Class 10 RD Sharma Solutions - Chapter 14 Coordinate Geometry - Exercise 14.3 | Set 2

The Coordinate Geometry is a crucial chapter in Class 10 Mathematics offering a blend of algebra and geometry. It provides students with the tools to analyze geometric shapes using the coordinate plane making it easier to study the properties and relationships of the points, lines, and figures. Exercise 14.3 focuses on applying the concepts of the distance formula section formula and area of the triangles helping students build a strong foundation in analytical geometry.

Coordinate Geometry

The Coordinate Geometry involves the study of geometric figures using the coordinate plane. It allows us to calculate distances between the points find midpoints and determine the area of the triangles and other figures using algebraic equations. This branch of geometry simplifies the analysis of shapes by converting geometric problems into algebraic ones making it an essential tool in mathematics.

Question 21. Find the ratio in which the point P(-1, y) lying on the line segment joining A(-3, 10) and B(6, -8) divides it. Also, find the value of y.

Solution:

Assume P divide A(-3, 10) and B(6, -8) in the ratio of k : 1

Given: coordinates of P as (-1, y)

After applying the section formula for x – coordinate,

We will get

-1 = \frac{6k – 3}{ k + 1}\\ -(k + 1) = 6k – 3\\ 7k = 2\\ k = \frac{2}{7}

Therefore, 

AB is divided by point P in the ratio of 2 : 7

By applying the value of k, to find the y-coordinate 

We will get

y = \frac{-8k + 10}{k + 1}\\ y = \left(\frac{-8\times\frac{2}{7} + 10)}{ \frac{2}{7} + 1}\right)

y = (-16 + 70)/(2 + 7) = 54/9

y = 6 

Hence, 

The y-coordinate of P is 6.

Question 22. Find the coordinates of a point A, where AB is the diameter of circle whose center is (2, -3) and B is (1, 4).

Solution:

Assume the coordinates of point A be (x, y)

Given: AB is the diameter, 

So the center in the mid-point of the diameter

Thus,

(2, -3) = (x + 1/ 2, y + 4/2)

2 = x + 1/2 and -3 = y + 4/2 

4 = x + 1 and -6 = y + 4

x = 3 and y = -10

Hence, the coordinates of A are (3, -10)

Question 23. If the points (-2, 1), (1, 0), (x, 3), and (1, y) form a parallelogram, find the values of x and y.

Solution:

Consider A(-2, 1), B(1, 0), C(x , 3) and D(1, y) are the given points of the parallelogram.

As we know that the diagonals of a parallelogram bisect each other.

Thus, 

Coordinates of mid-point of AC = Coordinates of mid-point of BD

((x - 2)/2, (3 - 1)/2) = (1+1)/2, (y + 0)/2 

((x - 2)/2, 1) = (1, y/2) 

(x - 2)/2 = 1

x - 2 = 2

x = 4

and y/2 = 1

y = 2

Hence, the value of x is 4 and the value of y is 2.

Question 24. The points A(2, 0), B(9, 1), C(11, 6), and D(4, 4) are the vertices of a quadrilateral ABCD. Determine whether ABCD is a rhombus or not.

Solution:

Given: A(2, 0), B(9, 1), C(11, 6) and D(4, 4).

Mid-point of AC coordinates are (11+\frac{2}{ 2},\ 6+\frac{0}{ 2}) = (\frac{13}{2},\ 3)

Mid-point of BD coordinates are (9+\frac{4}{2},\ 1+\frac{4}{2}) = (\frac{13}{2},\ \frac{5}{2})

Here, 

Coordinates of the mid-point of AC ≠ Coordinates of mid-point of BD, 

ABCD is not a parallelogram.

Hence, 

ABCD cannot be a rhombus too.

Question 25. In what ratio does the point (-4,6) divide the line segment joining the points A(-6,10) and B(3,-8)?

