Rajasthan Board RBSE Solutions for Class 10 Maths Chapter 7 Coordinate Geometry Ex 7.4 Textbook Exercise Questions and Answers.

## RBSE Class 10 Maths Solutions Chapter 7 Coordinate Geometry Exercise 7.4

Question 1.

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

Solution:

Let the line 2x + y – 4 = 0 divide the line segment joining the points A(2, – 2) and B(3, 7) in the ratio k : 1 at C.

∴ Coordinates of C are :

∴ k : 1 = \(\frac{2}{9}\) : 1 = 2 : 9

Hence the required ratio is 2 : 9

Question 2.

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

Solution:

According to the question, A(x, y), B(1, 2) and C(7, 0).

Here,

x_{1} = x, x_{2} = 1, x_{3} = 7

y_{1} = y, y_{2} = 2, y_{3} = 0

If the three points are collinear then

\(\frac{1}{2}\)[x_{1}(y_{2} – y_{3}) + x_{2}(y_{3} – y_{1}) + x_{3}(y_{1} – y_{2})] = 0

or \(\frac{1}{2}\)[x(2 – 0) + 1(0 – y) + 7(y – 2)] = 0

or 2x – y + 7y – 14 = 0

or 2x + 6y – 14 = 0

or x + 3y – 1 = 0

This is the required relation.

Question 3.

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

Solution:

Let 0(x, y) be the required centre of the circle which passes through the points P(6, – 6), Q(3, – 7) and R(3, 3).

∵ Radii of a circle are equal.

∴ OP = OQ = OR

or (OP)^{2} = (OQ)^{2} = (OR)^{2}

Now, (OP)^{2} = (OQ)^{2}

⇒ (x – 6)^{2} + (y + 6)^{2} = (x – 3)^{2} + (y + 1)^{2}

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

or – 12x + 12y + 72 = -6x + 14y + 58

or – 6x – 2y + 14 = 0

or 3x + y – 7 = 0 …(i)

Now (OQ)^{2} = (OR)^{2}

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

or (y + 7)^{2} = (y – 3)^{2}

or y^{2} + 49 + 14y = y^{2} + 9 – 6y

or 20y = – 40

y = \(\frac{-40}{20}\) = -2

Substituting this value of y in (i) we obtain :

3x – 2 – 7 = 0

or 3x – 9 = 0

or 3x = 9

or x = \(\frac{9}{3}\) = 3

∴ The required centre is (3, – 2).

Question 4.

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

Solution:

Let the two opposite vertices of a square ABCD be A(- 1, 2) and B(3, 2) and the coordinates of C will be (x, y).

∵ Length of each side of a square is the same.

∴ AC = BC

or (AC)^{2} = (BC)^{2}

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

or (x + 1)^{2} = (x – 3)^{2}

or x^{2} + 1 + 2x = x^{2} + 9 – 6x

or 8x = 8

or x = \(\frac{8}{8}\) = 1 …(i)

Now in right △ACB, by Pythagoras Theorem,

(AC)^{2} + (BC)^{2} = (AB)^{2}

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

= (3 + 1)^{2} + (2 – 2)^{2}

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

or 2x^{2} + 2y^{2} – 4x – 8y + 2 = 0

or x^{2} + y^{2} – 2x – 4y + 1 = 0

Substituting the value x = 1 of x obtained from (i)

(1)^{2} + (y)^{2} – 2(1) – 4y + 1 = 0

or y^{2} – 4y = 0

or y(y – 4) = o

or y = o or y – 4 = 0

or y = 0 or y = 4

∴ y = 0, 4

Hence the remaining two vertices of the square ABCD are (1, 0) and (1, 4)

Question 5.

The Class X students of a secondary school in Krishinagar have been allotted a rectangular plot of land for their gardening activity. Sapling 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 Fig. 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 areas of the triangles in these cases. What do you observe?

Solution :

Case First—When we take A as the origin then AD in the x-axis and AB in the y-axis.

Coordinates of the vertices of triangular grassy lawn PQR are : P(4, 6); Q(3, 2) and R(6, 5).

Now, area of △PQR

= \(\frac{1}{2}\)[x_{1}(y_{2} – y_{3}) + x_{2}(y_{3} – y_{2}) + x_{3}(y_{1} – y_{2})]

= \(\frac{1}{2}\)[4(2 – 5) + 3(5 – 6) + 6(6 – 2)]

= \(\frac{1}{2}\)[-12 – 3 + 24] =

= 4.5 square units

Case Second—When we take C as the origin then CB in the x-axis and CD in the y-axis.

