- Flash Back from Class IX notes
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- Distance formula
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- Section Formula
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- Area of triangle
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- How to Solve the line segment bisection ,trisection and four-section problem's
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- How to Prove three points are collinear
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- How to solve general Problems of Area in Coordinate geometry

- Coordinate Geometry Problem and Solutions
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- Coordinate Geometry Short questions
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- Coordinate Geometry 3 Marks Questions
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- Coordinate Geometry 5 Marks questions

Given below are the **Class 10 Maths** Important Questions for Coordinate Geometry

b) Calculation problems

c) Multiple choice questions

d) Long answer questions

e) Fill in the blank's

**Question 1**

Calculate the Following

- Distance between the point (1,3) and ( 2,4)
- Mid-point of line segment AB where A(2,5) and B( -5,5)
- Area of the triangle formed by joining the line segments (0,0) ,( 2,0) and (3,0)
- Distance of point (5,0) from Origin
- Distance of point (5,-5) from Origin
- Coordinate of the point M which divided the line segment A(2,3) and B( 5,6) in the ratio 2:3
- Quadrant of the Mid-point of the line segment A(2,3) and B( 5,6)
- 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)

**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 **

- False, As three point are A,B and C are collinear, So no quadrilateral can be formed
- False, As the distance between the point P and C is $\sqrt{26}$ which is less than 6.So point lies inside the circle
- True. Both the triangle are equilateral triangle with side 4 and 8 respectively
- True. As the Area formed by the triangle XYZ is not zero
- True, If we plot the point on the Coordinate system, it becomes clear that it is right angle at origin
- False. The radius of the circle is 5 and distance of the point B is more than 5,So it lies outside the circle
- 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

**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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