Solution:

Assume the line segment AB is divided by point (-4, 6) in the ratio of k : 1.

After applying the Section Formula, 

We will get

(-4,\ 6)=\left(\frac{3k-6}{k+1},\ \frac{-8k+10}{k+1}\right)\\ -4=\frac{3k-6}{k+1}

-4k -4 = 3k - 6

7k = 2

k : 1 = 2 : 7

We can also check for the y-coordinate also.

Hence, 

The ratio in which the line segment AB is divided by point (-4,6) is 2 : 7.

Question 26. Find the ratio in which the y-axis divides the line segment joining the points (5, -6) and (-1, -4). Also, find the coordinates of the point of division.

Solution:

Assume P(5, -6) and Q(-1, -4) be the given points.

Consider the line segment PQ is divided by y-axis in the ratio k : 1.

After applying the Section Formula for the x-coordinate (as it’s zero) 

We will get,

\frac{-k+5}{k+1}=0

-k + 5 = 0

k = 5

Therefore, 

The ratio in which the y-axis divides the given 2 points is 5 : 1

Now further, for finding the coordinates of the point of division

On putting k = 5, we will get

\left(\frac{-5+5}{5+1},\ \frac{-4\times5-6}{5+1}\right)=\left(0,\ \frac{-13}{3}\right)

Therefore, 

The coordinates of the point of division are (0, -13/3) 

Question 27. Show that A(-3, 2), B(-5, 5), C(2, -3), and D(4, 4) are the vertices of a rhombus.

Solution:

Given: A(-3, 2), B(-5, 5), C(2, -3) and D(4, 4)

Further,

Mid-point of AC coordinates are \left(-3+\frac{2}{ 2},\ 2\ -\ \frac{3}{ 2}\right) = \left(\frac{-1}{2},\ \frac{-1}{2}\right)

And,

Mid-point of BD coordinates are \left(-5+\frac{4}{ 2},\ -5+ \frac{4}{ 2}\right) = \left(\frac{-1}{2},\ \frac{-1}{2}\right)

Therefore, 

The mid-point for both the diagonals are the same. 

Thus, 

ABCD is a parallelogram.

Now, 

For the sides

AB=\sqrt{(-5+3)^2+(-5-2)^2}\\ AB=\sqrt{4+49}\\ AB=\sqrt{53}\\ BC=\sqrt{(-5-2)^2+(-5+3)^2}\\ BC=\sqrt{49+4}\\ BC=\sqrt{53}

AB = BC

We can see that ABCD is a parallelogram with adjacent sides equal.

Therefore, 

ABCD is a rhombus.

Question 28. Find the lengths of the medians of a ΔABC having vertices at A(0, -1), B(2, 1), and C(0, 3).

Solution:

Assume AD, BE and CF be the medians of ΔABC

Now,

Coordinates of D are \left(2+\frac{0}{ 2},\  1+\frac{3}{2}\right)   = (1, 2)

Coordinates of E are \left(\frac{0}{2},\ 3-\frac{1}{ 2}\right)   = (0, 1)

Coordinates of F are \left(2+\frac{0}{2},\ 1-\frac{1}{2}\right)   = (1, 0)

Further,

The length of the medians

Length of the median AD = \sqrt{(1-0)^2+(2+1)^2} = √10 units

Length of the median BE = \sqrt{(2-0)^2+(1-1)^2} = 2 units

Length of the median CF = \sqrt{(1-0)^2+(0-3)^2} = = √10 units

Question 29. Find the lengths of the median of a ΔABC having vertices at A(5, 1), B(1, 5), and C(-3, -1).

Solution:

Given: Vertices of ΔABC as A(5, 1), B(1, 5) and C(-3, -1).