Coordinates of the vertices of triangular grassy lawn PQR are : P(-12, -2), Q(-13, -6) and R(- 10, -3)

Here, x_{1} = -12, x_{2} = -13, x_{3} = -10

y_{1} = – 2, y_{2} = – 6, y_{3} = – 3

Now, area of △PQR

= \(\frac{1}{2}\)[x_{1}(y_{2} – y_{3}) + x_{2}(y_{3} – y_{1}) + x_{3}(y_{1} – y_{2})]

= \(\frac{1}{2}\)[(-12)(-6 + 3) + (-13)(-3 + 2) + (-10)(-2 + 6)]

= \(\frac{1}{2}\)[(-12) × (-3) + (-13) × (-1) + (-10) × 4]

= \(\frac{1}{2}\)[+36 + 13 – 40]

= \(\frac{9}{2}\)

= 4.5 square units.

From the above two cases it is clear that the areas of the triangular grassy lawn are the same in both the cases. Ans.

Question 6.

The vertices of AABC are A(4, 6), B(1l, 5) and C(7, 2). A line is drawn to intersect sides AB and AC at D and E respectively, such that \(\frac{AD}{AB}\) = \(\frac{AE}{AC}\) = \(\frac{1}{4}\). Calculate the area of the △ADE and compare it with the area of △ABC. (Recall Theorem 6.2 and Theorem 6.6).

Solution:

The vertices of △ABC are A(4, 6), B(1, 5) and C(7, 2)

Draw a line DE intersecting the sides AB and AC at respectively D(x_{1}, y_{1}) and E(x_{2}, y_{2}) such that

\(\frac{AD}{AB}\) = \(\frac{AE}{AC}\) = \(\frac{1}{4}\)

∴ D and E respectively divide AB and AC in the ratio 1 : 3.

Therefore by internal division formula

x = \(\frac{m_{1} x_{2}+m_{2} x_{1}}{m_{1}+m_{2}}\) and

= \(\frac{15}{32}\) square units

In △ABC,

x_{1} = 4, x_{2} = 1, x_{3} = 7

y_{1} = 6, y_{2} = 5, y_{3} = 2

Area of △ABC

= \(\frac{1}{2}\)[x_{1}(y_{2} – y_{3}) + x_{2}(y_{3} – y_{1}) + x_{3}(y_{1} – y_{2})]

= \(\frac{1}{2}\)[4(5 – 2) + 1(2 – 6) + 7(6 – 5)]

= \(\frac{1}{2}\)[4 × 3 + 1(-4) + 7 × 1]

= \(\frac{1}{2}\)[12 – 4 + 7] = \(\frac{15}{2}\) square units

Now, \(\frac{area of △ADE}{area of △ABC}\)

= \(\frac{\frac{15}{32}}{\frac{15}{2}}\) = \(\frac{15}{32}\) × \(\frac{2}{15}\) = \(\frac{1}{16}\)

⇒ Area of △ADE : Area of △ABC = 1 : 16

Question 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(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:

According to the question, the vertices of △ABC are A(4, 2), B(6, 5) and C(1, 4).

(i) AD in the median from vertex A.

∴ D is the mid-point of BC.

Hence the coordinates of D are (\(\frac{7}{2}\), \(\frac{9}{2}\))

(ii) Let P(x y) be any point on AD such that AP : PD = 2 : 1

Hence the coordinates of P are (\(\frac{11}{3}\), \(\frac{11}{3}\))

(iii) Let BE and CF be respectively the medians of AABC to AC and AB.

∴ E and F are respectively the mid-points of AC and AB.

Now, Q divides BE such that

BQ : QE = 2 : 1

Also, R divides CF such that

(iv) From the above discussion it is clear that the coordinates of P, Q and R are the same and the three medians are concurrent at a point. This point is called the centroid of the triangle, which divides each median in the ratio 2 : 1.

(v) The vertices of the given △ABC are A(x_{1}, y_{1}), B(x_{2}, y_{2}) and C(x_{3}, y_{3}).

Let AD be a median of △ABC.

∴ D is the mid-point of BC. Then, the coordinates of D are :

(\(\frac{x_{2}+x_{3}}{2}\), \(\frac{y_{2}+y_{3}}{2}\))

Question 8.

ABCD is a rectangle formed by 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.

Solution:

According to the question the vertices of rectangle ABC’D are : A(-1, -1), B(-1, 4), C(5, 4) and D(5, -1) D(5, -1) and C(5, 4)

∵ P is the mid-point of AB

∴ Coordinates of P = (\(\frac{-1-1}{2}\), \(\frac{-1+1}{2}\))

Here all the sides of the quadrilateral PQRS are equal. But its diagonals are not eaual.

∴ PQRS is a rhombus.

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