Coordinate Geometry : Exercise  7.4 (Mathematics NCERT Class 10th)

Class 10 : Maths + Science Video Lecture + Paper Solution+ TestMr. Kapil Sharma, GRS Academy, Mr. Neetin Agrawal
points A(2, –2) and B(3, 7).
Sol. Suppose the line 2x + y – 4 = 0 divides the line segment joining A(2, –2) and B(3, 7) in the ratio k : 1 at
point C. Then, the coordinates of C are
But C lies on 2x + y – 4 = 0 therefore,
9k – 2 = 0
9k = 2
So, the required ratio is 2 : 9 internally.
Q.2 Find a relation between x and y if the points (x, y), (1, 2) and (7, 0) are collinear.
Sol. The points A(x, y), B(1, 2) and C(7, 0) will be are collinear.
x(2 – 0) + 1(0 – y) + 7(y – 2) = 0
2x – y + 7y – 14 = 0
2x + 6y – 14 = 0
x + 3y – 7 = 0
which is the relation between x and y.
Q.3 Find the centre of a circle passing through the points (6, –6), (3, –7) and (3, 3).
Sol. Let P(x, y) be the centre of the circle passing through the points A(6, –6) B(3, –7) and C(3, 3).
Then, AP = BP = CP.
Now, AP = BP
= 0
....(1)
and, BP = CP
= 0
...(2)
Putting y = – 2 in (1), we get
3x – 2 – 7 = 0
3x = 9
x = 3
Thus, the centre of the circle is (3, –2).
Q.4 The two opposite vertices of a square are (–1, 2) and (3, 2). Find the coordinates of the other two vertices.
Sol. Let ABCD be a square and let A(–1, 2) and C(3, 2) be the given angular points. Let B(x, y) be the unknown vertex.
Then, AB = BC
8x = 8
x = 1 ...(1)
In ABC, we have
...(2)
Substituting the value of x from (1) into (2), we get
or 4
Hence, the required vertices of the square are (1, 0) and (1, 4).
Q.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 1m 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.
Sol. (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 whose vertices are and is given by
=
Therefore, area of PQR in the 1st case
sq. units
and, area of in the 2nd case
sq. units
Thus, we observe that the areas are the same in both the cases.
Q.6 The vertices of a 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 and compare it with the area of .
Sol. We know that if a line divides any two sides of a triangle in the same ratio, then the line is parallel to
the third side.
Therefore, DE  BC
Clearly, (AAA similarity)
We know that the ratio of the areas of two similar triangles is equal to the ratio of the squares of any
two corresponding sides.
Therefore,
...(1)
Now, Area
sq. units ...(2)
From (1) and (2),
Area
= sq. units
sq. units
Also, form (1), area : area = 1 : 16.
ALITER : (Using only Coordinate Geometry)
Since
Since D divides AB in the ratio 1 : 3
Therefore, coordinates of D are
, i.e.,
From , we find that
Since E divides AC in the ratio 1 : 3
Therefore, coordinates of E are
,i.e.,
Now, Area
=
sq. units
Also find the area
Area sq. units
Area : Area ()
Q.7 Let A(4, 2), B(6, 5) and C(1, 4) be the vertices of .
(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 and are the vertices of , find the coordinates of the centroid of the triangle.
Sol. Let A(4, 2), B(6, 5) and C(1, 4) be the vertices of .
(i) Since AD is the median of , Therefore, D is the midpoint of BC.
Its coordinates are ,i.e., .
(ii) Since P divides AD in the ratio 2 : 1 , so its coordinates are
or i.e.,
(iii) Since BE is the median of , so E is the midpoint of AC and its coordinates are i.e.,
Since Q divides BE in the ratio 2 : 1, so its coordinates are
or , i.e.,
Since CF is the median of , so F is the midpoint of AB. Therefore, its coordinates are
, i.e., .
Since R divides CF in the ratio 2 : 1, so its coordinates are
or or
(iv) We observe that the points P,Q and R coincide, the medians AD, BE and CF are concurrent at the point
. This point is known as the centroid of the triangle.
(v) Let and be the vertices of whose medians are AD, BE, and CF respectively. So, D,E and F are respectively the midpoints of BC,CA and AB.
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
Therefore, the medians of a triangle are concurrent and the coordinates of the centroid are
Q.8 ABCD is a rectangle formed by joining the points A(– 1,–1), B(– 1, 4), C(5, 4) and D(5,–1). P, Q, R and S are the midpoints of AB, BC, CD and DA respectively. Is the quadrilateral PQRS a square ? a rectangle ? or a rhombus ? Justify your answer.
Sol. Various points are marked in the adjoining figure.
=
=
=
=
and,
PQ = QR = RS = SP
Now, =
and
Since all the sides are equal but the diagonals are not equal,
therefore, PQRS is a rhombus.
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