NCERT Solutions for Class 10 Maths Chapter 7 Exercise 7.1
The Distance formula in Coordinate Geometry
If there are two points \(A(x_1, \;y_1)\) and \(B(x_2, \;y_2)\) on a cartesian plane,
then their distance can be calculated with the following formula :
Distance between two points
\[ \quad\Large{ \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2} } \quad\]
This is known as distance formula in the world of Coordinate Geometry.
It has been derived from the famous Pythagorean theorem. All problems in this page are related to
finding distance between two points.
To find the distance of a point, say \((x, \;y)\), from the origin \((0, \;0)\), the formula becomes:
Distance between origin and one point
\[ \Large{ \sqrt{x^2+y^2} }\]
which is the simplified form of \( \sqrt{(x-0)^2 + (y-0)^2} \)
Solved problems
Class 10 Maths Chapter 7 Exercise 7.1
Q. 1(i): Find the distance between the pairs of points \(\;(2, 3)\) and \(\;(4, 1)\).
Q.2. Find the distance between the pairs of points \(\;(0, 0)\) and \(\;(36, 15)\).
Solution
Here,
\[ \quad x_1 = 0,\quad y_1 = 0,\quad x_2 = 36,\quad y_2 = 15 \quad \]
So, the distance between those two points
$$ \begin{align}
& = \sqrt{(36 - 0)^2 + (15 - 0)^2} \\
& = \sqrt{36^2 + 15^2} \\
& = \sqrt{1296 + 225} \\
& = \sqrt{1521} \\
& = \sqrt{3^2 \times 13^2 } \\
& = 3 \times 13 \\
& = \color{limegreen} {39}\text{ unit Ans.}\\
\end{align} $$
Following is the graph of the straight line:
Q.3. Determine if the points \(\;(1, 5), (2, 3)\) and \(\;(-2, -11)\) are collinear.
Solution
Let us name the given points as
\(\;P(1, 5), Q(2, 3)\) and \(\;R(-2, -11)\).
Distance between P & Q, Q & R and P & R are as follows:
For PQ,
\[ \quad x_1 = 1,\quad y_1 = 5,\quad x_2 = 2,\quad y_2 = 3 \quad \]
$$ \begin{align}
PQ & = \sqrt{(2 - 1)^2 + (3 - 5)^2} \\
& = \sqrt{1^2 + (-2)^2} \\
& = \sqrt{1 + 4} \\
& = \color{royalblue} {\sqrt{5}}\text{ unit } \\\\
\end{align} $$
For QR,
\[ \quad x_1 = 2,\quad y_1 = 3,\quad x_2 = -2,\quad y_2 = -11 \quad \]
$$ \begin{align}
QR & = \sqrt{(-2 - 2)^2 + (-11 - 3)^2} \\
& = \sqrt{(-4)^2 + (-14)^2} \\
& = \sqrt{16 + 196} \\
& = \sqrt{212} \\
& = \sqrt{2^2 \times 53} \\
& = \color{royalblue} {2\sqrt{53}}\text{ unit } \\\\
\end{align} $$
For PR,
\[ \quad x_1 = 1,\quad y_1 = 5,\quad x_2 = -2,\quad y_2 = -11 \quad \]
$$ \begin{align}
PR & = \sqrt{(-2 - 1)^2 + (-11 - 5)^2} \\
& = \sqrt{(-3)^2 + (-16)^2} \\
& = \sqrt{9 + 256} \\
& = \color{royalblue} {\sqrt{265}}\text{ unit } \\\\
\end{align} $$
If the 3 points P, Q and R are collinear,
then PR(the whole line) must be equal to
sum of PQ and QR.
But in the given case, \( PQ + QR \ne PR \).
So, the three points are not collinear. Ans.
Following is the graph of the three points P, Q and R.
Q. 4.Check whether \(\;(5, -2), (6, 4)\) and \(\;(7, -2)\) are
the vertices of an isosceles triangle.
Solution
Let us name the given points as
\(\;P(5, -2), Q(6, 4)\) and \(\;R(7, -2)\).
We calculate the distances between
P and Q, Q and R, P and R.
