# NCERT Solutions For Class 10 Maths Chapter 7 Exercise 7.2

Coordinate Geometry

Exercise 7.2

## Chapter 7 Ex.7.2 Question 1

Find the coordinates of the point which divides the join of \((-1, 7)\) and \((4, -3)\) in the ratio \(2:3\).

**Solution**

**Video Solution**

**Reasoning:**

The coordinates of the point \(P(x, y)\) which divides the line segment joining the points \(A(x_1, y_1)\) and \(B(x_2, y_2)\), internally, in the ratio \(m_1 : m_2\) is given by the Section Formula

\[\begin{align}\!P(x,y) \!=\! \left[ {\frac{{mx_2} \!+\! {nx_1}}{{m \!+ n}},\frac{{my_2} \!+\! {ny_1}}{{m} +\! {n}}}\! \right]\! \end{align}\]

**What is Known?**

The \(x\) and \(y\) co-ordinates of the points which is to be divided in the ratio \(2:3\)

**What is Unknown?**

The coordinates of the point which divides the join of \((-1, 7)\) and \((4, -3)\) in the ratio \(2:3\)

**Steps:**

Given,

- Let \(P(x, y)\) be the required point.

- Let \(A(-1, 7)\) and \(B(4, -3)\)
- \(m: n = 2:3\)
- Hence
- \(x_1 = -1\)
- \(y_1 = 7\)
- \(x_2 = 4\)
- \(y_2 = -3\)

By Section formula

\[\begin{align}\!P(x,y) \!=\! \left[ {\frac{{mx_2} \!+\! {nx_1}}{{m \!+ n}},\frac{{my_2} \!+\! {ny_1}}{{m} +\! {n}}}\! \right]\!\;\;\dots(1) \end{align}\]

By substituting the values in the Equation (1)

\[\begin{align}x &= \frac{{2 \times 4 + 3x( - 1)}}{{2 + 3}} \\x &= \frac{{8 - 3}}{5}\\x &= \frac{5}{5} = 1\\\\ y &= \frac{{2 \times ( - 3) + 3 \times 7}}{{2 + 3}} \\y &= \frac{{ - 6 + 21}}{5} \\y &= \frac{{15}}{5} = 3\end{align}\]

Therefore, the co-ordinates of point \(P\) are \((1, 3)\).

## Chapter 7 Ex.7.2 Question 2

Find the coordinates of the points of trisection of the line segment joining \((4, -1)\) and \((-2, -3)\).

**Solution**

**Video Solution**

**Reasoning:**

The coordinates of the point \(P(x, y)\) which divides the line segment joining the points \(A(x1, y1)\) and \(B(x2, y2)\), internally, in the ratio \(m1 : m2\) is given by the Section Formula.

**What is Known?**

The \(x\) and \(y\) co-ordinates of the line segment joining \((4, -1)\) and \((-2, -3)\).

**What is Unknown?**

The coordinates of the points of trisection of the line segment joining \((4, -1)\) and \((-2, -3)\).

**Steps:**

Given,

- Let line segment joining the points be \(A(4, -1)\) and \(B(-2, -3)\).
- Let \(P (x_1, y_1)\) and \(Q(x_2, y_2)\) be the points of trisection of the line segment joining the given points i.e., \(AP = PQ = QB\)

By Section formula

\[\begin{align}P(x,y)\! =\! \left[\! {\frac{mx_2 + nx_1}{m+n},\!{\frac{my_2 +\! ny_1}{m+ n}}}\right]\;\;\dots(1)

\end{align}\]

Therefore, by observation point \(P\) divides \(AB\) internally in the ratio \(1:2\).

