# PRACTICE QUESTIONS ON COORDINATE GEOMETRY FOR GRADE 10

Practice Questions on Coordinate Geometry for Grade 10

Here we are going to see, some practice questions on coordinate geometry.

## Practice Questions on Coordinate Geometry for Grade 10

Question 1 :

PQRS is a rectangle formed by joining the points P(-1,-1), Q(-1, 4), R(5, 4) and S(5,-1) . A, B, C and D are the mid-points of PQ, QR, RS and SP respectively. Is the quadrilateral ABCD a square, a rectangle or a rhombus? Justify your answer.

Solution :

Midpoint of the side PQ  =  A

Midpoint of the side QR  =  B

Midpoint of the side RS  =  C

Midpoint of the side SP  =  D

A = (x1 + x2)/2, (y1 + y2)/2

P(-1,-1), Q(-1, 4)

A  =  (-1 + (-1))/2, (-1 + 4)/2

A  =  (-1, 3/2)

Q(-1, 4) R(5, 4)

B  =  (-1 + 5)/2, (4 + 4)/2

B  =  (2, 4)

R(5, 4) and S(5,-1)

C  =  (5 + 5)/2, (4 + (-1))/2

C  =  (5, 3/2)

S(5,-1) and P(-1,-1)

D  =  (5 + (-1))/2, (-1 + (-1))/2

D  =  (2, -1)

Distance between AB  =  √(x2 - x1)+ (y2 - y1)2

 A (-1, 3/2) B (2, 4) =  √(2 + 1)2 + (4 - 3/2)2 =  √9 + (25/4) =  √61/2 B(3, 4) C (5, 3/2) =  √(3-5)2 + (4-(3/2))2 =  √4 + (25/4) =  √61/2 C (5, 3/2) D (2, -1) =  √(2-5)2 + (-1-(3/2))2 =  √9 + (25/4) =  √61/2 D (2, -1) A (-1, 3/2) =  √(-1-2)2 + ((3/2)+1)2 =  √9 + (25/4) =  √61/2

Since the length of all sides are equal, it may be a square or rhombus not a rectangle.

I order to check if it is a square, let us find the slope of the sides AB and BC. If they are perpendicular, then it is square otherwise it is rhombus.

A (-1, 3/2) B (2, 4)

m1  =  (y2 - y1)/(x2 - x1)

=  (4 - (3/2))/(2 + 1)

=  (5/2)/3

m1  =  5/6 ----(1)

B(3, 4) C (5, 3/2)

m2  =  ((3/2) - 4)/(5 - 3)

=  (-5/2)/2

m2  =  -5/4 ----(2)

Since they are not perpendicular, it is not square. Hence it is rhombus.

Question 2 :

The area of a triangle is 5 sq.units. Two of its vertices are (2, 1) and (3, –2). The third vertex is (x, y) where y = x + 3 . Find the coordinates of the third vertex.

Solution :

Let the missing vertex be (x, y)

Area of triangle  =  5

A (2, 1) B (3, -2) and C (x, y) (1/2)[(-4 + 3y + x) - (3 - 2x + 2y)]  =  5

x + 3y - 4 - 3 + 2x - 2y  =  10

3x + y - 7  =  10

3x + y  =  17  -----(1)

y = x + 3 (Given)

By applying the value of y in (1), we get

3x + x + 3  =  17

4x  =  17 - 3

4x  =  14

x  =  7/2

By applying the value of x in the given equation, we get

y  =  (7/2)  + 3

y = (7 + 6)/3

y = 13/3

Hence the third vertex is (7/2, 13/3) After having gone through the stuff given above, we hope that the students would have understood, "Practice Questions on Coordinate Geometry for Grade 10".

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