NCERT Solutions for Class 10 Maths Exercise 7.4 Chapter 7 Coordinate Geometry – FREE PDF Download
NCERT Class 10 Maths Ch 7 is one of the most important ones in the NCERT syllabus. Duly following NCERT Solutions for Class 10 Maths Chapter
7 ensures you that there will be no hindrance when you opt for more advanced branches of Maths. This is where CoolGyan comes in. Our free Class 10 Coordinate Geometry solutions will help you understand the chapter thoroughly.
NCERT Solutions for Class 10 Maths Chapter 7 – Coordinate Geometry
1. Determine the ratio in which the line 
divides the line segment joining the points A
and B
divides the line segment joining A and B (3, 7) in the ratio 
at point C. Then, the coordinates of C are 
But C lies on , therefore
Hence, the required ratio if 2: 9 internally.
2. Find a relation between 
and 
if the points 
and 
are collinear.
B (1, 2) and C (7, 0) will be collinear if Area of triangle = 0
3. Find the centre of a circle passing through the points 
and 
be the centre of the circle passing through the points A
B
and C (3, 3). Then AP = BP = CP. Taking AP = BP
……….(i)
Again, taking BP = CP
Putting the value of in eq. (i),
Hence, the centre of the circle is 
4. The two opposite vertices of a square are 
and 
Find the coordinates of the other two vertices.
be the unknown vertex. AB = BC
x2+1+2x+y2+4−4y=x2+9−6x+y2+4−4yx2+1+2x+y2+4−4y=x2+9−6x+y2+4−4y
……….(i)
In 
ABC,
⇒⇒ 2(y−2)2+[(x+1)2+(x−3)2]=16+02(y−2)2+[(x+1)2+(x−3)2]=16+0
⇒⇒ 2(y−2)2+[(1+1)2+(1−3)2]=162(y−2)2+[(1+1)2+(1−3)2]=16
⇒⇒ 2(y2−4y+4)+8=162(y2−4y+4)+8=16
y2−4y+4=4y2−4y+4=4
= 0 or 4
Hence, the required vertices of the square are (1, 0) and (1, 4).
5. The class X students of a secondary school in Krishinagar have been allotted a rectangular plot of land for their gardening activity. Saplings of Gulmohar are 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?
(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.
6. The vertices of a ABC are A (4, 6), B (1, 5) and C (7, 2). A line is drawn to intersect sides AB and AC at D and E respectively such that 
Calculate the area of the ADE and compare it with the area of ABC.
DE 
BC[By Thales theorem]
ADE 
ABC
……….(i)
Now, Area (ABC) = 
= 
sq. units……….(ii)
From eq. (i) and (ii),
Area (ADE) = 
Area (ABC) = 
sq. units
Area (ADE): Area (ABC) = 1: 16
7. 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.
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 .
8. 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.
Similarly, the coordinates of Q are (2, 4)., the coordinates of R are (5,32)(5,32) and the coordinates of S are (2,−1)(2,−1).
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.