Unit-4:Co-ordinate geometry
Q1.
Find the coordinates of the vertices of an equilateral triangle of side 4a as shown in figure (OA = 4a)
(a) (3√3a,a) (b) (a√3,a) (c) (a,2a) (d) (2a,2√3a)
Solution.
In triangle OMB
Using Pythagoras theorem
OB2 = OM2 + BM2
BM2 = OB2
- OM2
= (4a)2 - (2a)2
BM2 = 16a2 - 4a2 = 12a2
BM = 2√3a
Co-ordinate of B = (2a,2√3a)
Solution
Q2.
If three points to be collinear then the sum of the distances between two pairs of points is equal to the third pair of points .
(a) Yes (b) No (c) Can't say
Solution.
(a) Yes
Solution
Q3.
Point (-8,0) lie on which axis ?
(a) x-axis (b) y-axis (c) both (d) none of these
Solution.
(a) x-axis
Solution
Q4.
Find a point on x-axis which is equi-distance from A(2,-5) and B(-2,9).
(a) (5,-6) (b) (-7,0) (c) (-2,8) (d) (-2,0)
Solution.
We know that a point on x-axis is of the form R(x,0)
So, RA = RB
√(x - 2)2 + (0 + 5)2 =
√(x + 2)2 + (0 - 9)2
(x - 2)2 + 25 = (x + 2)2 + 81
x2 + 4 - 4x + 25 = x2 + 4 + 4x + 81
-8x = 56
x = -7
So, point is R (x,0) = R(-7,0)
Solution
Q5.
Which type of triangle is formed by the points A(3,2) B(-2,-3) C(2,3)
(a) obtuse triangle (b) right triangle
(c) acute triangle (d) none of these
Solution.
(b) right triangle
AB = √50
BC = √52
AC = √2
AB2 + AC2 = BC 2
Solution
Q6.
Intersection point of all three medians of an triangle is known as __________.
(a) In-centre (b) Orthocentre
(c) Centroid (d) None of these
Solution.
(c) Centroid
Solution
Q7.
Determine the ratio in which the line 3x + y - 9 = 0 divides the segment. Joining the points (1,3) (2,7)
(a) 4:3 (b) 1:2 (c) 2:3 (d) 3:4
Solution.
(d) 3:4
The line 3x + y - 9 = 0 divides the line segment joining A(1,3) and B(2,7) in the ratio k:1 at point C,
Co-ordinate of C are
[(2k + 1)/(k + 1) , [(7k + 3)/(k + 1)]
and C lies on line 3x + y - 9 = 0
3[(2k + 1)/(k + 1)] + [(7k + 1)/(k + 1)] - 9 = 0
k = 3/4
k:1 = 3/4 : 1 = 3:4
Solution
Q8.
The diagonals of a rhombus are perpendicular to each other
(a) Yes (b) No (c) Can't say
Solution.
(a) Yes
Solution
Q9.
Two vertices of a triangle are (3,-5) and (-7,4). If its centriod is (3,-2) find the third vertex.
(a) (13,2)
(b) (-5,12) (c) (-2,1) (d) (13,-5)
Solution.
(d) (13,-5)
Let the co-ordinate of third vertex is (x,y)
x + 3 - 7 /3
= 3
x - 4 /3
= 3
x - 4 = 9
x = 13
y - 5 + 4 /
3
= -2
y -1 = -6
y = -5
Solution
Q10.
The mid-point of the hypotenuse of a right angled triangle is equidistance from its vertices .
(a) No (b) Yes (c) Can't say
Solution.
(b) Yes
Solution