Consider AD, BE and CF be the medians

Coordinates of D are \left(1-\frac{3}{ 2},\  5-\frac{1}{2}\right)   = (-1, 2)

Coordinates of E are \left(5-\frac{3}{ 2},\  1-\frac{1}{2}\right)   = (1, 0)

Coordinates of F are \left(5+\frac{1}{ 2},\  1+\frac{5}{2}\right)   = (3, 3)

Further, 

The length of the medians

Length of the median AD = \sqrt{(5+1)^2+(1-2)^2} = √37 units

Length of the median BE = \sqrt{(1-1)^2+(5-0)^2} = 5 units

Length of the median CF = \sqrt{(3+3)^2+(3+1)^2} = √52 units

Question 30. Find the coordinates of the point which divide the line segment joining the points (-4, 0) and (0, 6) in four equal parts.

Solution:

Consider A(-4, 0) and B(0, 6) as they are the given points

And, 

Assume P, Q and R be the points which divide AB is four equal points, as shown in the fig.

Thus, 

As we know that AP : PB = 1 : 3

By applying the Section Formula the coordinates of P are

\left(\frac{1\times0+3(-4)}{1+3},\ \frac{1\times6+3\times0}{1+3}\right)=\left(-3,\ \frac{3}{2}\right)

And,

We can see that Q is the mid-point of AB

Thus, the coordinates of Q are

\left(\frac{-4+0}{2},\ \frac{0+6}{2}\right)=(-2, 3)

Finally, 

The ratio of AR : BR is 3 : 1

Then, after applying the Section Formula the coordinates of R are

\left(\frac{3\times0+1\times(-4)}{3+1},\ \frac{3\times6+1\times0}{3+1}\right)=\left(-1,\ \frac{9}{2}\right)

Question 31. Show that the mid-point of the line segment joining the points (5, 7) and (3, 9) is also the mid-point of the line segment joining the points (8, 6) and(0, 10).

Solution:

Assume M be the mid-point of AB. Coordinates of the mid-point of this line segment joining two points A (5, 7) and B (3, 9).

=\left(\frac{x_1+x_2}{2},\ \frac{y_1+y_2}{2}\right)\\ =\left(\frac{5+3}{2},\ \frac{7+9}{2}\right)\\ =\left(\frac{8}{2},\ \frac{16}{2}\right)\\ =(4,\ 8)

Now coordinates of the mid-point of the line segment joining the points (8, 6) and (0, 10) are;

=\left(\frac{8+0}{2},\ \frac{6+10}{2}\right)\\ =\left(\frac{8}{2},\ \frac{16}{2}\right)\\ =(4,\ 8)

Thus, this is the same as the first case.

Question 32. Find the distance of the point (1, 2) from the mid-point of the line segment joining the points (6, 8) and (2, 4).

Solution:

Assume M be the mid-point of the line segment joining the points (6, 8) and (2, 4)

Now 

Coordinates of M will be

=\left(\frac{x_1+x_2}{2},\ \frac{y_1+y_2}{2}\right)\\ =\left(\frac{6+3}{2},\ \frac{8+4}{2}\right)\\ =\left(\frac{8}{2},\ \frac{12}{2}\right)\\ =(4,\ 6)

Now, 

Distance between the points (4, 6) and (1, 2)

=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\\ =\sqrt{(1-4)^2+(2-6)^2}\\ =\sqrt{(-3)^2+(-4)^2}\\ =\sqrt{9+16}\\ =\sqrt{25} = 5 units

Question 33. If A and B are (1, 4) and (5, 2) respectively, find the co-ordinates of P When \frac{AP}{BP}=\frac{3}{4}

Solution:

Here, Point P divides the line segment joining the points (1, 4) and (5, 2) in the ratio of AP : PB = 3 : 4

Coordinates of P will be

=\left(\frac{mx_1+nx_2}{m+n},\ \frac{my_1+ny_2}{m+n}\right)\\ =\left(\frac{3\times5+4\times1}{3+4},\ \frac{3\times2+4\times4}{3+4}\right)\\ =\left(\frac{15+4}{7},\ \frac{6+16}{7}\right)\\ =\left(\frac{19}{7},\ \frac{22}{7}\right)

Question 34. Show that the points A (1, 0), B (5, 3), C (2, 7), and D (-2, 4) are the vertices of a parallelogram.

