Home / Class 10 Math / Ch 7 Coordinate Geometry- MCQ Online Test 1 Class 10 Maths
Chapter 7 Coordinate Geometry- MCQ Online Test 1 Class 10 Maths
1.The distance between A (1, 3) and B (x, 7) is 5. The value of x > 0 is 1 2 3 4
2. Mid-point of the line-segment joining the points (– 5, 4) and (9, – 8) is : (– 4, 3) ( 2, 1) (– 2, 2) (2, – 2)
3. The co-ordinates of the points which divides the join of (– 2, – 2) and (– 5, 7) in the ratio 2 : 1 is (-4, 4) (4, – 4) (-4, – 4) none of these
4. The co-ordinates of the point on x-axis which is equidistant from the points (5, 4) and (– 2, 3) are : (5, 0) (0, 2) (2, 0) (3, 0)
5. The coordinates of the center of a circle are (– 6, 1.5). If the ends of a diameter are (– 3, y) and (x, – 2) then: x = – 8, y = 1 x = – 5, y = 7 x = – 9, y = 5 x = – 9, y = 3
6. The points (– 2, 2), (8, – 2) and (– 4, – 3) are the vertices of a : obtuse Δ equilateral Δ isosceles Δ right Δ
7. The points (1, 7), (4, 2), (– 1, 1) and (– 4, 4) are the vertices of a : square. rectangle parallelogram rhombus
8. The line segment joining (2, – 3) and (5, 6) is divided by x-axis in the ratio: 2 : 1 3 : 2 3 : 1 2 : 5
9. The line segment joining the points (3, 5) and (– 4, 2) is divided by y-axis in the ratio: 2: 5 5: 4 3: 4 2: 5
10. If (3, 2), (4, k) and (5, 3) are collinear then k is equal to : 5/7 3/2 5/2 5/9
Class 10 Coordinate Geometry Quiz 1
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Question 1 of 10
1. Question
The distance between A (1, 3) and B (x, 7) is 5. The value of x > 0 is
Correct
A (1,3), B(x,7)
AB=5 units
Distance between two points (x1,y1) and (x2,y2)
Squaring both sides
(AB)2=(x−1)2+(4)2
(5)2=x2−2x+1+16
25=x2−2x+17
x2−2x−8=0
x2−4x+2x−8=0
(x−4)(x+2)
x=−2 or 4
Thus, x = 4
Incorrect
A (1,3), B(x,7)
AB=5 units
Distance between two points (x1,y1) and (x2,y2)
Squaring both sides
(AB)2=(x−1)2+(4)2
(5)2=x2−2x+1+16
25=x2−2x+17
x2−2x−8=0
x2−4x+2x−8=0
(x−4)(x+2)
x=−2 or 4
Thus, x = 4
Question 2 of 10
2. Question
Mid-point of the line-segment joining the points (– 5, 4) and (9, – 8) is :
Correct
Midpoint of two points (x1,y1) and (x2,y2) is calculated by the formula
Using this formula, mid-point of the line-segment joining the points (5,4) and (9,8) is:
Incorrect
Midpoint of two points (x1,y1) and (x2,y2) is calculated by the formula
Using this formula, mid-point of the line-segment joining the points (5,4) and (9,8) is:
Question 3 of 10
3. Question
The co-ordinates of the points which divides the join of (– 2, – 2) and (– 5, 7) in the ratio 2 : 1 is
Correct
We know that, the coordinates of the point dividing the line segment joining the points x1,y1 and x2,y2 in the ratio m:n is given by
Given (x1,y1)=(−2,-2); (x2,y2)=(-5,7)
(m1,m2)=2:1
Coordinates of point of intersection of line segment is
= (-12/3, 12/3)
= (-4, 4)
Incorrect
We know that, the coordinates of the point dividing the line segment joining the points x1,y1 and x2,y2 in the ratio m:n is given by
Given (x1,y1)=(−2,-2); (x2,y2)=(-5,7)
(m1,m2)=2:1
Coordinates of point of intersection of line segment is
= (-12/3, 12/3)
= (-4, 4)
Question 4 of 10
4. Question
The co-ordinates of the point on x-axis which is equidistant from the points (5, 4) and (– 2, 3) are :
Correct
Let that point be C. Thus the coordinate of C(x,0)
Given points,
A(5, 4) and B(– 2, 3)
AC=BC (Equidistance)
AC2=BC2
(x−5)2+(0−4)2 = (x+2)2+(0−3)2
x2−10x+25+16 = x2+4+4x+9
−14x+41−13=0
−14x+28=0
=14x = −28
x =28/14
x= 2
Thus, C(2,0)
Incorrect
Let that point be C. Thus the coordinate of C(x,0)
Given points,
A(5, 4) and B(– 2, 3)
AC=BC (Equidistance)
AC2=BC2
(x−5)2+(0−4)2 = (x+2)2+(0−3)2
x2−10x+25+16 = x2+4+4x+9
−14x+41−13=0
−14x+28=0
=14x = −28
x =28/14
x= 2
Thus, C(2,0)
Question 5 of 10
5. Question
The coordinates of the center of a circle are (– 6, 1.5). If the ends of a diameter are (– 3, y) and (x, – 2) then:
Correct
(-6, 1.5) would be the mid point of line joining (-3,y) and (x,-2) as the centre of a circle is the mid point of the diameter.