- Hence \(m: n = 1:2\)

By substituting the values in the Equation (1)

\[\begin{align}x_1 & \! = \! \frac{{1 \! \times \! ( - 2) \! + \! 2 \! \times \! 4}}{{1 \! + \! 2}}\\x_1 & \! = \! \frac{{ - 2 \! + \! 8}}{3} \\& \! = \! \frac{6}{3} \\& \! = \! 2\\\\ y_1 & \! = \! \frac{{1 \! \times \! ( - 3) \! + \! 2 \! \times \! ( - 1)}}{{1 \! + \! 2}}\\ y_1 & \! = \! \frac{{ - 3 - 2}}{3} \\& \! = \! \frac{{ - 5}}{3}\end{align}\]

Therefore,

\(\begin{align}{\text{P}}\left( x_1,y_1 \right) = \left( {2,\frac{{ - 5}}{3}} \right)\end{align}\)

Therefore, by observation point \(Q\) divides \(AB\) internally in the ratio \(2:1\).

- Hence \(m:n = 2:1\)

By substituting the values in the Equation (1)

\[\begin{align}x_2 & \! = \! \frac{{2 \! \times \! ( - 2) \! + \! 1 \! \times \! 4}}{{2 \! + \! 1}}\\x_2 & \! = \! \frac{{ - 4 \! + \! 4}}{3} \\& \! = \! 0\\\\y_2 & \! = \! \frac{{2x( - 3) \! + \! 1 \! \times \! ( - 1)}}{{2 \! + \! 1}}\\ y_2 & \! = \! \frac{{ - 6 - 1}}{3} \\& \! = \! \frac{{ - 7}}{3}\end{align}\]

Therefore,

\[\begin{align}Q( x_2,y_2) = \left( {0, - \frac{7}{3}} \right)\end{align}\]

Hence the points of trisection are

\[\begin{align}P( x_1,y_1) &= \left( {2,\frac{{ - 5}}{3}} \right) \\&\text{and}\;\\ Q(x_2,y_2) &= \left( {0, - \frac{7}{3}} \right)\end{align}\]

## Chapter 7 Ex.7.2 Question 3

To conduct Sports Day activities, in your rectangular shaped school ground \(ABCD\), lines have been drawn with chalk powder at a distance of \(1\rm\,m \) each. \(100\) flower pots have been placed at a distance of \(1\rm\,m \) from each other along \(AD\), as shown in the following figure. Niharika runs \(\begin{align}\frac{1}{4}\end{align}\)th the distance \(AD\) on the \(2^\rm {nd}\) line and posts a green flag. Preet runs \(\begin{align}\frac{1}{5}\end{align}\)th the distance \(AD\) on the eighth line and posts a red flag. What is the distance between both the flags? If Rashmi has to post a blue flag exactly halfway between the line segments joining the two flags, where should she post her flag?

**Solution**

**Video Solution**

**Reasoning:**

The coordinates of the point \(P(x, y)\) which divides the line segment joining the points \(A(x_1, y_1)\) and \(B(x_2, y_2)\), internally, in the ratio \(m_1 : m_2\) is given by the Section Formula.

**What is Known?**

- The school ground \(ABCD\) is rectangular shaped.
- Lines are drawn at a distance of \(1\rm\,m \) each and \(100\) flower pots have been placed at a distance of \(1\rm\,m \) each along \(AD\).
- The distance covered by Niharika and Preet on line \(AD\).

**What is Unknown?**

- The distance between the flags posted by Niharika and Preet.
- The position on the line segment where Rashmi has to post the flag.

**Steps:**

From the Figure,

Given,

- By observation, that Niharika posted the green flag at of the distance \(P\) i.e., \(\begin{align}\left( {\frac{1}{4} \times 100} \right)m = 25\,m\end{align}\) from the starting point of \(2^\rm{nd}\) Therefore, the coordinates of this point \(P\) is \((2, 25)\).

- Similarly, Preet posted red flag at \(\frac{{1}}{5}\) of the distance \(Q\) i.e., \(\begin{align}\left( {\frac{1}{5} \times 100} \right)m = 20\,m\end{align}\) from the starting point of \(8^\rm {th}\) Therefore, the coordinates of this point \(Q\) are \((8, 20)\)

We know that the distance between the two points is given by the Distance Formula,

\[\begin{align}&\sqrt {{{\left( x_1 - x_2 \right)^2 + \left(y_1 - y_2 \right)}^2}} \;\;\dots(1)\end{align}\]

To find the distance between these flags \(PQ\) by substituting the values in Equation (1),

\(\begin{align}PQ &= \sqrt {{{(8 - 2)}^2} + {{(25 - 20)}^2}} \\ &= \sqrt {36 + 25} \\ &= \sqrt {61}\text{ m} \end{align}\)

- The point at which Rashmi should post her blue flag is the mid-point of the line joining these points.
- Let this point be \(M \;(x, y)\).

