# Free Geometry Set I - 03 Practice Test - CAT

The coordinates of circumcentre of the triangle with vertices (2,3), (4, - 1) and (4,3) are

A. (2, 3)
B. (1,3)
C. (3,1)
D. (3,2)

#### SOLUTION

Solution : C

1st method: - In a right angled triangle

The coordinates of circumcentre = (3,1) 2nd method: - If O(x1, y1) is the circumcentre, then OA2=OB2.
(x12)2+(Y13)2=(X14)2+(Y1+1)2

4X16Y1+8X12Y1=1713=4
i.e., 4X18Y1=4 or X12Y1=1 ..................(I)
OB2=OC2
X14)2+(Y1+1)2=(X14)2+(Y13)2
2Y1+6Y1=8
8Y1=8 ....................(II)
Y1=1
From (i) X1=3
X1,Y1)=(3,1) Find the area of a triangle whose vertices are (0,0), (12, 9) and (14, 6).

A. 27
B. 54
C. 84
D. 168

#### SOLUTION

Solution : A

The simplest way to find this is to put a rectangle around the given triangle and then subtract off the areas of the right triangles that surround it. (See Figure) The area of the entire rectangle is 126, but after subtracting the areas of the 3 right triangles, whose areas are 54, 3, and 42, we are left with an area of 27. Two opposite sides of a square are given by the lines 5x + 12y + 18 = 0 and 5x + 12y – 21 = 0. Find the length of diagonal of the square.

A. 32
B. 62
C. 23
D. 43

#### SOLUTION

Solution : A Given lines are 5x + 12y + 18 = 0……………….(i) and 5x + 12y – 21 = 0………………….(ii)

Length of the side = distance between lines (i) and (ii) is given by

D = C2C1a2+b2=211825+144=3913=3

Length of the diagonal = 32; Option(a).

Find the area of quadrilateral formed by joining the points (- 4, 2), (- 1, 0), (4, 1) and (2, 5):

A.

20

B.

40

C.

36

D.

21 .5

#### SOLUTION

Solution : D

1st method: - The simplest way to find this is to put a rectangle around the given quadrilateral and then subtract off the areas of the right triangles that surround it. (See Figure) The area of the entire rectangle is 40, but after subtracting the areas of the 4 right triangles, whose areas are 52, 4, 9 and 3. we are left with an area of 21.5. 2nd method: - Area of quadrilateral = 2(x1y2x2y1)+(x2y3x3y2)+(x3y4x4y3)+(x4y1x1y4)

=12(4×0(1)×2)+(1×14×0)+(4×52×1)+(2×2(4)×5)=21.5

The lines 2x + 2y + 1 = 0 and 8x – 2y + 9 = 0 intersect in the

#### SOLUTION

Solution : B

2x + 2y + 1 = 0 …………….(i)

8x – 2y + 9 = 0 …………………………(ii)

After solving equation (i) and (ii)

x = - 1 and y = 12

(1,12); which is in 2nd quadrant.

Find the slope of the line joining the points (7, 5) and (9, 7):

A. 1
B. 2
C. 3
D. 4

#### SOLUTION

Solution : A

Option(a)

Slope of the line = y2y1x2x1=7597=1

Here (x1,y1) = (7, 5) and (x2,y2) = (9, 7)

Which of the following points is not on the line y = 5x + 3?

A. (12,112)
B. (13,143)
C. (8,3+102)
D. (2,313)

#### SOLUTION

Solution : D

Even without calculations it is clear that (2,313) cannot lie on the line y = 5x + 3. If x is irrational, y must also be irrational. Option(d)

Three points A, B and C are collinear such that AB = 2BC. If the coordinates of the points A and B are (1, 7) and (6, -3) respectively, then the coordinates of the point C can be

A. (72;2)
B.  (172;8)
C. (172;8)
D. (172;2)

#### SOLUTION

Solution : B

Given, AB = 2BC

AB : BC = 2 : 1

Let C BE (h, k), then

(2h±12±1;2k±72±1) = (6, - 3) [B may divide AC internally or externally]

