**Maths Class 10
Exercise 7.4**

**1. Determine the ratio in which the line 2 x + y
– 4 = 0 divides the line segment joining the points A(2, –2) and B(3,7).**

**Solution: **Let the line 2*x* + *y*
– 4 = 0 divides the line segment joining the points A(2, –2) and B(3, 7) in the
ratio *k* : 1 at point C. Then,
the coordinates of C are

⇒ 6*k*
+ 4 + 7*k* – 2 –
4*k* – 4 = 0

⇒ 9*k* – 2 = 0

⇒ *k* = 2/9

Hence,
the required ratio is 2 : 9 internally.

**2. Find a relation between ***x*** and ***y*** if
the points ( x, y), (1, 2) and (7, 0) are collinear.**

**Solution:
**We
know that the
points A(*x*, *y*), B(1, 2) and C(7, 0) will be collinear only if the area
of triangle ABC = 0.

*x*–

*y*+ 7

*y*– 14 = 0

⇒ 2*x* + 6*y* – 14 = 0

⇒ *x* + 3*y* – 7 = 0

**3. Find the centre of a circle passing through
the points (6, –6), (3, –7) and (3, 3).**

**Solution:
**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.

Taking
AP = BP

⇒ AP^{2} = BP^{2 }

⇒ (*x*
– 6)^{2} + (*y* + 6)^{2} = (*x* – 3)^{2}
+ (*y* + 7)^{2}

⇒ *x*^{2} – 12*x* + 36 + *y*^{2 }+ 12*y*
+ 36 = *x*^{2} – 6*x*
+ 9 + *y*^{2} + 14*y* + 49

⇒ –12*x* + 6*x*
+ 12*y* – 14*y* + 72 –58 = 0

⇒ –6*x* – 2*y*
+ 14 = 0

⇒ 3*x* + *y* – 7 = 0 ………. (i)

Again,
taking BP = CP

⇒ BP^{2} = CP^{2}

⇒ (*x* – 3)^{2} + (*y* + 7)^{2} = (*x* –
3)^{2} + (*y* – 3)^{2}

⇒ *x*^{2} – 6*x* + 9 + *y*^{2} + 14*y* + 49
= *x*^{2} – 6*x* + 9 + *y*^{2} – 6*y* + 9

⇒ –6*x* + 6*x*
+ 14*y* + 6*y* + 58 – 18 = 0

⇒ 20*y* + 40 = 0

⇒ *y* = –2

Putting
the value of *y* in equation (i), we get

3*x* + *y*
– 7 = 0

⇒ 3*x* – 2 – 7 =
0

⇒ 3*x* = 9

⇒ *x* = 3

Hence,
the centre of the circle is (3, –2).

**4. The two opposite vertices of a square are
(–1, 2) and (3, 2)****.**** Find the coordinates of
the other two vertices.**

**Solution: **Let ABCD be a square and B(*x*, *y*) be
the unknown vertex. A(–1,
2) and B(3, 2) are the known vertices.

Here,
AB = BC

⇒ AB^{2} = BC^{2}

⇒ (*x* + 1)^{2} + (*y* – 2)^{2 } = (*x*
– 3)^{2} + (*y* – 2)^{2}

⇒ *x*^{2} + 2*x* + 1 + *y*^{2 }– 4*y* + 4 = *x*^{2} – 6*x* + 9 + *y*^{2 }– 4*y* + 4

⇒ 2*x* + 1 = –6*x* + 9

⇒ 8*x* = 8

⇒ *x* = 1 ………. (i)

In
right-angled ∆ABC, AB^{2} + BC^{2} = AC^{2 }[Using Pythagoras
theorem]

⇒ (*x* + 1)^{2} + (*y* – 2)^{2} + (*x* – 3)^{2} + (*y* –
2)^{2} = (3 + 1)^{2} + (2 – 2)^{2}

⇒ 2*x*^{2} + 2*y*^{2}
+ 2*x* – 4*y* – 6*x*
– 4*y* + 1 + 4 + 9 + 4 = 16

⇒ 2*x*^{2} + 2*y*^{2}
– 4*x*
– 8*y* + 2 = 0

⇒ *x*^{2} + *y*^{2}
– 2*x*
– 4*y* + 1 = 0 ………. (ii)

Putting
the value of *x* in equation
(ii), we get

1
+ *y*^{2} – 2 – 4*y* + 1 = 0

⇒ *y*^{2} – 4*y* = 0

⇒ *y*(*y* –4) = 0

⇒ *y* = 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?**

**Solution:
(i)** Taking A as
the origin, AD and AB are the coordinate axes. Clearly, the coordinates of the points
P, Q and R are (4, 6), (3, 2) and (6, 5), respectively.