Solution:

If ABCD is a parallelogram, 

Then its diagonal AC and BD will bisect each other at O

Consider O is the mid-point of AC, 

Then coordinates of O will be;

=\left(\frac{x_1+x_2}{2},\ \frac{y_1+y_2}{2}\right)\\ =\left(\frac{1+2}{2},\ \frac{0+7}{2}\right)\\ =\left(\frac{3}{2},\ \frac{7}{2}\right)

And assume O is the mid-point of BD,

Then coordinates of O will be;

=\left(\frac{5+(-2)}{2},\ \frac{3+4}{2}\right)\\ =\left(\frac{5-2}{2},\ \frac{3+4}{2}\right)\\ =\left(\frac{3}{2},\ \frac{7}{2}\right)

We see that coordinates of the mid-points of AC and BD are same

Therefore, AC and BD  bisect each other at O

Now, length of AC 

=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\\ =\sqrt{(2-1)^2+(7-0)^2}\\ =\sqrt{1^2+7^2}\\ =\sqrt{1+49}\\ =\sqrt{50}

and length of BD = =\sqrt{(-2-5)^2+(4-3)^2}\\ =\sqrt{(-7)^2+(1)^2}\\ =\sqrt{49+1}\\ =\sqrt{50}

We can see that AC = BD

Therefore, ABCD is a rectangle.

Question 35. Determine the ratio in which the point P (m, 6) divides the join of A (-4, 3) and B (2, 8). Also, find the value of m. 

Solution:

Assume the ratio be r : s in which P (m, 6) divides the line segment joining the points A (-4, 3) and B (2, 8)

Therefore,

m=\frac{rx_2+sx_1}{r+s}

and 6=\frac{ry_2+sy_1}{r+s}\\ =\frac{r\times8+s\times3}{r+s}\\ =\frac{8r+3s}{r+s}=6

⇒ 8r + 3s = 6r + 6s

⇒ 8r - 6r = 6s - 3s

⇒ 2r = 3s

⇒ \frac{r}{s}=\frac{3}{2}

Therefore,

Ratio is 3 : 2

Now, 

m=\frac{3\times2+2\times(-4)}{3+2}\\ =\frac{6-8}{5}\\ =\frac{-2}{5}

Hence, m = -2/5 

Question 36. Determine the ratio in which the point (-6, a) divides the join of A (-3, -1) and B (-8, 9). Also, find the value of a.

Solution:

Assume the point P (-6, a) divides the join of A (-3, -1) and B (-8, 9) in the ratio m : n

Therefore,

-6=\frac{mx_2+nx_1}{m+n}\\ =\frac{m(-8)+n(-3)}{m+n}

-6 = (-8m -3n)/(m + n)

-6m - 6n = -8m - 3n

8m - 6m = 6n - 3n

2m = 3n

m/n = 3/2

Therefore,

Ratio = 3 : 2

and

a=\frac{my_2+ny_1}{m+n}=\frac{3\times9+2\times(-1)}{3+2}\\ =\frac{27-2}{5}\\ =\frac{25}{5}\\ =5

Question 37. 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 mid-points of sides AB, BC, CD, and DA respectively. Is the quadrilateral PQRS a square ? a rectangle ? or a rhombus? Justify your answer.

Solution:

ABCD is a rectangle whose vertices are A (-1,-1), B (-1,4), C (5, 4) and D (5, -1). 

P, Q, R, and S are the mid-points of the sides AB, BC, CD and DA respectively 

And are joined PR and QS are also joined.