By Section formula

\[\begin{align} P(x,y)& \!=\! \left[ {\frac{{mx_2 \! + \! nx_1}}{{m \! + \! n}}\!,\!\frac{{my_2 \! + \! ny_1}}{{m \! + \! n}}} \!\right]\;\;\dots(2) \; \end{align}\]

\(\begin{align}x &= \frac{{2 + 8}}{2}, \qquad y = \frac{{25 + 20}}{2}\\x &= \frac{{10}}{2}, \qquad\quad\; y = \frac{{45}}{2}\\x &= 5, \qquad\qquad \;y = 22.5\end{align}\)

Therefore, Rashmi should post her blue flag at \(22.5\;\rm m\) on \(5^\rm {th}\) line

## Chapter 7 Ex.7.2 Question 4

Find the ratio in which the line segment joining the points \((-3, 10)\) and \((6, -8)\) is divided by \((-1, 6)\).

**Solution**

**Video Solution**

**Reasoning:**

The coordinates of the point \(P(x, y)\) which divides the line segment joining the points \(A(x_1, y_1)\) and \(B(x_2, y_2)\), internally, in the ratio \(m_1 : m_2\) is given by the Section Formula.

\[\begin{align} P(x,y)& \!=\! \left[ {\frac{{mx_2 \! + \! nx_1}}{{m \! + \! n}}\!,\!\frac{{my_2 \! + \! ny_1}}{{m \! + \! n}}} \!\right]\end{align}\]

**What is Known?**

The \(x\) and \(y\) co-ordinates of the line segment which is divided by the point \((-1, 6)\).

**What is Unknown?**

The ratio in which the line segment joining the points \((-3, 10)\) and \((6, -8)\) is divided by \((-1, 6).\)

**Steps:**

From the figure,

Given,

- Let the ratio in which the line segment joining \(A(-3, 10)\) and \(B(6, -8)\) is divided by point \(P(-1, 6)\) be \(k:1\).

By Section formula

\[\begin{align} P(x,y)& \!=\! \left[ {\frac{{mx_2 \! + \! nx_1}}{{m \! + \! n}}\!,\!\frac{{my_2 \! + \! ny_1}}{{m \! + \! n}}} \!\right]\;\;\dots(2) \; \end{align}\]

Therefore,

\[\begin{align} - 1 &= \frac{{6k - 3}}{{k + 1}}\\ - k - 1 &= 6k - 3\\7k &= 2\end{align}\]

By Cross Multiplying & Transposing

\[\begin{align}k &= \frac{2}{7}\end{align}\]

Hence the point \(P\) divides \(AB\) in the ratio \(2:7\)

## Chapter 7 Ex.7.2 Question 5

Find the ratio in which the line segment joining \(A(1, -5)\) and \(B(-4, 5)\) is divided by the \(x\)-axis. Also find the coordinates of the point of division.

**Solution**

**Video Solution**

**Reasoning:**

The coordinates of the point \(P(x, y)\) which divides the line segment joining the points \(A(x_1, y_1)\) and \(B(x_2, y_2)\), internally, in the ratio \(m_1 : m_2\) is given by the Section Formula.

**What is Known?**

The \(x\) and \(y\) co-ordinates of the line segment which is divided by the \(x\)-axis.