For PQ,
\[ \quad x_1 = 5,\quad y_1 = -2,\quad x_2 = 6,\quad y_2 = 4 \quad \]
$$ \begin{align}
PQ & = \sqrt{(6 - 5)^2 + (4 - (-2))^2} \\
& = \sqrt{1^2 + (6)^2} \\
& = \sqrt{1 + 36} \\
& = \color{royalblue} {\sqrt{37}}\text{ unit } \\\\
\end{align} $$
For QR,
\[ \quad x_1 = 6,\quad y_1 = 4,\quad x_2 = 7,\quad y_2 = 2 \quad \]
$$ \begin{align}
QR & = \sqrt{(7 - 6)^2 + (-2 - 4)^2} \\
& = \sqrt{1^2 + (-6)^2} \\
& = \sqrt{1 + 36} \\
& = \color{royalblue} {\sqrt{37}}\text{ unit } \\\\
\end{align} $$
For PR,
\[ \quad x_1 = 5,\quad y_1 = -2,\quad x_2 = 7,\quad y_2 = -2 \quad \]
$$ \begin{align}
PR & = \sqrt{(7 - 5)^2 + (-2 - (-2))^2} \\
& = \sqrt{2^2 + (-2 + 2)^2} \\
& = \sqrt{4 + 0^2} \\
& = \sqrt{4} \\
& = \color{royalblue} 2\text{ unit } \\
\end{align} $$
Since side PQ = side QR, PQR is an isosceles triangle.
So, points \(\;(5, -2), (6, 4)\) and \(\;(7, -2)\) are the
vertices of an isosceles triangle. Ans.
An isosceles triangle has two equal sides.
Following is the graph of the 3 lines formed
by the given coordinates.
It is clear that \(\;PQ = QR\).
So the three given points form an ISOSCELES triangle.
Q.5. In a classroom, 4 friends are seated at the points
A, B, C and D as shown in the given figure. Champa and
Chameli walk into the class and after observing for a
few minutes Champa asks Chameli, "Don’t you think ABCD
is a square?" Chameli disagrees. Using distance formula,
find which of them is correct.
Solution
According to the box, let us name the given points as A, B, C and D.
So the coordinates are \(\;A(3, 4), B(6, 7), C(9, 4)\) and \(\;D(6, 1)\).
Let us now find the distances AB, BC, CD and DA.
$$ \begin{align}
AB & = \sqrt{(6 - 3)^2 + (7 - 4)^2} \\
& = \sqrt{3^2 + 3^2} \\
& = \sqrt{9 + 9} \\
& = \sqrt{18} \\
& = \sqrt{3^2 \times 2} \\
& = \color{royalblue} {3\sqrt{2}}\text{ unit } \\\\
BC & = \sqrt{(9 - 6)^2 + (4 - 7)^2} \\
& = \sqrt{3^2 + (-3)^2} \\
& = \sqrt{9 + 9} \\
& = \sqrt{18} \\
& = \sqrt{3^2 \times 2} \\
& = \color{royalblue} {3\sqrt{2}}\text{ unit } \\\\
CD & = \sqrt{(6 - 9)^2 + (1 - 4)^2} \\
& = \sqrt{(-3)^2 + (-3)^2} \\
& = \sqrt{9 + 9} \\
& = \sqrt{18} \\
& = \sqrt{3^2 \times 2} \\
& = \color{royalblue} 3\sqrt{2}\text{ unit } \\\\
AD & = \sqrt{(6 - 3)^2 + (1 - 4)^2} \\
& = \sqrt{3^2 + (-3)^2} \\
& = \sqrt{9 + 9} \\
& = \sqrt{18} \\
& = \sqrt{3^2 \times 2} \\
& = \color{royalblue} {3\sqrt{2}}\text{ unit } \\\\
\end{align} $$
Till now we find that \( AB=BC=CD=AD=3\sqrt{2} \).
So, this quadrilateral is either a square or a rhombus.
Note: The diagonals of a square are always equal to each other.
But diagonals of a rhombus are unequal.
To be sure, let us compare the lengths of diagonals AC and BD ...
\begin{align}
AC & = \sqrt{(9 - 3)^2 + (4 - 4)^2} \\
& = \sqrt{6^2 + 0^2} \\
& = \sqrt{36} \\
& = \color{royalblue} 6 \text{ unit }\\\\
BD & = \sqrt{(6 - 6)^2 + (1 - 7)^2} \\
& = \sqrt{0^2 + (-6)^2} \\
& = \sqrt{0 + 36} \\
& = \sqrt{36} \\
& = \color{royalblue} 6 \text{ unit }\\\\
\end{align}
Since diagonals are equal to each other, and the four sides are also
equal, \(ABCD\) is a square, and
Champa was correct. Ans.
Q.6(i). Name the type of quadrilateral formed, if any,
by the following points, and give reasons for your answer:
\(\;(-1, -2), (1, 0), (-1, 2), (-3, 0)\)
Solution
Let us name the given points as
\(\;P(-1, -2), Q(1, 0), R(-1, 2)\) and \(\;S(-3, 0)\).
