CBSE Class 10 Maths Chapter-7 Coordinate Geometry – Free PDF Download
Free PDF download of Important Questions with Answers for CBSE Class 10 Maths Chapter 7 – Coordinate Geometry prepared by expert Maths teachers from latest edition of CBSE(NCERT) books only by CoolGyan to score more marks in CBSE board examination.
CBSE Class 10 Maths Chapter-7 Coordinate Geometry Important Questions
CBSE Class 10 Maths Important Questions Chapter 7 – Coordinate Geometry
4 Mark Questions
1. If the points 
is equidistant from the points 
and 
prove that 
Ans. Let 
2. 
and
are three concyclic points whose centre is 
. Find the possible value of 
and y.
Ans. 
Radius of circle
3. Find the vertices of the triangle, the mid-points of whose sides are 
and 
Ans. Let vertices of 
be 
and 
By mid-points formula
Adding (i), (iii) and (v)
Adding (ii), (iv)and (vi)
Subtracting (i), (iii) and (v) from (vii)
We get, 
Subtracting (ii), (iv) and (vi) from eq. (viii)
We get, 
4. The two opposite vertices of a square are 
and
. Find the coordinates of the other two vertices.
Ans. 
In right 
Put the value of y in eq. (i)
Now 
And 
5. Find the coordinates of the circumcentre of a triangle whose vertices are A(4,6), B(0,4) and C(6,2). Also find its circum-radius.
Ans. Let P be the circum-centre of then PA = PB = PC
On solving equations (i) and (ii), 
Circum-radius (PA) = 
6. If two vertices of an equilateral triangle are 
Find the third vertex.
Ans. 
Put the value of y in eq. (i),
7. If P and Q are two points whose coordinates are 
and 
respectively and S is the point 
show that 
is independent of 
.
Ans. 
8. Find the area of the quadrilateral whose vertices taken in order are (-4,-2), (-3,5), (3,-2) and (2,3).
Ans. 
= 10.5 sq. units
area of quadrilateral = 10.5 + 17.5 = 28 sq. units.
9. The vertices of are 
and 
. A line is drawn to intersect sides AB and AC at D and E respectively such that 
. Calculate the area of the 
and compare it with the area of .
Ans.
Now coordinate of D and E are
and 
10. Prove that the points 
and 
are the vertices of an equilateraltriangle. Calculate the area of this triangle.
Ans. Let 
11. 
are the vertices of a 
is the mid-point of BC and P is a point on AD joined such that 
find the coordinates of P.
Ans. Let 
and 
are the vertices of 
is the mid- point of BC
Coordinate of D 
i.e., 
Coordinate of P are 
12. The coordinates of the vertices of are
. Given that the area of is 12, find the value of K.
Ans. 
Value of 
13. Find the lengths of the medians of the triangle whose vertices are 
and 
.
Ans. Coordinates of points D, E and F are
Length of the median AD
Length of the median BE
And length of the median CF
14. The area of a triangle is 5. Two of its vertices are 
and
. The third vertex lies on
. Find the third vertex.
Ans. Let the third vertex be
. Other two vertices of the 
are 
and 
lies on eq.
On solving eq. 
We get 
Similarly, on solving eq. 
We get 
15. Prove that the point 
and 
are collinear, if 
Ans. Since 
are collinear
Area = 0
Dividing by 
,
16. If, Q (0, 1) is equidistant from P (5, –3) and R (x, 6), find the values of x. Also, find the distances QR and PR.
Ans. It is given that Q is equidistant from P and R. UsingDistance Formula, we get
PQ = RQ
PQ2 = RQ2
⇒ 
⇒ 
Squaring both sides, we get
⇒ 25+16=x2+25
⇒ x2=16
⇒ x=4,−4
Thus, Q is (4, 6) or (–4, 6).
Using Distance Formula to find QR, we get
Using value of x = 4
QR=
Using value of x = –4
QR=
Therefore, QR=
Using Distance Formula to find PR, we get
Using value of x = 4
PR=
Using value of x =–4
PR=
Therefore, x = 4, –4
QR=, PR=
17. Find the coordinates of the points which divides the line segment joining A(–2, 2) and B(2, 8) into four equal parts.
Ans. A = (–2, 2) and B = (2, 8)
Let P, Q and R are the points which divide line segment AB into 4 equal parts.
Let coordinates of point P =(x1, y1), Q =(x2, y2) and R =(x3, y3)
We know AP = PQ = QR = RS.
It means, point P divides line segment AB in 1:3.
Using Section formula to find coordinates of point P, we get
Since, AP = PQ = QR = RS.
It means, point Q is the mid-point of AB.
Using Section formula to find coordinates of point Q, we get
Because, AP = PQ = QR = RS.
It means, point R divides line segment AB in 3:1
Using Section formula to find coordinates of point P, we get
Therefore, P=(–1, 
), Q= (0, 
)and R =(1, 
)
18. The two opposite vertices of a square are 
and 
Find the coordinates of the other two vertices.
Ans. Let ABCD be a square and B be the unknown vertex.
AB = BC
AB2 = BC2
……….(i)
In 
ABC, AB2 + BC2 = AC2
……….(ii)
Putting the value of in eq. (ii),
= 0 or 4
19. The class X students of a secondary school in Krishinagar have been allotted a rectangular plot of land for their gardening activity. Saplings of Gulmoharare planted on the boundary at a distance of 1 m from each other. There is a triangular grassy lawn in the plot as shown in the figure. The students are to sow seeds of flowering plants on the remaining area of the plot.
(i) Taking A as origin, find the coordinates of the vertices of the triangle.