According to Section formula,
(-6,1.5) = [(-3+x)/2, {y+ (-2)}/2]
So -6 = (-3+x)/2 and 1.5 = (y-2)/2
-12 = -3+x and 3 = y-2
-9 = x and 5 = y
So x and y are -9 and 5.
Incorrect
(-6, 1.5) would be the mid point of line joining (-3,y) and (x,-2) as the centre of a circle is the mid point of the diameter.
According to Section formula,
(-6,1.5) = [(-3+x)/2, {y+ (-2)}/2]
So -6 = (-3+x)/2 and 1.5 = (y-2)/2
-12 = -3+x and 3 = y-2
-9 = x and 5 = y
So x and y are -9 and 5.
Question 6 of 10
6. Question
The points (– 2, 2), (8, – 2) and (– 4, – 3) are the vertices of a :
Correct
Let the points A(-2,2), B(8,-2)and C(-4,3) be the vertices of a triangle ABC.
AB2 = (8+2)2 + (-2-2)2 = 102 + 42 = 100 + 16 = 116
BC2 = (-4-8)2 + (3+2)2 = (-12)2 + 52 = 144 + 25 = 169
AC2 = (-4+2)2 + (3-2)2 = (-2)2 + 1 = 4 + 1 = 5
AB2 + AC2 ≠ BC2
The above vertices are not points of a right angled triangle.
Incorrect
Let the points A(-2,2), B(8,-2)and C(-4,3) be the vertices of a triangle ABC.
AB2 = (8+2)2 + (-2-2)2 = 102 + 42 = 100 + 16 = 116
BC2 = (-4-8)2 + (3+2)2 = (-12)2 + 52 = 144 + 25 = 169
AC2 = (-4+2)2 + (3-2)2 = (-2)2 + 1 = 4 + 1 = 5
AB2 + AC2 ≠ BC2
The above vertices are not points of a right angled triangle.
Question 7 of 10
7. Question
The points (1, 7), (4, 2), (– 1, 1) and (– 4, 4) are the vertices of a :
Correct
Given points are (1,7),(4,2),(-1,-1),(-4,4)
Let the points are A,B,C,D.
Distance formula = √(x₂-x₁)²+(y₂-y₁)²
AB = √(4-1)²+(2-7)² = √9+25 = √34
BC = √(-1-4)²+(-1-2)² = √25+9 = √34
CD = √(-4-(-1))²+(4-(-1))² = √9+25 = √34
DA = √(1-(-4))²+(7-4)² = √25+9 = √34
We know that the all sides of the square are equal.
Here by the distance formula, we got the four sides equal.
So, from this we can say that these points are the vertices of a square.
Incorrect
Given points are (1,7),(4,2),(-1,-1),(-4,4)
Let the points are A,B,C,D.
Distance formula = √(x₂-x₁)²+(y₂-y₁)²
AB = √(4-1)²+(2-7)² = √9+25 = √34
BC = √(-1-4)²+(-1-2)² = √25+9 = √34
CD = √(-4-(-1))²+(4-(-1))² = √9+25 = √34
DA = √(1-(-4))²+(7-4)² = √25+9 = √34
We know that the all sides of the square are equal.
Here by the distance formula, we got the four sides equal.
So, from this we can say that these points are the vertices of a square.
Question 8 of 10
8. Question
The line segment joining (2, – 3) and (5, 6) is divided by x-axis in the ratio:
Correct
Given x-axis divide the segment joining points
(2,−3) and (5,6)
Let the ratio be k:1
⇒ 6k=3
⇒ k = 1/2
the ratio is 1:2
Incorrect
Given x-axis divide the segment joining points
(2,−3) and (5,6)
Let the ratio be k:1
⇒ 6k=3
⇒ k = 1/2
the ratio is 1:2
Question 9 of 10
9. Question
The line segment joining the points (3, 5) and (– 4, 2) is divided by y-axis in the ratio:
Correct
Using the section formula, if a point (x,y) divides the line joining the points (x1,y1) and (x2,y2) in the ratio m:n, then
Substituting (x1,y1)=(3,5) and (x2,y2)=(−4,2) in the section formula, we get the point
Since the point of intersection lies on the y-axis, x-coordinate =0\
⇒−4m+3n=0
⇒4m=3n
⇒ m:n=3:4
Incorrect
Using the section formula, if a point (x,y) divides the line joining the points (x1,y1) and (x2,y2) in the ratio m:n, then
Substituting (x1,y1)=(3,5) and (x2,y2)=(−4,2) in the section formula, we get the point
Since the point of intersection lies on the y-axis, x-coordinate =0\
⇒−4m+3n=0
⇒4m=3n
⇒ m:n=3:4
Question 10 of 10
10. Question
If (3, 2), (4, k) and (5, 3) are collinear then k is equal to :
Correct
As the points are collinear, the slope of the line joining any two points, should be same as the slope of the line joining two other points.
Slope of the line passing through points (x1,y1) and (x2,y2) = (y2-y1)/(x2-x1)
So, Slope of the line joining (3,2),(4,k)= Slope of the line joining (3,2) and (5,3)
(k−2)/(4-3)= (3−2)/(5−3)
k−2= 1/2
k= 5/2
Incorrect
As the points are collinear, the slope of the line joining any two points, should be same as the slope of the line joining two other points.
Slope of the line passing through points (x1,y1) and (x2,y2) = (y2-y1)/(x2-x1)
So, Slope of the line joining (3,2),(4,k)= Slope of the line joining (3,2) and (5,3)