**What is Unknown?**

The ratio in which the line segment joining \(A(1, -5)\) and \(B(-4, 5)\) is divided by the \(x\)-axis and the coordinates of the point of division

**Steps:**

From the Figure,

Given,

- Let the ratio be \(k : 1\).
- Let the line segment joining \(A(1, -5)\) and \(B(-4, 5)\)

By Section formula

\[\begin{align}{{P(x,}}\,{{y)}} & \!=\! \left[\! {\frac{{{{m}}{{{x}}_2} \!+\! {{n}}{{{x}}_1}}}{{{{m}} \!+\! {{n}}}},\frac{{{{m}}{{{y}}_2} \!+\! {{n}}{{{y}}_1}}}{{{{m}} \!+\! {{n}}}}}\! \right] \;\;\dots(1) \\ &\begin{bmatrix}\text{By substituting the values} \\ \text{in Equation (1)}\end{bmatrix} \end{align}\]

Therefore, the coordinates of the point of division is \(\begin{align}\left[{\frac{{ - 4k + 1}}{{k + 1}},\;\frac{{5k - 5}}{{k + 1}}} \right]\end{align}\)

We know that \(y\)-coordinate of any point on \(x\)-axis is \(0\).

\[\begin{align}∴\; \frac{{5{{k}} - 5}}{{{{k}} + 1}} &= 0\\\;\;\;5{{k}} - 5 &= 0\\\;\;\;\, \to 5{{k}}& = 5

\\&\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\! \begin{bmatrix}\text{By Cross Multiplying}\\\text { & Transposing}\end{bmatrix} \\\\{{k}} &= 1\end{align}\]

Therefore, \(x\)-axis divides it in the ratio \(1:1\).

Division point

\[\begin{align}&= \left[ {\frac{{ - 4(1) + 1}}{{1 + 1}},\frac{{5(1) + 5}}{{1 + 1}}} \right]\\ &= \left[ {\frac{{ - 4 + 1}}{2},\frac{{5 + 5}}{2}} \right]\\ &= \left[ {\frac{{ - 3}}{2},0} \right]\end{align}\]

## Chapter 7 Ex.7.2 Question 6

If \((1, 2)\), \((4, y)\), \((x, 6)\) and \((3, 5)\) are the vertices of a parallelogram taken in order, find \(x\) and \(y\).

**Solution**

**Video Solution**

**Reasoning:**

\[\begin{align}{{P(x,}}\,{{y)}} & \!=\! \left[\! {\frac{{{{m}}{{{x}}_2} \!+\! {{n}}{{{x}}_1}}}{{{{m}} \!+\! {{n}}}},\frac{{{{m}}{{{y}}_2} \!+\! {{n}}{{{y}}_1}}}{{{{m}} \!+\! {{n}}}}}\! \right]\end{align}\]

**What is the known?**

The \(x\) and \(y\) co-ordinates of the vertices of the parallelogram.

**What is the unknown?**

The missing \(x\) and \(y\) co-ordinate.

**Steps:**

From the Figure,

Given,

- Let \(A (1, 2)\), \(B (4, y)\), \(C(x, 6)\), and \(D(3, 5)\) are the vertices of a parallelogram \(ABCD\).
- Since the diagonals of a parallelogram bisect each other, Intersection point \(O\) of diagonal \(AC\) and \(BD\) also divides these diagonals

Therefore, \(O\) is the mid-point of \(AC\) and \(BD\).

If \(O\) is the mid-point of \(AC\), then the coordinates of \(O\) are

\(\begin{align}\left( {\frac{{1 + x}}{2},\;\frac{{2 + 6}}{2}} \right) \Rightarrow \left( {\frac{{x + 1}}{2},\;4} \right)\end{align}\)

If \(O\) is the mid-point of \(BD\), then the coordinates of \(O\) are

\(\begin{align}\left( {\frac{{4 + 3}}{2},\;\frac{{5 + y}}{2}} \right) \Rightarrow \left( {\frac{7}{2},\;\frac{{5 + y}}{2}} \right)\end{align}\)

Since both the coordinates are of the same point \(O\),

\(\begin{align} &\therefore \frac{{x + 1}}{2} = \frac{7}{2}\;\;{\text{ and }}\;\;4 = \frac{{5 + y}}{2} \end{align}\)

\(\begin{align} &\Rightarrow x + 1 = 7\;\;{\text{ and }}\;\;5 + y = 8 \, \end{align}\) (By cross multiplying & transposing)

\(\begin{align} &\Rightarrow x = 6\;\;{\text{ and }}\;\; y = 3\end{align}\)

## Chapter 7 Ex.7.2 Question 7

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**

**Video Solution**

**Reasoning:**

**What is the known?**

The \(x\) and \(y\) co-ordinates of the center of the circle and one end of the diameter \(B\).