(ii) What will be the coordinates of the vertices of PQR if C is the origin? Also calculate the area of the triangle in these cases. What do you observe?
Ans. (i) Taking A as the origin, AD and AB as the coordinate axes. Clearly, the points P, Q and
R are (4, 6), (3, 2) and (6, 5) respectively.
(ii) Taking C as the origin, CB and CD as the coordinate axes. Clearly, the points P, Q and R are given by (12, 2), (13, 6) and (10, 3) respectively.
We know that the area of the triangle = 
Area of PQR (First case) = 
= 
= 
= 
sq. units
And Area of PQR (Second case) = 
= 
= 
= sq. units
Hence, the areas are same in both the cases.
20. ABCD is a rectangle formed by joining points A
B
C and D
P, Q, R and S are the mid-points of AB, BC, CD and DA respectively. Is the quadrilateral PQRS a square? Or a rhombus? Justify your answer.
Ans. Using distance formula, PQ = 
= 
= 
QR = 
= =
RS = 
= =
SP = 
= =
PQ = QR = RS = SP
Now, PR = 
= 
= 6
And SQ = 
= 
= 5
PR 
SQ
Since all the sides are equal but the diagonals are not equal.
PQRS is a rhombus.
21. In a classroom, 4 friends are seated at the points A (3, 4), B (6, 7), C (9, 4) and D (6, 1). Champa and Chameli walk into the class and after observing for a few minutes Champa asks Chameli. “Don’t you think ABCD is a square?”Chameli disagrees. Using distance formula, find which of them is correct.
Ans. We have A = (3, 4), B = (6, 7), C = (9, 4) and D = (6, 1)
Using Distance Formula to find distances AB, BC, CD and DA, we get
AB = 
BC = 
CD = 
DA = 
Therefore, All the sides of ABCD are equal here. … (1)
Now, we will check the length of its diagonals.
AC = 
BD = 
So, Diagonals of ABCD are also equal. … (2)
From (1) and (2), we can definitely say that ABCD is a square.
Therefore, Champa is correct.
22. Name the type of quadrilateral formed, if any, by the following points, and give reasons for your answer.
(i) (–1, –2), (1, 0), (–1, 2), (–3, 0)
(ii) (–3, 5), (3, 1), (0, 3) , (–1, –4)
(iii) (4, 5), (7, 6), (4, 3), (1, 2)
Ans. (i) Let A = (–1, –2), B = (1, 0), C= (–1, 2) and D = (–3, 0)
Using Distance Formula to find distances AB, BC, CD and DA, we get
AB = 
BC=
CD=
DA=
Therefore, all four sides of quadrilateral are equal. … (1)
Now, we will check the length of diagonals.
AC=
BD=
Therefore, diagonals of quadrilateral ABCD are also equal. … (2)
From (1) and (2), we can say that ABCD is a square.
(ii) Let A = (–3, 5), B= (3, 1), C= (0, 3) and D= (–1, –4)
Using Distance Formula to find distances AB, BC, CD and DA, we get
AB=
BC=
CD=
DA=
We cannot find any relation between the lengths of different sides.
Therefore, we cannot give any name to the quadrilateral ABCD.
(iii) Let A = (4, 5), B= (7, 6), C= (4, 3) and D= (1, 2)
Using Distance Formula to find distances AB, BC, CD and DA, we get
AB=
BC=
CD=
DA=
Here opposite sides of quadrilateral ABCD are equal. … (1)
We can now find out the lengths of diagonals.
AC=
BD=
Here diagonals of ABCD are not equal. … (2)
From (1) and (2), we can say that ABCD is not a rectangle therefore it is a parallelogram.
23. Let A (4, 2), B (6, 5) and C (1, 4) be the vertices of 
ABC.
(i) The median from A meets BC at D. Find the coordinates of the point D.
(ii) Find the coordinates of the point P on AD such that AP : PD = 2 : 1.
(iii) Find the coordinates of points Q and R on medians BE and CF respectively such that BQ : QE = 2 : 1 and CR : RF = 2 : 1.
(iv) What do you observe?
(Note: The point which is common to all the three medians is called centroid and this point divides each median in the ratio 2 : 1)
(v) If A
B
and C
are the vertices of ABC, find the coordinates of the centroid of the triangle.
Ans. Let A (4, 2), B (6, 5) and C (1, 4) be the vertices of ABC.
(i) Since AD is the median of ABC.
D is the mid-point of BC.
Its coordinates are 
= 
(ii) Since P divides AD in the ratio 2 : 1
Its coordinates are 
= 
(iii) Since BE is the median of ABC.
E is the mid-point of AD.
Its coordinates are 
= 
Since Q divides BE in the ratio 2 : 1.
Its coordinates are 
=
Since CF is the median of ABC.
F is the mid-point of AB.
Its coordinates are 
= 
Since R divides CF in the ratio 2 : 1.
Its coordinates are 
=
(iv) We observe that the points P, Q and R coincide, i.e., the medians AD, BE and CF are concurrent at the point . This point is known as the centroid of the triangle.
(v) According to the question, D, E, and F are the mid-points of BC, CA and AB respectively.
Coordinates of D are 
Coordinates of a point dividing AD in the ratio 2 : 1 are
= 
The coordinates of E are 
.
The coordinates of a point dividing BE in the ratio 2 : 1 are
=
Similarly the coordinates of a point dividing CF in the ratio 2 : 1 are
Thus, the point is common to AD, BE and CF and divides them in the ratio 2 : 1.
The median of a triangle are concurrent and the coordinates of the centroid are .