\begin{align}
PQ & = \sqrt{(1 - (-1))^2 + (0 - (-2))^2} \\
& = \sqrt{2^2 + 2^2} \\
& = \sqrt{4 + 4} \\
& = \sqrt{8} \\
& = \sqrt{2^2 \times 2} \\
& = \color{royalblue} {2\sqrt{2}}\text{ unit } \\\\
QR & = \sqrt{(-1 - 1)^2 + (2 - 0)^2} \\
& = \sqrt{(-2)^2 + 2^2} \\
& = \sqrt{4 + 4} \\
& = \sqrt{8} \\
& = \sqrt{2^2 \times 2} \\
& = \color{royalblue} {2\sqrt{2}}\text{ unit } \\\\
RS & = \sqrt{(-3 - (-1))^2 + (0 - 2)^2} \\
& = \sqrt{(-2)^2 + (-2)^2} \\
& = \sqrt{4 + 4} \\
& = \sqrt{8} \\
& = \sqrt{2^2 \times 2} \\
& = \color{royalblue} {2\sqrt{2}}\text{ unit } \\\\
PS & = \sqrt{(-3 - (-1))^2 + (0 - (-2))^2} \\
& = \sqrt{(-2)^2 + (0 + 2)^2} \\
& = \sqrt{4 + 4} \\
& = \sqrt{8} \\
& = \sqrt{2^2 \times 2} \\
& = \color{royalblue} {2\sqrt{2}}\text{ unit } \\
\end{align}
As all the 4 sides are equal, the quadrilateral should
either be a square, or a rhombus.
The two diagonals of a square are equal to each other.
But a rhombus has unequal diagonals.
Let us measure the diagonals ...
\begin{align}
\text{Diagonal }PR & = \sqrt{(-1 - (-1))^2 + (2 - (-2))^2} \\
& = \sqrt{(-1 + 1)^2 + (2 + 2)^2} \\
& = \sqrt{0^2 + 4^2} \\
& = \sqrt{16} \\
& = \color{royalblue} 4 \text{ unit } \\\\
\text{Diagonal }QS & = \sqrt{(-3 - 1)^2 + (0 - 0)^2} \\
& = \sqrt{(-4)^2 + 0^2} \\
& = \sqrt{16 + 0} \\
& = \sqrt{16} \\
& = \color{royalblue} 4 \text{ unit } \\\\
\end{align}
Since all sides are equal, and the two diagonals are also
equal to each other, the shape is a square.
Ans.
Please check the following graph ...
Q.6(ii). Name the type of quadrilateral formed, if any,
by the following points, and give reasons for your answer:
\(\;(-3, 5), (3, 1), (0, 3), (-1, -4)\)
Solution
Let us name the given points as
\(\;A(-3, 5), B(3, 1), C(0, 3)\) and \(\;D(-1, –4)\).
$$ \begin{align}
AB & = \sqrt{(3 - (-3))^2 + (1 - 5)^2} \\
& = \sqrt{(3 + 3)^2 + (-4)^2} \\
& = \sqrt{6^2 + 16} \\
& = \sqrt{36 + 16} \\
& = \sqrt{52} \\
& = \sqrt{2^2 \times 13} \\
& = \color{royalblue} {2\sqrt{13}}\text{ unit } \\\\
BC & = \sqrt{(0 - 3)^2 + (3 - 1)^2} \\
& = \sqrt{(-3)^2 + 2^2} \\
& = \sqrt{9 + 4} \\
& = \color{royalblue} {\sqrt{13}} \text{ unit } \\\\
CD & = \sqrt{(-1 - 0)^2 + (-4 - 3)^2} \\
& = \sqrt{(-1)^2 + (-7)^2} \\
& = \sqrt{1 + 49} \\
& = \sqrt{50} \\
& = \sqrt{5^2 \times 2} \\
& = \color{royalblue} {5\sqrt{2}}\text{ unit } \\\\
AD & = \sqrt{(-1 - (-3))^2 + (-4 - 5)^2} \\
& = \sqrt{2^2 + (-9)^2} \\
& = \sqrt{4 + 81} \\
& = \color{royalblue} \sqrt{85}\text{ unit } \\\\
AC & = \sqrt{(0 - (-3))^2 + (3 - 5)^2} \\
& = \sqrt{3^2 + (-2)^2} \\
& = \sqrt{9 + 4} \\
& = \color{royalblue} \sqrt{13} \text{ unit }\\\\
BD & = \sqrt{(-1 - 3)^2 + (-4 - 1)^2} \\
& = \sqrt{(-4)^2 + (-5)^2} \\
& = \sqrt{16 + 25} \\
& = \color{royalblue} {\sqrt{41}} \text{ unit } \\
\end{align} $$
Since \( AC + BC = \sqrt 13 + \sqrt 13 = 2\sqrt 13 = AB \),
points \(A, B, C\;\) are collinear.
So no quadrilateral is possible with the given four points.
Ans.
Q.6(iii). Name the type of quadrilateral formed, if any,
by the following points, and give reasons for your answer:
\(\;(4, 5), (7, 6), (4, 3), (1, 2)\)
Solution
Let us name the given points as
\(\;A(4, 5),\; B(7, 6),\; C(4, 3)\;\) and \(\;D(1, 2)\).
