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Class 10 Maths Important Questions for Coordinate Geometry



Given below are the Class 10 Maths Important Questions for Coordinate Geometry
a) Concepts questions
b) Calculation problems
c) Multiple choice questions
d) Long answer questions
e) Fill in the blank's
 Question 1
Calculate the Following
  1. Distance between the point  (1,3) and ( 2,4)
  2. Mid-point of line segment  AB where A(2,5) and B( -5,5)
  3. Area of  the triangle  formed by joining the line segments (0,0)  ,( 2,0) and (3,0)
  4. Distance of point (5,0) from Origin
  5. Distance of point (5,-5) from Origin
  6. Coordinate of the point M which divided the line segment A(2,3) and B( 5,6) in the ratio 2:3
  7. Quadrant of the Mid-point of the line segment A(2,3) and B( 5,6)
  8.  the coordinates of a point A, where AB is the diameter of circle whose center is (2,−3) and B is (1, 4)
Solution
a) $D=\sqrt{(1-2)^{2}+(3-4)^{2}}=\sqrt{2}$
b)  Mid-point is given by (2-5)/2,(5+5)/2 or (-3/2, 5)
c) $A=\frac{1}{2}[0(0-0)+2(0-0)+3(0-0)]=0$
Since the three points are collinear, the area is zero
d) $D=\sqrt{5^{2}+0^{2}}=5$
e)$D=\sqrt{(-5)^{2}+0^{2}}=5$
f) Coordinates of point M is given by
$x=\frac{2X3+3X2}{2+3}=\frac{12}{5}$
$y=\frac{2X6+3X5}{2+3}=\frac{27}{5}$
g) Mid point is given by (7/2, 9/2) which lies in First quadrant
h) We know that center is mid point of AB, So
$2=\frac{1+x}{2}$
$-3=\frac{4+y}{2}$
Solving these, we get (3,-10)

True or False statement

Question 2
a) Point A( 0,0) B( 0,3) ,C( 0,7) and D( 2,0) formed a quadrilateral
b) The point P (–2, 4) lies on a circle of radius 6 and center C (3, 5)
c) Triangle PQR with vertices P (–2, 0), Q (2, 0) and R (0, 2) is similar to Δ XYZ with
Vertices X (–4, 0) Y (4, 0) and Z (0, 4).
 d) Point X (2, 2) Y (0, 0) and Z (3, 0) are not collinear
e) The triangle formed by joining the point A( -3,0) , B( 0,0) and C( 0,2) is a right angle triangle
f) A circle has its center at the origin and a point A (5, 0) lies on it. The point B (6, 8) lies inside the circle
g) The points A (–1, –2), B (4, 3), C (2, 5) and D (–3, 0) in that order form a rectangle
Solution
  1. False, As three point are A,B and C are collinear, So no quadrilateral can be formed
  2. False, As the distance between the point P and C is $\sqrt{26}$ which is less than 6.So point lies inside the circle
  3. True. Both the triangle are equilateral triangle with side 4 and 8 respectively
  4. True. As the Area formed by the triangle XYZ is not zero
  5. True, If we plot the point on the Coordinate system, it becomes clear that it is right angle at origin
  6. False. The radius of the circle is 5 and distance of the point B is more than 5,So it lies outside the circle
  7. True. If we calculate the distance between two points, it becomes clear that opposite side are equal, also the diagonal are equal. So it is a rectangle

Multiple choice Questions

Question 3
Find the centroid of the triangle XYZ whose vertices are X (3, - 5) Y (- 3, 4) and Z (9, - 2).
a) (0, 0)
b) (3, 1)
c) (2, 3)
d) (3,-1)
Solution (d)
Centroid of the triangle is given by
$x=\frac{x_{1}+x_{2}+x_{3}}{3}=\frac{3-3+9}{3}=3$
$y=\frac{y_{1}+y_{2}+y_{3}}{3}=\frac{-5+4-2}{3}=-1
 $
Question 3
The area of the triangle ABC with coordinates as A (1, 2) B (2, 5) and C (- 2, - 5)
a)  -1
b) .4
c)  2
d) 1
Solution (d)
$A=\frac{1}{2}[1(5+5)+2(-5-2)-2(2-5)]=1$
Question 4
Find the value of  p for which these point are collinear  (7,-2) , (5,1) ,(3,p)?
a) 2
b) 4
c)  3
d) None of these
Solution a
For these points to be collinear
A=0
Or
$\frac{1}{2}[7(1-p)+5(p+2)+3(-2-1)]=0$
7-7p+5p+10-9=0
P=2
Question 5
Determine the ratio in which the line 2x + y – 4 = 0 divides the line segment joining the points A (2, – 2) and B (3, 7).
a) 2:9
b) 1:9
c)1:2
d) 2:3
Solution (a)
Let the ratio be m: n
Now
Coordinate of the intersection
$x=\frac{3m+3n}{m+n}$
$y=\frac{7m-2n}{m+n}$
Now these points should lie of the line, So
$2(\frac{3m+2n}{m+n})+(\frac{7m-n}{m+n})-4=0$
  • m:n=2:9
Question 6
If the mid-point of the line segment joining the points A (3, 4) and
B (a, 4) is P (x, y) and x + y – 20 = 0,then find the value of a
a) 0
b) 1
c) 40
d)  45
Solution (d)
Mid point (3+a)/2, 4
Now
(3+a)/2 -4 -20=0
3+a=48
A=45


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