**What is the unknown?**

The coordinates of a point \(A\).

**Steps:**

From the Figure,

Given,

- Let the coordinates of point \(A\) be \((x, y)\).
- Mid-point of \(AB\) is \(C\) \((2, -3)\), which is the center of the circle.

\(\begin{align} \therefore \; (2, - 3) &= \left( {\frac{{x + 1}}{2},\frac{{y + 4}}{2}} \right) \end{align}\)

\(\begin{align} \Rightarrow \frac{{x + 1}}{2}& = 2\;\;{\text{ and }}\;\;\frac{{y + 4}}{2} = - 3 & & \end{align}\) (By Cross multiplying & transposing)

\(\begin{align} \Rightarrow x + 1 &= 4\;\;{\text{ and }}\;\;y + 4 = - 6 \end{align}\)

\(\begin{align} \Rightarrow x &= 3\;\;{\text{ and }}\;\;y = - 10\end{align}\)

Therefore, the coordinates of \(A\) are \((3, -10)\)

## Chapter 7 Ex.7.2 Question 8

If \(A\) and \(B\) are \((-2, -2)\) and \((2, -4)\), respectively, find the coordinates of \(P\) such that \(\begin{align}AP= \frac{3}{7}AB\end{align}\) and \(P\) lies on the line segment \(AB\).

**Solution**

**Video Solution**

**Reasoning:**

\[\begin{align}{{P}}({{x}},{{y}}) \!=\! \left[\! {\frac{{{{m}}{{{x}}_2} \!+\! {{n}}{{{x}}_1}}}{{{{m}} \!+ \!{{n}}}}\!,\!\frac{{{{m}}{{{y}}_2} \!+\! {{n}}{{{y}}_1}}}{{{{m}} \!+\! {{n}}}}} \!\right] \! \end{align}\]

**What is the known?**

The \(x\) and \(y\) co-ordinates of the points \(A\) and \(B\).

The ratio in which \(P\) divides \(AB\).

**What is the unknown?**

Co-ordinates of \(P\)

**Steps:**

From the Figure,

Given,

- The coordinates of point \(A\) and \(B\) are \((-2, -2)\) and \((2, -4)\) respectively.
- \(\begin{align}{{AP}} = \frac{3}{7}{{AB}}\end{align}\)

Hence \(\begin{align}\frac{{{{AB}}}}{{{{AP}}}} = \frac{7}{3}\end{align}\)

We know that \(\begin{align}AB = AP + PB\end{align}\)

from figure,

\[\begin{align}\frac{{{{AP}} + {{PB}}}}{{{{AP}}}} &= \frac{{3 + 4}}{3}\\1 + \frac{{{{PB}}}}{{{{AP}}}}& = 1 + \frac{4}{3}\\\frac{{{{PB}}}}{{{{AP}}}}& = \frac{4}{3}\end{align}\]

Therefore, \(AP:PB = 3:4\)

Point \(P(x, y)\) divides the line segment \(AB\) in the ratio \(3:4.\) Using Section Formula,

Coordinates of

\[\begin{align}P({{x}},{{y}}) & \! = \! \begin{bmatrix}\left({\frac{{3 \times 2 + 4 \times ( - 2)}}{{3 + 4}}}\right), \\ \left(\frac{{3 \times ( - 4) + 4 \times ( - 2)}}{{3 + 4}}\right) \end{bmatrix}\\ & \! = \! \left[ {\frac{{6 - 8}}{7},\;\frac{{ - 12 - 8}}{7}} \right]\\ & \! = \! \left[ { - \frac{2}{7},\; - \frac{{20}}{7}} \right]\end{align}\]

## Chapter 7 Ex.7.2 Question 9

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

**Solution**

**Video Solution**

**Reasoning:**

**What is the known?**

The \(x\) and \(y\) co-ordinates of the points \(A\) and \(B\).