2h + 1 = 18 and 2k + 7 = - 9

(or) 2h - 1 = 6 , 2k - 7 = - 3

2h = 17 and 2k = - 16 (or) 2h = 7, 2k = 4

h = 172 and k = - 8 (or) h = 72; k = 2

C = (172;8) or (172;2)

Option(b)

The circumradius of a triangle with its vertices at (1, -2), (-3, 0) and (5, 6) is

A. 5
B. 10
C. 5
D. 102

#### SOLUTION

Solution : A

A(1, - 2) ,B(- 3, 0) and C(5, 6)

AB = 20, BC = 100 , AC = 80

BC2=AB2+AC2

ABC is a right angled triangle with the right angle at A.

Circumradius is half the length of the hypotenuse.

R = 12× BC, BC = 5. Option(a). Find the area of pentagon formed by joining the points (4, 4), (2, 2), (6, 2), (2, - 3) and (6, -3):

A. 24
B. 20
C. 16
D. 12

#### SOLUTION

Solution : A Area of pentagon = Area of triangle ABE + Area of rectangle BCDE

12×4×2+5×4=24

Determine the radius of the circle, two of whose tangents are the lines 2x+3y-9 = 0  and 4x+6y+19 =0

A. 13437
B.  35413
C.  11413
D.  37413

#### SOLUTION

Solution : D

The given tangents are

2x+3y-9 = 0

4x+6y+19 =0

Or 2x + 3y + 192 = 0

Which are parallel, than distance between parallel tangents must be diameter of the circle than diameter.
192(9)22+32=37213

Radius = 12 diameter = 37413

What is the equation of a line which has an x-intercept of 4 units and passes through the point (-2,3)?

A. x+4y-4=0
B. x+2y-4=0
C. x+6y-16=0
D. none of these

#### SOLUTION

Solution : B

option (b)

To have an x-intercept 4, the value at y=0 should give x=4.

Put y=0 in all 3 answer options.

Option a and b give x=4

Find the equation of the straight line passing through the point (2, 1) and through the point of intersection of the lines x + 2y = 3 and 2x – 3y= 4.

A. 5x + 3y – 13 = 0
B. 4x – 7y – 1= 0
C. 2x – 7y – 20=0
D. x – 7y + 13 =0

#### SOLUTION

Solution : A

1st method: - equation of any straight line passing through the intersection of the lines x + 2y = 3 and 2x – 3y= 4 is

λ(x + 2y – 3) + (2x – 3y – 4) = 0

Since it passes through the point (2, 1)

λ(2 + 2 – 3) + (4 – 3 – 4) = 0

λ - 3 = 0

λ = 3

Now substituting this value of λ  in (i), we get

3(x + 2y – 3) + (2x – 3y – 4) = 0

5x + 3y – 13 = 0

2nd method: - The straight line passing through the point (2, 1), put x = 2 and y = 1only option (a) and (b) will satisfy. Now, intersection point of the lines x + 2y = 3 and 2x – 3y= 4 is (177,27), Put x = 177 and y = 27 only option(a) will satisfy. Option(a).

Find the equation of circle whose radius is 5 and which touches the circle x2+y22x4y20=0 at the point (5, 5).

A. x2+y218x16y+120=0
B. x2+y2+6x3y54=0;
C.  x2+y218x17y+50=0
D. x2+y218x8y+60=0

#### SOLUTION

Solution : A

The given circle is x2+y22x4y20=0 with centre (1, 2) and radius = 1+4+20=5.

And required circle has radius 5 hence circles touch each other externally.

Since point of contact is P(5, 5).

P is the mid point of C1 and C2, let co – ordinate of centre C2 is (h, k) then

1+h2=5;h=9

2+k2=5; k = 8 And, Equation of required circle is (x9)2+(y8)2=52

x2+y218x16y+120=0; Option(a)

Find the co-ordinates of the foot of the perpendicular from (2, 4) upon the line x + y = 4

A. (1, 3)
B. (3, - 1)
C. (2, 1)
D. (4, 0)

#### SOLUTION

Solution : A From the given figure, we can easily determine that the co-ordinates of the foot of the perpendicular from (2, 4) upon the line x + y = 4 are (1, 3)