Find the equation of circumcentre of the triangle whose vertices are (5 7 6 6 and 2)2

I worked out the general case here .

I'm going to ignore the details of the general answer, and state the main result as: The squared radius of the circumcircle equals the product of the squared sides of the triangle divided by sixteen times the squared area of the triangle. Given triangle with sides #a,b,c# and area #A#, the circumradius #r# satisfies

# r^2 = {a^2b^2c^2}/{16A^2} #

It's tempting to take the square root, but experience shows much smoother sailing if we don't. We can get the area from the coordinates using the Shoelace Theorem, or get #16A^2# directly from the squared sides using Archimedes' Theorem:

# 16A^2 = 4a^2b^2 - (c^2-a^2-b^2)^2#

Since we need the squared sides for the numerator anyway, let's do it this way. We'll label the vertices #A(5,7),## B(2,1),## C(1,6) # and calculate

# a^2 = BC^2 = (2-1)^2+(1-6)^2=1^2+5^2=26#

#b^2 = AC^2 = 4^2+1^2=17#

#c^2 = AB^2 = 3^2+6^2=45#

#16 A^2 = 4(26)(17)-(45-26-17)^2 = 1764#

# r^2 = {a^2b^2c^2}/{16A^2} = { (26)(17)(45) }/ 1764 = 1105/98 #

Circumcircle area # = pi r^2 = {1105 pi}/98#

The equations in the above paragraph may look scary, but you don't need to worry, it's not that difficult! Let's check how to find the orthocenter with an example, where our triangle ABC has the vertex coordinates: A = (1, 1), B = (3, 5), C = (7, 2).

1. Find the slope:

AB side slope = (5 - 1) / (3 - 1) = 2

1. Calculate the slope of the perpendicular line:

perpendicular slope to AB side = - 1/2

1. Find the line equation:

y - 2 = - 1/2 * (x - 7) so y = 5.5 - 0.5 * x

1. Repeat for another side, e.g., BC;

BC side slope = (2 - 5) / (7 - 3) = - 3/4

perpendicular slope to BC side = 4/3

y - 1 = 4/3 * (x - 1) so y = -1/3 + 4/3 * x

1. Solve the system of linear equations:

y = 5.5 - 0.5 * x and
y = -1/3 + 4/3 * x

so

5.5 - 0.5 * x = -1/3 + 4/3 * x

35/6 = x * 11/6

x = 35/11 ≈ 3.182.

Substituting x into either equation will give us:

y = 43/11 ≈ 3.909

Of course, you'll obtain the same result from our orthocenter calculator💪! Just type the three triangle vertices and we'll calculate the orthocenter coordinates for you.

Rd Sharma Xi 2020 2021 _volume 2 Solutions for Class 11 Commerce Maths Chapter 22 Brief Review Of Cartesian System Of Rectangular Co Ordinates are provided here with simple step-by-step explanations. These solutions for Brief Review Of Cartesian System Of Rectangular Co Ordinates are extremely popular among Class 11 Commerce students for Maths Brief Review Of Cartesian System Of Rectangular Co Ordinates Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rd Sharma Xi 2020 2021 _volume 2 Book of Class 11 Commerce Maths Chapter 22 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rd Sharma Xi 2020 2021 _volume 2 Solutions. All Rd Sharma Xi 2020 2021 _volume 2 Solutions for class Class 11 Commerce Maths 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:

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-b22a c

⇒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+x3 y1-y2
∴Area of ∆DBC=12-3-2-3x+4 3x-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+0-1 2=50=52

BC=11-92+6-12=29

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

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

∵AB≠BC≠CD≠AD, 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, B x2,y2 and Cx3,y3 are ax1+bx2+cx3 a+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.

⇒a2+x2=2a2       ∵BC=2a⇒x2 =3a2⇒x=±3a

So, the vertices of the triangle are 0,-a, 0,a and 3 a,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-x1 2+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-x 2+4-02⇒49+x2-14x+36=9+x2-6x+16⇒85-14x=25 -6x⇒60=8x⇒152=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-42⇒h2-4h+4+k2-8k+16=h2 ⇒k2-4h-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=54 ⇒h-22+k-02h-12+k-32=54

Squaring both sides, we get:

 16h2-4h+4+k2=25h2-2h+1+k2-6k+9 ⇒9h2+9k2+64h-50h-150k-64+250=0⇒9h2+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=2 a⇒h+ae2+k2=2a+h-ae2+k2

Squaring both sides, we get:

h2+a2e2+2aeh+k2=4a2+h2+a2e2-2aeh+k2 +4ah-ae2+k2⇒aeh=2a2-aeh+2ah-ae2+ k2⇒eh-a=h-ae2+k2    ∵a≠0

Squaring both sides again, we get:

e2h2+a2-2aeh=h2+a2e2-2aeh+k2    ⇒e2h2+a2=h2+a2e2+k2 ⇒a2e2-1= h2e2-1-k2⇒h2a2-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-h 2+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+4k 2+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=k2⇒ h2-2h+1+k2-6k+9=k2⇒h2-2h-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=3k2⇒h2+k2=9k2 ⇒h2=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=1 2x1y2-y3+x2y3-y1+x3y1-y2 ⇒9=125-2-k+3k-3+h3+2⇒5h-2k-19=18⇒5h-2k-19=18 or 5h-2k-19=-18 ⇒5h-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.