**What is the unknown?**

The coordinates of the points which divide the line segment joining \(A \;(-2, 2)\) and \(B\;(2, 8)\) into four equal parts.

**Steps:**

From the Figure,

By observation, that points \(P\), \(Q\), \(R\) divides the line segment \(A(-2, 2)\) and \(B(2, 8)\) into four equal parts

Point \(P\) divides the line segment \(AQ\) into two equal parts

Hence, Coordinates of

\[\begin{align}\!\! P \!&\!=\!\! \!\left[ \!{\frac{{\!1\! \times \!2 \!+\! 3 \!\times \!( - 2)}}{{1 + 3}}\!,\!\frac{{1 \times \!8\! + \!3\! \times\!2}}{{1 + 3}}}\!\! \right]\!\! \\&= \left[ { - 1,\;\frac{7}{2}} \right]\end{align}\]

Point \(Q\) divides the line segment \(AB\) into two equal parts

Coordinates of

\[\begin{align} Q &= \left[ {\frac{{2 + ( - 2)}}{2},\;\frac{{2 + 8}}{2}} \right]\\ &= (0,\;5)\end{align}\]

Point \(R\) divides the line segment \(BQ\) into two equal parts

Coordinates of

\[\begin{align}\!\! R &\!=\!\left[\! {\frac{{3\! \times\! 2\! + \!1 \!\times\! ( -\! 2)}}{{3 + 1}}\!,\!\frac{{3\! \times\! 8\! + \!1\! \times\! 2}}{{3 + 1}}}\! \right]\;\\ &=\left[1, \frac{13}{2}\right]\end{align}\]

## Chapter 7 Ex.7.2 Question 10

Find the area of a rhombus if its vertices are \((3, 0)\), \((4, 5)\), \((-1, 4)\) and \((-2, -1)\) taken in order.

[Hint: Area of a rhombus \(=\) (product of its diagonals)]

**Solution**

**Video Solution**

**Reasoning:**

A rhombus has all sides of equal length and opposite sides are parallel.

**What is the known?**

The \(x\) and \(y\) co-ordinates of the vertices of the rhombus.

**What is the unknown?**

The area of the rhombus

**Steps:**

From the Figure,

Given,

- Let \(A(3, 0)\), \(B(4, 5)\), \(C(-1, 4)\) and \(D(-2, -1)\) are the vertices of a rhombus \(ABCD\).

We know that the distance between the two points is given by the Distance Formula,

\[\begin{align}&\sqrt {{{\left( {{{{x}}_1} - {{{x}}_2}} \right)}^2} + {{\left( {{{{y}}_1} - {{y}_2}} \right)}^2}}\;\;\dots(1)\end{align}\]

Therefore, distance between \(A(3, 0)\) and \(C (-1, 4)\) is given by

Length of diagonal

\[\begin{align} AC &= \sqrt {{{[3 - ( - 1)]}^2} + {{(0 - 4)}^2}} \\ &= \sqrt {16 + 16} = 4\sqrt 2 \end{align}\]

Therefore, distance between \(B(4, 5)\) and \(D(-2, -1)\) is given by

Length of diagonal

\[\begin{align} BD &= \sqrt {{{[4 - ( - 2)]}^2} + \left( {5 - {{( - 1)}^2}} \right.} \\& = \sqrt {36 + 36} \\&= 6\sqrt 2 \end{align}\]

\[\begin{align} &\text{Area of the rhombus} ABCD\\ &=\! \frac{1}{2}\!\! \times\!\! \text{(Product of lengths of diagonals)} \\&= \frac{1}{2} \times {\text{AC}} \times {\text{BD}}\end{align}\]

Therefore, area of rhombus

\[\begin{align}ABCD &= \frac{1}{2} \times 4\sqrt 2 \times 6\sqrt 2 \\ &= 24 \;\rm square\; units\end{align}\]