Exercise 7.1 (Revised) - Chapter 7 - Coordinate Geometry - Ncert Solutions class 10 - Maths

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NCERT Class 10 Maths: Chapter 7 Coordinate Geometry Solutions

Ex 7.1 Question 1.

Find the distance between the following pairs of points:
(i) $(2,3),(4,1)$
(ii) $(-5,7),(-1,3)$
(iii) (a, b), (-a, -b)

Answer.

(i) Applying Distance Formula to find distance between points $(2,3)$ and $(4,1)$, we get
$
\mathrm{d}=\sqrt{(4-2)^2+(1-3)^2}=\sqrt{(2)^2+(-2)^2}=\sqrt{4+4}=\sqrt{8}=2 \sqrt{2} \text { units }
$
(ii) Applying Distance Formula to find distance between points $(-5,7)$ and $(-1,3)$, we get
$
d=\sqrt{[-1-(-5)]^2+(3-7)^2}=\sqrt{(4)^2+(-4)^2}=\sqrt{16+16}=\sqrt{32}=4 \sqrt{2} \text { units }
$
(iii) Applying Distance Formula to find distance between points $(a, b)$ and $(-a,-b)$, we get
d
$
\sqrt{(-a-a)^2+(-b-b)^2}=\sqrt{(-2 a)^2+(-2 b)^2}=\sqrt{4 a^2+4 b^2}=\sqrt{4\left(a^2+b^2\right)}=2 \sqrt{a^2+b^2}
$
Ex 7.1 Question 2.

Find the distance between the points $(0,0)$ and $(36,15)$. Also, find the distance between towns $A$ and $B$ if town $B$ is located at $36 \mathrm{~km}$ east and15 $\mathbf{k m}$ north of town $A$.

Answer.

Applying Distance Formula to find distance between points $(0,0)$ and $(36,15)$, we get
$
d=\sqrt{(36-0)^2+(15-0)^2}=\sqrt{(36)^2+(15)^2}=\sqrt{1296+225}=\sqrt{1521}=39 \text { units }
$

Town B is located at $36 \mathrm{~km}$ east and15 km north of town A. So, the location of town A and B can be shown as:

Clearly, the coordinates of point A are $(0,0)$ and coordinates of point $B$ are $(36,15)$.
To find the distance between them, we use Distance formula:
$
d=\sqrt{[36-0]^2+(15-0)^2}=\sqrt{(36)^2+(15)^2}=\sqrt{1296+225}=\sqrt{1521}=39 \mathrm{Km}
$
Ex 7.1 Question 3.

Determine if the points $(1,5),(2,3)$ and $(-2,-11)$ are collinear.

Answer.

Let $A=(1,5), B=(2,3)$ and $C=(-2,-11)$

Using Distance Formula to find distance $\mathrm{AB}, \mathrm{BC}$ and $\mathrm{CA}$.
$
\begin{aligned}
& \mathrm{AB}=\sqrt{[2-1]^2+(3-5)^2}=\sqrt{(1)^2+(-2)^2}=\sqrt{1+4}=\sqrt{5} \\
& \mathrm{BC}=\sqrt{[-2-2]^2+(-11-3)^2}=\sqrt{(-4)^2+(-14)^2}=\sqrt{16+196}=\sqrt{212}=2 \sqrt{53} \\
& \mathrm{CA}=\sqrt{[-2-1]^2+(-11-5)^2}=\sqrt{(-3)^2+(-16)^2}=\sqrt{9+256}=\sqrt{265}
\end{aligned}
$

Since $\mathrm{AB}+\mathrm{AC} \neq \mathrm{BC}, \mathrm{BC}+\mathrm{AC} \neq \mathrm{AB}$ and $\mathrm{AC} \neq \mathrm{BC}$.

Therefore, the points $\mathrm{A}, \mathrm{B}$ and $\mathrm{C}$ are not collinear.
Ex 7.1 Question 4.

Check whether $(5,-2),(6,4)$ and $(7,-2)$ are the vertices of an isosceles triangle.

Answer.