∴A B2=AP2+BP2

∴2+22+0=h-22+k2+h+ 22+k2⇒16=h2+4-4h+k2+h2+4+4h+k2⇒h2 +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=12 x1y2-y3+x2y3-y1+x3y1-y2 ⇒8×2=-13-k+2k-1+h2-3⇒16=- 3+k+2k-2-2h⇒16=2h-3k+5⇒2h-3k+5=16 or  2h-3k+5=-16⇒2h-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+2b3⇒a=3h, b=3k2          ... (1)

The length of the given rod is l.

∴AB=l⇒a2+ b2=l⇒a2+b2=l2

Using equation (1), we get:

⇒9h2+3k22= l2⇒h2+k24=l29

Hence, the locus of (h, k) is x2+y24=l2 9 .

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=p⇒y=pcosecα

at y = 0, xcosα+0= p⇒x=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+pcosecα2⇒cosα=p2h  and sinα=p2k

We know that sin2α+cos2α=1.

∴p2h2+ p2k2=1⇒1h2+1k2=4p2

Hence, the locus of (h, k) is 1x 2+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+b2⇒a=2h and b=2k

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

2k2=2h⇒2k2=h

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

Page No 22.21:

Question 1:

What does the equation (x − a)2 + (y − b)2 = r2 become when the axes are transferred to parallel axes through the point (a − c, b)?

Answer:

Substituting x=X+a-c,  y=Y+b in the given equation, we get:
X+a-c-a2+Y+b-b2=r2⇒X -c2+Y2=r2⇒X2+Y2-2cX=r2-c2

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

Page No 22.21:

Question 2:

What does the equation (a − b) (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+a ba-b=0⇒a-bX2+a2b2a-b2+2ab Xa-b+Y2-2abX-2a2b2a-b=0⇒a-bX2 +Y2+a2b2a-b+2abX-2abX-2a2b2a-b=0 ⇒a-bX2+Y2-a2b2a-b=0⇒a-b 2X2+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 − 3x − y + 2 = 0
(ii) x2 − y2 − 2x +2y = 0
(iii) xy − x − y + 1 = 0
(iv) xy − y2 − x + y = 0

Answer:

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

X+12+X+1 Y+1-3X+1-Y+1+2=0⇒X2+2X+1+XY+X+Y+1 -3X-3-Y-1+2=0⇒X2+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=0⇒X2+2X+1-Y2-2Y-1-2X-2+2Y+2=0 ⇒X2-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=0⇒XY+ X+Y+1-X-1-Y-1+1⇒XY=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=0⇒XY+X+Y+1-Y2-1-2Y-X-1+Y+1=0⇒XY-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+h Y+k-3X+h-Y+k+2=0⇒X2+2hX+h2+XY+kX+ hY+hk-3X-3h-Y-k+2=0⇒X2+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=12x1 y2-y3+x2y3-y1+x3y1-y2                                 =12 27+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 − 3y2 − y + 2 = 0
(ii) xy − y2 − x + y = 0
(iii) xy − x − y + 1 = 0
(iv) x2 − y2 − 2x + 2y = 0

Answer:

(i) The given equation is x2 + xy − 3y2 − y + 2 = 0.

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

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

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

(ii) The given equation is xy − y2 − x + y = 0.

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

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

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

(iii) The given equation is xy − x − y + 1 = 0.

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

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

Hence, the transformed equation is xy=0.

(iv) The given equation is x2 − y2 − 2x + 2y = 0.

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

X+12-Y+12- 2X+1+2Y+1=0⇒X2+2X+1-Y2-2Y-1-2X-2+2Y+2 =0⇒X2-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+h2-4 X+h-8Y+k+3=0⇒Y2+2kY+k2+X2+2hX+h2-4X- 4h-8Y-8k+3=0⇒X2+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=0⇒X2+2hX+h2+Y2+ 2kY+k2-5X-5h+2Y+2k-5=0⇒X2+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=0⇒X2+2hX+ h2-12X-12h+4=0⇒X2+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, k∈R.

