Q 1 :

Let O(0, 0), P(3, 4), Q(6, 0) be the vertices of the triangle OPQ. The point R inside the triangle OPQ is such that the triangles OPR, PQR, OQR are of equal area. The coordinates of R are                [2007]

  • (43,3)

     

  • (3,23)

     

  • (3,43)

     

  • (43,23)

     

(3)

  Ar(OPR)=Ar(PQR)=Ar(OQR)

[IMAGE 287]

 By simply geometry, R should be the centroid of PQO

co-ordinate of R=(3+6+03,4+0+03)=(3,43)



Q 2 :

Area of the triangle formed by the line x+y=3 and angle bisectors of the pair of straight lines x2-y2+2y=1 is                [2004]

  • 2 sq. units

     

  • 4 sq. units

     

  • 6 sq. units

     

  • 8 sq. units

     

(1)

x2-y2+2y=1   x=±(y-1)

[IMAGE 288]

 Area between x=0, y=1 and x+y=3 is the shaded region shown in figure.

 Area=12×2×2=2 sq. units



Q 3 :

Orthocentre of triangle with vertices (0, 0), (3, 4) and (4, 0) is                       [2003]

  • (3,54)

     

  • (3, 12)

     

  • (3,34)

     

  • (3, 9)

     

(3)

We know that point of intersection of altitudes of a triangle is the orthocentre of the triangle.

[IMAGE 289]

Equation of altitude AD i.e., line parallel to y-axis through (3,4) is

x=3                       (i)

Now, equation of OEAB is

y=-3-44-0x  y=x4                   (ii)

Solving (i) and (ii), we get orthocentre as (3,34)



Q 4 :

The number of integral points (integral point means both the coordinates should be integer) exactly in the interior of the triangle with vertices (0, 0), (0, 21) and (21, 0) is    [2003]

  • 133

     

  • 190

     

  • 233

     

  • 105

     

(2)

Total number of points within the square OACB=20×20=400

[IMAGE 290]

Points on line AB=20{(1,20),(2,19),(3,18),,(10,11),(11,10),,(20,1)}

 Points within OBC and ABC=400-20=380

By symmetry, points within OAB=3802=190



Q 5 :

A straight line through the origin O meets the parallel lines 4x+2y=9 and 2x+y+6=0 at points P and Q respectively. Then the point O divides the segment PQ in the ratio                  [2002]

  • 1 : 2

     

  • 3 : 4

     

  • 2 : 1

     

  • 4 : 3

     

(2)

The given lines are

         2x+y=92                                               (i)

and  2x+y=-6                                                 (ii)

Signs of constants on R.H.S. show that two lines lie on opposite sides of origin. Let a line through origin meets these lines in P and Q respectively. Then required ratio is OP:OQ.

[IMAGE 291]

In OPA and OQC,

    POA=QOC  (vertically opposite angles)

    PAO=OCQ  (alternate interior angles)

 OPA~OQC  (AA similarity)

 OPOQ=OAOC=943=34

 Required ratio=3:4



Q 6 :

Area of the parallelogram formed by the lines y=mx, y=mx+1, y=nx and y=nx+1 equals                     [2001]

  • |m+n|(m-n)2

     

  • 2|m+n|

     

  • 1|m+n|

     

  • 1|m-n|

     

(4)

[IMAGE 292]

The vertices, O(0,0), A(1m-n,mm-n), B(0,1)

Area (parallelogram OABC)=2×area (OAB)

=2×12|[0(mm-n-1)+1m-n(1-0)+0(0-mm-n)]|

=1|m-n|



Q 7 :

The number of integer values of m, for which the x-coordinate of the point of intersection of the lines 3x+4y=9 and y=mx+1 is also an integer, is           [2001]

  • 2

     

  • 0

     

  • 4

     

  • 1

     

(1)

3x+4y=9 and y=mx+1 are two lines.

On equating the value of y from both equations to get the x-coordinate of the point of intersection,

     3x+4(mx+1)=9(3+4m)x=5

x=53+4m

For x to be an integer, 3+4m should be a divisor of 5 i.e., 1,-1,5  or -5.

3+4m=1m=-12   (not an integer)

3+4m=-1m=-1 (integer)

3+4m=5m=12 (not an integer)

3+4m=-5m=-2 (integer)

 There are 2 integral values of m.



Q 8 :

The incentre of the triangle with vertices (1,3), (0,0) and (2, 0) is                            [2000]

  • (1, 32)

     

  • (23, 13)

     

  • (23, 32)

     

  • (1, 13)

     

(4)

Let (1,3), (0,0) and (2,0) are the coordinates of vertices A, O, B of ABC.

AO = OB = AB. So, it is an equilateral triangle and the incentre coincides with centroid.

Incentre =(0+1+23,0+0+33)=(1,13)