For distance A to B,
\[x_1= 4,\, y_1= 5,\, x_2= 7,\, y_2= 6\]
\[\sqrt {( 7 - 4 )^2 + (6 - 5)^2 }\]
\begin{align}
& = \sqrt{3^2 + 1^2} \\\\
& = \sqrt{9 + 1} \\\\
& = \color{royalblue} \sqrt{10}\text{ unit } \\\\
\end{align}
For distance B to C,
\[x_1= 7,\, y_1= 6,\, x_2= 4,\, y_2= 3\]
\[\sqrt {( 4 - 7 )^2 + (3 - 6)^2 }\]
\begin{align}
& = \sqrt{(-3)^2 + (-3)^2} \\\\
& = \sqrt{9 + 9} \\\\
& = \sqrt{18} \\\\
& = \sqrt{3^2 \times 2} \\\\
& = \color{royalblue} {3\sqrt{2}} \text{ unit } \\\\
\end{align}
For distance C to D,
\[x_1= 4,\, y_1= 3,\, x_2= 1,\, y_2= 2\]
\[\sqrt {( 1 - 4 )^2 + (2 - 3)^2 }\]
\begin{align}
& = \sqrt{(-3)^2 + (-1)^2} \\\\
& = \sqrt{9 + 1} \\\\
& = \color{royalblue} \sqrt{10}\text{ unit } \\\\
\end{align}
For distance A to D,
\[x_1= 4,\, y_1= 5,\, x_2= 1,\, y_2= 2\]
\[\sqrt {( 1 - 4 )^2 + (2 - 5)^2 }\]
\begin{align}
& = \sqrt{(-3)^2 + (-3)^2} \\\\
& = \sqrt{9 + 9} \\\\
& = \sqrt{18} \\\\
& = \sqrt{3^2 \times 2} \\\\
& = \color{royalblue} {3\sqrt{2}} \text{ unit } \\\\
\end{align}
For diagonal AC,
\[x_1= 4,\, y_1= 5,\, x_2= 4,\, y_2= 3\]
\[\sqrt {( 4 - 4 )^2 + (3 - 5)^2 }\]
\begin{align}
& = \sqrt{0^2 + (-2)^2} \\\\
& = \sqrt{0 + 4} \\\\
& = \color{royalblue} 2 \text{ unit }\\\\
\end{align}
For diagonal BD,
\[x_1= 7,\, y_1= 6,\, x_2= 1,\, y_2= 2\]
\[\sqrt {( 1 - 7 )^2 + (2 - 6)^2 }\]
\begin{align}
& = \sqrt{(-6)^2 + (-4)^2} \\\\
& = \sqrt{36 + 16} \\\\
& = \sqrt{52} \\\\
& = \sqrt{2^2 \times 13} \\
& = \color{royalblue} {2\sqrt{13}} \text{ unit } \\\\
\end{align}
Since \( AB = CD = \sqrt 10\), and \( BC = AD = 3\sqrt 2\),
the opposite sides of this quadrilateral are equal.
The quadrilateral is either a rectangle or a parallelogram.
Note: The diagonals of a rectangle are always equal to each other.
But diagonals of a parallelogram are unequal.
In our case, the diagonals are not equal; \(AC \ne BD, \text{ since } 2 \ne 2\sqrt{13}, \).
So, the given quadrilateral is a parallelogram. Ans.
Q.7. Find the point on the x-axis which is equidistant
from \(\;(2, -5)\) and \(\;(-2, 9)\).
Solution
Given that an unkonwn point is equidistant from \(\;(2, -5)\) and \(\;(-2, 9)\).
It is also given that the said point lies on the X-axis.
So, from the 2nd condition we know that y coordinate of the
unknown point is 0.
Let the unknown point be \(P(x, \;0)\).
Q.9. If \(\;Q(0, \;1)\) is equidistant from \(\;P(5, \;-3)\), and \(\;R(x, \;6)\),
find the value of \(\;x.\) Also, find the distances \(QR\) and \(PR\).
Solution
For distance between \(\; Q(0, \;1) \) and \(\; P(5, \;-3) \),
\( x_1 = 0,\quad y_1 = 1,\quad x_2 = 5,\quad y_2 = -3 \)
For distance between \(\; Q(0, \;1) \) and \(\; R(x, \;6) \),
\( x_1 = 0,\quad y_1 = 1,\quad x_2 = x,\quad y_2 = 6 \)
In the diagram, points A, B and C are equidistant from points P and Q.
Their coordinates \(\;(0,5), (1,2)\;\) and \(\;(2,-1)\) have been calculated from the equation
\( \color{red}{\,y = -3x + 5}\quad \)
When \(\;x=0,\; y=5,\) when \(\;x=1,\; y=2,\) when \(\;x=2,\; y=-1\) ...