Let $\mathrm{A}=(5,-2), \mathrm{B}=(6,4)$ and $\mathrm{C}=(7,-2)$
Using Distance Formula to find distances $\mathrm{AB}, \mathrm{BC}$ and $\mathrm{CA}$.
$
\begin{aligned}
& \mathrm{AB}=\sqrt{[6-5]^2+[4-(-2)]^2}=\sqrt{(1)^2+(6)^2}=\sqrt{1+36}=\sqrt{37} \\
& \mathrm{BC}=\sqrt{[7-6]^2+(-2-4)^2}=\sqrt{(1)^2+(-6)^2}=\sqrt{1+36}=\sqrt{37} \\
& \mathrm{CA}=\sqrt{[7-5]^2+[-2-(-2)]^2}=\sqrt{(2)^2+(0)^2}=\sqrt{4+0}=\sqrt{4}=2
\end{aligned}
$

Since $A B=B C$.
Therefore, $\mathrm{A}, \mathrm{B}$ and $\mathrm{C}$ are vertices of an isosceles triangle.
Ex 7.1 Question 5.

In a classroom, 4 friends are seated at the points A $(3,4)$, B $(6,7), C(9,4)$ and D $(6,1)$. Champa and Chameli walk into the class and after observing for a few minutes Champa asks Chameli. "Don't you think $\mathrm{ABCD}$ is a square?"Chameli disagrees. Using distance formula, find which of them is correct.

Answer.

$\text {We have } \mathrm{A}=(3,4), \mathrm{B}=(6,7), \mathrm{C}=(9,4) \text { and } \mathrm{D}=(6,1)$

Using Distance Formula to find distances $\mathrm{AB}, \mathrm{BC}, \mathrm{CD}$ and $\mathrm{DA}$, we get
$
\begin{aligned}
& \mathrm{AB}=\sqrt{[6-3]^2+[7-4]^2}=\sqrt{(3)^2+(3)^2}=\sqrt{9+9}=\sqrt{18}=3 \sqrt{2} \\
& \mathrm{BC}=\sqrt{[9-6]^2+[4-7]^2}=\sqrt{(3)^2+(-3)^2}=\sqrt{9+9}=\sqrt{18}=3 \sqrt{2} \\
& \mathrm{CD}=\sqrt{[6-9]^2+[1-4]^2}=\sqrt{(-3)^2+(-3)^2}=\sqrt{9+9}=\sqrt{18}=3 \sqrt{2} \\
& \mathrm{DA}=\sqrt{[6-3]^2+[1-4]^2}=\sqrt{(3)^2+(-3)^2}=\sqrt{9+9}=\sqrt{18}=3 \sqrt{2}
\end{aligned}
$

Therefore, All the sides of $A B C D$ are equal here.
Now, we will check the length of its diagonals.
$
\begin{aligned}
& \mathrm{AC}=\sqrt{[9-3]^2+[4-4]^2}=\sqrt{(6)^2+(0)^2}=\sqrt{36+0}=6 \\
& \mathrm{BD}=\sqrt{[6-6]^2+[1-7]^2}=\sqrt{(0)^2+(-6)^2}=\sqrt{0+36}=\sqrt{36}=6
\end{aligned}
$

So, Diagonals of ABCD are also equal. ... (2)
From (1) and (2), we can definitely say that $\mathrm{ABCD}$ is a square.
Therefore, Champa is correct.

Ex 7.1 Question 6.

Name the type of quadrilateral formed, if any, by the following points, and give reasons for your answer.
(i) $(-1,-2),(1,0),(-1,2),(-3,0)$
(ii) $(-3,5),(3,1),(0,3),(-1,-4)$
(iii) $(4,5),(7,6),(4,3),(1,2)$

Answer.