Now coordinates of P will be

=\left(\frac{x_1+x_2}{2},\ \frac{y_1+y_2}{2}\right)\\ =\left(\frac{-1-1}{2},\ \frac{-1+4}{2}\right)\\ =\left(\frac{-2}{2},\ \frac{3}{2}\right)\\ =(-1,\ \frac{3}{2})

Similarly, the coordinates of Q, will be:

=\left(\frac{x_1+x_2}{2},\ \frac{y_1+y_2}{2}\right)\\ =\left(\frac{-1+5}{2},\ \frac{4+4}{2}\right)\\ =\left(\frac{4}{2},\ \frac{8}{2}\right)\\ =(2,\ 4)

Coordinates of R will be:

=\left(\frac{x_1+x_2}{2},\ \frac{y_1+y_2}{2}\right)\\ =\left(\frac{5+5}{2},\ \frac{4-1}{2}\right)\\ =\left(\frac{10}{2},\ \frac{3}{2}\right)\\ =(5,\ \frac{3}{2})

Coordinates of S will be:

=\left(\frac{x_1+x_2}{2},\ \frac{y_1+y_2}{2}\right)\\ =\left(\frac{5-1}{2},\ \frac{-1-1}{2}\right)\\ =\left(\frac{4}{2},\ \frac{-2}{2}\right)\\ =(2,\ -1)

Coordinates of P (-1, 3/2), Q (2, 4), R(5, 3/2) and S (2, -1)

PQ=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\\ =\sqrt{(2+1)^2+(4-\frac{3}{2})^2}\\ =\sqrt{(3)^2+(\frac{5}{3})^2}\\ =\sqrt{9+\frac{25}{4}}\\ =\sqrt{\frac{36+25}{4}}\\ =\sqrt{\frac{61}{4}}\\ =\frac{\sqrt{61}}{2}

Now, assume the diagonals PQ and QS intersect each other at O

Assume O is the mid-point of PR,

Then coordinates of O will be

\left(\frac{-1+5}{2},\ \left(\frac{3}{2}+\frac{3}{2}\right)\frac{1}{2}\right)\\ =\left(\frac{4}{2},\ \frac{3}{2}\right)\\ =\left(2,\ \frac{3}{2}\right)

Similarly, of O is the mid-point of QS, then the coordinates of O will be

\left(\frac{2+2}{2},\ \frac{4+(-1)}{2}\right)\\ =\left(\frac{4}{2},\ \frac{3}{2}\right)\\ =\left(2,\ \frac{3}{2}\right)

Now, we see that the coordinates of O in both case is same and adjacent sides are also equal

Then it may be a square or a rhombus

Now length of PR = \sqrt{(5+1)^2+\left(\frac{3}{2}-\frac{3}{2}\right)^2}\\ =\sqrt{(6)^2+(0)^2}\\ =\sqrt{36+0}\\ =\sqrt{36}\\ =6

And length of OS

\sqrt{(2-2)^2+(-1-4)^2}\\ =\sqrt{(0)^2+(-5)^2}\\ =\sqrt{0+25}\\ =\sqrt{25}\\ =5

Because diagonal are not equal

Hence, PQRS is a rhombus.

Question 38. Points P, Q, R, and S divide the line segment joining the points A (1, 2) and B (6, 7) in 5 equal parts. Find the coordinates of the points P, Q, and R.

Solution:

Points P, Q, R and S divides AB in 5 equal parts and assume coordinates of P, Q, R and S are,

(x₁, y₁), (x₂, y₂), (x₃, y₃), (x₄, y₄) 

x=\frac{m_1x_2+m_2x_1}{m_1+m_2}

⇒ P divides AB in ratio 1 : 4

Therefore,

x_1=\frac{1\times6+4\times1}{1+4}\\ =\frac{6+4}{5}\\ =\frac{10}{5}\\ =2\\ y_1=\frac{1\times7+4\times2}{1+4}\\ =\frac{7+8}{5}\\ =\frac{15}{5}\\ =3

Hence, Coordinates of P are (2, 3)

