COORDINATE GEOMETRY Test
COORDINATE GEOMETRY
This is COORDINATE GEOMETRY Test-04 for CBSE class 10 Maths.. There are 15 questions in this test with each question having around four answer choices.
Questions & Answers
1
Three consecutive vertices of a parallelogram ABCD are A(1, 2), B(1, 0) and C(4, 0). The co – ordinates of the fourth vertex D are
- A(4, 2)Correct
- B( – 4, 2)
- C(4, – 2)
- D( – 4, – 2)
2
The length of the median through A of \(\) \(\Delta ABC\) with vertices A(7, – 3), B(5, 3) and C(3, – 1) is
- A5 unitsCorrect
- B3 units
- C25 units
- D7 units
3
If the line segment joining the points A\(({x_1},{y_1})\) and B\(({x_2},{y_2})\) is divided by a point P in the ratio 1 : k internally, then the co – ordinates of the point P are
- A\(\left( {\frac{{{x_2} - k{x_1}}}{{1 + k}},\frac{{{y_2} - k{y_1}}}{{1 + k}}} \right)\)
- B\(\left( {\frac{{{x_1} + k{x_2}}}{{1 + k}},\frac{{{y_1} + k{y_2}}}{{1 + k}}} \right)\)
- C\(\left( {\frac{{{x_2} + k{x_1}}}{{1 + k}},\frac{{{y_2} + k{y_1}}}{{1 + k}}} \right)\)Correct
- D\(\left( {\frac{{{x_2} + k{x_1}}}{{1 - k}},\frac{{{y_2} + k{y_1}}}{{1 - k}}} \right)\)
4
The base of an equilateral triangle ABC lies on the y – axis. The co – ordinates of the point C is (0, – 3). If origin is the midpoint of BC, then the co – ordinates of B are
- A(0, 3)Correct
- B( – 3, 0)
- C(0, – 3)
- D(3, 0)
5
The point where the perpendicular bisector of the line segment joining the points A(2, 5) and B(4, 7) cuts is:
- A(6, 3)
- B(3, 6)Correct
- C(0, 0)
- D(2, 5)
6
If the point P(2, 1) lies on the line segment joining points A(4, 2) and B(8, 4), then
- A\(AP = \frac{1}{4}AB\)
- B\(AP = \frac{1}{2}AB\)Correct
- CAP = PB
- D\(AP = \frac{1}{3}AB\)
7
The centroid of a triangle divides the median in the ratio
- A3 : 1
- Bnone of these
- C2 : 1Correct
- D1 : 3
8
If the mid – point of the line segment joining the points (a, b – 2) and ( – 2, 4) is (2, – 3), then the values of ‘a’ and ‘b’ are
- A6, – 8Correct
- B6, 8
- C4, – 5
- D– 6, 8
9
If the points (2, 3), (4, k) and (6, – 3) are collinear, then the value of ‘k’ is
- A1
- B3
- C4
- D0Correct
10
If the points (x, y), (1, 2) and (7, 0) are collinear, then the relation between ‘x’ and ‘y’ is given by
- A3x + y + 7 = 0
- Bx + 3y – 7 = 0Correct
- Cx – 3y + 7 = 0
- D3x – y – 7 = 0
11
If the vertices of a triangle are (1, 1), ( – 2, 7) and (3, – 3), then its area is
- A12 sq. units
- B2 sq. units
- C24 sq. units
- D0 sq. unitsCorrect
12
The area of the triangle with vertices (a, b+c), (b, c+a) and (c, a+b) is
- A\({a^2} + {\text{ }}{b^2} + {\text{ }}{c^2}\)
- B\({\left( {a{\text{ }} + {\text{ }}b{\text{ }} + {\text{ }}c} \right)^2}\)
- Ca + b + c
- D0Correct
13
If points (a, 0), (0, b) and (1, 1) are collinear, then \(\frac{1}{a} + \frac{1}{b}\) is
- A1Correct
- B0
- C– 1
- Dnone of these
14
The area of the triangle formed by the vertices \(({x_1},{y_1})\), \(({x_2},{y_2})\) and \(({x_3},{y_3})\) is given by
- A\(\frac{1}{2}\left[ {{x_1}({y_2} - {y_3}) + {x_2}({y_3} - {y_1}) + {x_3}({y_1} - {y_2})} \right]sq.units\)Correct
- B\(\frac{1}{2}\left[ {{x_1}({y_2} - {y_3}) - {x_2}({y_3} - {y_1}) - {x_3}({y_1} - {y_2})} \right]sq.units\)
- C\(\frac{1}{2}\left[ {{x_1}({y_2} + {y_3}) + {x_2}({y_3} + {y_1}) + {x_3}({y_1} + {y_2})} \right]sq.units\)
- D\(\frac{1}{2}\left[ {{x_1}({y_2} + {y_3}) - {x_2}({y_3} + {y_1}) - {x_3}({y_1} + {y_2})} \right]sq.units\)
15
Three given points will be collinear, if the area of the triangle formed by these points is
- A1 sq. units
- B2 sq. units
- C– 1 sq. units
- D0 sq. unitsCorrect