle with vertices at (0, 0), (5, 0) and (0, 12).

Answer:

Let A(0,0), B(5, 0) and C(0, 12) be the vertices of the given triangle.
In-centre I of a triangle with vertices Ax1,y1, Bx2,y2 and Cx3,y3 is given by:

Iax1+bx2+cx3a+b+c, ay1+by2+cy3a+b+c, where a = BC, b = AC and c = AB.

Now,

a=BC=5-02+0-122=13b=AC=0+122=12c=AB=0+52=5

I13×0+12×5+5×013+12+5, 13×0+12×0+5×1213+12+5I6030, 6030=2, 2

Hence, the coordinates of the in-centre of the triangle with vertices at (0, 0), (5, 0) and (0, 12) is (2, 2).

Page No 22.22:

Question 9:

If the points (1, −1), (2, −1) and (4, −3) are the mid-points of the sides of a triangle, then write the coordinates of its centroid.

Answer:

Let P(1, −1), Q(2, −1) and R(4, −3) be the mid-points of the sides AB, BC and CA, respectively, of ABC.
Let Ax1,y1, Bx2,y2 and Cx3,y3 be the vertices of ABC.
Since, P is the mid-point of AB,

x1+x22=1,  y1+y22=-1              ... (1)

Q is the mid-point of BC.

x2+x32=2,  y2+y32=-1              ... (2)

R is the mid-point of AC.

x1+x32=4,  y1+y32=-3              ... (3)

Adding equations (1), (2) and (3), we get:

x1+x2+x3=1+2+4=7y1+y2+y3=-1-1-3=-5

Centroid of ABC=x1+x2+x33, y1+y2+y33=73, -53

Hence, the coordinates of the centroid of the triangle is 73, -53.

Page No 22.22:

Question 10:

  Write the area of the triangle with vertices at (a, b + c), (b, c + a) and (c, a + b).

Answer:

Let A(a, b + c), B(b, c + a) and C(c, a + b) be the vertices of the the given triangle.

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

Hence, area of the triangle with vertices at (a, b + c), (b, c + a) and (c, a + b) is 0.



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