⇒ Q divides AB in the ratio 2 : 3

Therefore,

x_2=\frac{2\times6+3\times1}{2+3}\\ =\frac{12+3}{5}\\ =\frac{15}{5}\\ =3\\ y_2=\frac{2\times7+3\times2}{2+3}\\ =\frac{14+6}{5}\\ =\frac{20}{5}\\ =4

Hence, Coordinates of 3, 4

⇒ R divides AB in ration 3 : 2

Therefore,

x_3=\frac{3\times6+2\times1}{3+2}\\ =\frac{18+2}{5}\\ =\frac{20}{5}\\ =4\\ y_2=\frac{3\times7+2\times2}{3+2}\\ =\frac{21+4}{5}\\ =\frac{25}{5}\\ =5

Hence, Coordinates of R are (4, 5).

Question 39. If A and B are two points having coordinates (-2, -2) and (2, -4) respectively, find the coordinates of P such that AP = 3/7 AB

Solution:

AP = 3/7 AB

7AP = 3AB

7AP = 3(AP + BP)

⇒ 7AP = 3AP + 3BP

⇒ 7AP - 3AP = 3BP

⇒ 4 AP = 3 BP

⇒ \frac{AP}{BP}=\frac{3}{4}

Therefore, 

AP : BP = 3 : 4

Because P divides AB in the ratio of 3 : 4 whose end points are A(-2, -2) and B(2, -4)

Therefore, Coordinates of P will be

x=\frac{mx_2+nx_1}{m+n}\\ y=\frac{my_2+ny_1}{m+n}\\ x=\frac{3(2)+4(-2)}{3+4}\\ =\frac{6-8}{7}\\ =\frac{-2}{7}

\\ y=\frac{3(-4)+4(-2)}{3+4}\\ =\frac{-12-8}{7}\\ =\frac{-20}{7}

Therefore,

Coordinates of P will be \left(\frac{-2}{7},\ \frac{-20}{7}\right)

Question 40. Find the coordinates of the points which divide the line segment joining A (-2, 2) and B (2, 8) into four equal parts.

Solution:

Assume P, Q and R divides the line segment AB in four equal parts

Co-ordinates of A are (-2, 2) and of B are (2, 8)

It can be seen that Q divides AB in two equal parts while P bisects AQ and R, bisect QB.

Now,

Coordinates of Q will be :

\frac{x_1+x_2}{2},\ \frac{y_1+y_2}{2}\\ =\left(\frac{-2+2}{2},\ \frac{2+8}{2}\right)\\ =\left(\frac{0}{2},\ \frac{10}{2}\right)\\ =(0, 5)

Similarly, coordinates of P will be:

\frac{x_1+x_2}{2},\ \frac{y_1+y_2}{2}\\ =\left(\frac{-2+0}{2},\ \frac{2+5}{2}\right)\\ =\left(\frac{-2}{2},\ \frac{7}{2}\right)\\ =(-1, \frac{7}{2})

Coordinates of R will be:

\frac{x_1+x_2}{2},\ \frac{y_1+y_2}{2}\\ =\left(\frac{0+2}{2},\ \frac{5+8}{2}\right)\\ =\left(\frac{2}{2},\ \frac{13}{2}\right)\\ =(1, \frac{13}{2})

Hence, Coordinates of P are(1, 7/2) 

Coordinates of Q are (0, 5)

Coordinates of R are (1, 13/2)

Read More:

• Coordinate Geometry
• Uses of Co-Ordinate Geometry

Conclusion

Exercise 14.3 | Set 2 of Chapter 14 in RD Sharma's Class 10 Mathematics textbook provides the students with an opportunity to deepen their understanding of the coordinate geometry by the applying key formulas to the solve various problems. This exercise not only reinforces the concepts learned but also prepares students for more complex geometrical problems. The Mastery of these topics is crucial for success in both the board examinations and future mathematical studies.