**(ii)
**Taking C as the
origin, CB and CD are the coordinate axes. Clearly, the coordinates of the
points P, Q and R are (12, 2), (13, 6) and (10, 3), respectively.

We know that the area of the triangle =

= ½ [4(–3) + 3(–1) + 6(4)]

= ½ [–12 – 3 + 24]

= 9/2 sq. units

And Area of ∆PQR (Second case) = ½ [12(6 – 3) + 13(3 – 2) + 10(2 – 6)]

= ½ [12(3) + 13(1) +
10(–4)]

= ½ [12(3) + 13(1) +
10(–4)]

= ½ [36 + 13 – 40]

= 9/2 sq. units

Hence,
the areas are the 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 AD/AB = AE/AC = ¼. Calculate the area of the ****∆****ADE and compare it with the
area of ****∆****ABC.**

**Solution: **Since, AD/AB
= AE/AC = ¼

Therefore, ∆ADE ~
∆ABC

⇒ (AD/AB)^{2} = (1/4)^{2}
= 1/16 ……….(i)

Now, Area (∆ABC) = ½ [4(5 – 2)
+ 1(2 – 6) + 7(6 – 5)]

= ½ [4(3)
+ 1(– 4) + 7(1)]

= ½ [12 – 4 + 7] =15/2 sq. units ………. (ii)

From
equations (i) and (ii), we get

Area
(∆ADE) = 1/16 × Area (∆ABC) = 1/16 × 15/2 = 15/32 sq. units

Therefore, 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 the centroid and
this point divides each median in the ratio 2 : 1.]**

**(v)
If A( x_{1}, y_{1})**,

**B(**

*x*_{2},*y*_{2}) and C(*x*_{3},*y*_{3}) are the vertices of**∆**

**ABC, find the coordinates of the centroid of the triangle.**

**Solution: **It is given that A(4, 2), B(6, 5) and
C(1, 4) be the vertices of ∆ABC.

**(i)** Since AD is the median of ∆ABC from the vertex A.

Therefore, D
is the mid-point of BC.

**(ii)** Since P divides AD in the ratio 2 : 1.

So, the coordinates of point P are = (11/3, 11/3)

**(iii)** Since BE is the median of ∆ABC from the vertex B.

Therefore, E
is the mid-point of AC.

So, its
coordinates are (5/2, 3).

Since
Q divides BE in the ratio 2 : 1.

So, the coordinates of the point Q are = (11/3, 11/3)

Therefore, F
is the mid-point of AB.

So, its
coordinates are = (5, 7/2).

Since
R divides CF in the ratio 2 : 1.

So, the coordinates of the point R are = (11/3, 11/3)

**(iv)** We observe that the points P, Q
and R coincide, i.e., the medians AD, BE and CF are concurrent points. This
point is known as the centroid of the triangle. Its coordinates are (11/3,
11/3).

**(v)** According to the question, D, E
and F are the mid-points of BC, CA and AB respectively.

Therefore, the coordinates of D are

Therefore,
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 median of a
triangle are concurrent and the coordinates of the centroid are

**8. ABCD is a rectangle formed by joining
points A(–****1,****
–1), B(–1, 4), C(5, 4) and D(5, –1). P, Q, R and S are the
mid-points of AB, BC, CD and DA respectively. Is the quadrilateral PQRS a
square? a rectangle? or a rhombus? Justify your answer.**

**Solution: **

⇒ PQ = QR = RS = SP

⇒ PR ≠ SQ

Since
all the sides are equal but the diagonals are not equal.

Therefore, PQRS
is a rhombus.