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:

If the points (a cosα, a sinα) and (a cosβ, a sinβ) are at a distance k sinα-β2 apart, then k  = __________.

Answer:

Given (a cos α, a sin α) and (a cos β, a sin β) are at a distance k sin α-β2 a part
Using distance formula,

 (a cosβ- a cosα)2+(a sinβ-a  sinα)2 =a2cosβ-cosα2+a2sinβ-sinα2=a cos2β+cos2α-2 cosα cosβ+sin2α+sin2β-2 sinα sinβ=a1+1 -2 cosα cosβ-2 sinα sinβ=a21-cosα cosβ-sinα sinβ=2a 1-cosα-β=2 a2 sin2α-β2=  2a sinα-β2i.e K=2a

Page No 22.21:

Question 2:

If two vertices of a triangle are (6, 4), (2, 6) and its centroid is (4, 6), then the coordinates of its third vertex are _________.

Answer:

Let us suppose the co-ordinates of third vertex is given by (x, y)
then centroid = sum of respective co-ordinates3
i.e for (6, 4), (2, 6) and (x, y) centroid is (4, 6)

i.e 4=6+2+x3
i.e 12 = 8 + x
i.e x = 4 and 6 = 4+6+y3
i.e 18 = 10 + y
i.e y = 8
∴ Co-ordinate of third vertex is (4, 8)

Page No 22.21:

Question 3:

The coordinates of the orthocentre of the triangle with vertices 2, 3-12, 12, - 12 and 2, -12 are ________.

Answer:

Given vertices are
 2, 3-12 , 12, -12 and 2, -12

Page No 22.22:

Question 4:

The coordinates of the in centre of the triangle whose vertices are (0, 0), (4, 0) and (0, 3) are __________.

Answer:

In a triangle with vertices  (0, 0), (4, 0) and (0, 3)

Page No 22.22:

Question 5:

A(–3, 0), B(4,–1) and C(5, 2) are vertices of ΔABC. The length of altitude through A  is __________.

Answer:

In ΔABC, Let D be such that AD defines altitude of ΔABC.
A = (–3, 0), B = (4, –1) and C = (5, 2)

Here Base=  BC= 5-42+2+12                            = 1+9= 10⇒11= 12×10× ADi.e  2210is the length of altitude AD.

Page No 22.22:

Question 6:

A(a, 1), B(b, 3) and C(4, c) are the vertices of ΔABC. If its centroid lies on x–axis, then __________.

Answer:

In ΔABC,
A = (a, 1) 
B = (b, 3)          
C = (4, c) 
Since centroid lies an x-axis
⇒ y-cordinate of centroid is 0

i.e  0 = 1+3+c3   (since for A(x1, y1)  B(x2, y2)  C(x3, y3)
i.e  c = –4             (centroid is given by x1+x2+x33 , y1+y2+y33

Page No 22.22:

Question 7:

The distance between the circumcentre and centroid of a triangle whose vertices are (6, 0), (0, 6) and (6, 6), is _______.

Answer:

In ∆ABC
A = (6, 0)
B = (0, 6)
C = (6, 6)

Page No 22.22:

Question 8:

The distance between the orthocentre and circumcentre of the triangle whose vertices are at -12, 32, -12, - 32 and (1, 0),
 is __________.

Answer:

In ∆ABC
A=-12 , 32    B = -12 , -32   C= 1,0AB= 3 2+322+-12+122i.e AB= 3 2      = 3     BC= 322+32 2i.e BC= 34+94  =124= 3 and AC = 322+1+122= 3

Since AB = BC = AC 
i.e ∆ABC is an equilateral triangle 
⇒ Orthocentre and circumcentre of  ∆ABC coincide 
⇒  Distance between orthocentre and circumcentre of  ∆ABC is 0.

Page No 22.22:

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.

Page No 22.22:

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.

∴ Distance between the circumcentre and orthocentre of ∆OAB = OC

⇒OC=a2-0 2+b2-02=a2+b22

Page No 22.22:

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.

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 22.22:

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+2 2=-5+y2⇒x=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=0 ⇒a2at1-2at2-0+12a2t12t2-2a2t1t22 =0⇒2a2t1-t2+2a2t1t2t1-t2=0⇒2a2t1-t21+t1t2=0

⇒t1-t2 =0 or 1+t1t2=0      a≠0⇒1+t1t2= 0      ∵t1≠t2⇒t1t2=-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)

∴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+2y3⇒x=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:

I≡ax1+ 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

∴I≡13×0+12×5+5×013+ 12+5, 13×0+12×0+5×1213+12+5⇒I≡6030, 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+x3 2=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=7 3, -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+x3y 1-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.

View NCERT Solutions for all chapters of Class 13

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