(i) Let $\mathrm{A}=(-1,-2), \mathrm{B}=(1,0), \mathrm{C}=(-1,2)$ and $\mathrm{D}=(-3,0)$
Using Distance Formula to find distances $\mathrm{AB}, \mathrm{BC}, \mathrm{CD}$ and $\mathrm{DA}$, we get

$
\begin{aligned}
& \mathrm{AB}=\sqrt{[1-(-1)]^2+[0-(-2)]^2}=\sqrt{(2)^2+(2)^2}=\sqrt{4+4}=\sqrt{8}=2 \sqrt{2} \\
& \mathrm{BC}=\sqrt{[-1-1]^2+[2-0]^2}=\sqrt{(-2)^2+(2)^2}=\sqrt{4+4}=\sqrt{8}=2 \sqrt{2} \\
& \mathrm{CD}=\sqrt{[-3-(-1)]^2+[0-2]^2}=\sqrt{(-2)^2+(-2)^2}=\sqrt{4+4}=\sqrt{8}=2 \sqrt{2} \\
& \mathrm{DA}=\sqrt{[-3-(-1)]^2+[0-(-2)]^2}=\sqrt{(-2)^2+(2)^2}=\sqrt{4+4}=\sqrt{8}=2 \sqrt{2}
\end{aligned}
$

Therefore, all four sides of quadrilateral are equal. ... (1)
Now, we will check the length of diagonals.
$
\begin{aligned}
& \mathrm{AC}=\sqrt{[-1-(-1)]^2+[2-(-2)]^2}=\sqrt{(0)^2+(4)^2}=\sqrt{0+16}=\sqrt{16}=4 \\
& \mathrm{BD}=\sqrt{[-3-1]^2+[0-0]^2}=\sqrt{(-4)^2+(0)^2}=\sqrt{16+0}=\sqrt{16}=4
\end{aligned}
$

Therefore, diagonals of quadrilateral $\mathrm{ABCD}$ are also equal. ... (2)
From (1) and (2), we can say that $\mathrm{ABCD}$ is a square.
(ii) Let $\mathrm{A}=(-3,5), \mathrm{B}=(3,1), \mathrm{C}=(0,3)$ and $\mathrm{D}=(-1,-4)$

Using Distance Formula to find distances $\mathrm{AB}, \mathrm{BC}, \mathrm{CD}$ and $\mathrm{DA}$, we get
$
\begin{aligned}
& \mathrm{AB}=\sqrt{[3-(-3)]^2+[1-5]^2}=\sqrt{(6)^2+(-4)^2}=\sqrt{36+16}=\sqrt{52}=2 \sqrt{13} \\
& \mathrm{BC}=\sqrt{[0-3]^2+[3-1]^2}=\sqrt{(-3)^2+(2)^2}=\sqrt{9+4}=\sqrt{13} \\
& \mathrm{CD}=\sqrt{[-1-0]^2+[-4-3]^2}=\sqrt{(-1)^2+(-7)^2}=\sqrt{1+49}=\sqrt{50}=5 \sqrt{2} \\
& \mathrm{DA}=\sqrt{[-1-(-3)]^2+[-4-5]^2}=\sqrt{(2)^2+(-9)^2}=\sqrt{4+81}=\sqrt{85}
\end{aligned}
$

We cannot find any relation between the lengths of different sides.
Therefore, we cannot give any name to the quadrilateral $A B C D$.

(iii) Let $\mathrm{A}=(4,5), \mathrm{B}=(7,6), \mathrm{C}=(4,3)$ and $\mathrm{D}=(1,2)$

Using Distance Formula to find distances AB, BC, CD and DA, we get
$
\begin{aligned}
& \mathrm{AB}=\sqrt{[7-4]^2+[6-5]^2}=\sqrt{(3)^2+(1)^2}=\sqrt{9+1}=\sqrt{10} \\
& \mathrm{BC}=\sqrt{[4-7]^2+[3-6]^2}=\sqrt{(-3)^2+(-3)^2}=\sqrt{9+9}=\sqrt{18}=3 \sqrt{2} \\
& \mathrm{CD}=\sqrt{[1-4]^2+[2-3]^2}=\sqrt{(-3)^2+(-1)^2}=\sqrt{9+1}=\sqrt{10} \\
& \mathrm{DA}=\sqrt{[1-4]^2+[2-5]^2}=\sqrt{(-3)^2+(-3)^2}=\sqrt{9+9}=\sqrt{18}=3 \sqrt{2}
\end{aligned}
$

Here opposite sides of quadrilateral $A B C D$ are equal. ... (1)

We can now find out the lengths of diagonals.
$
\begin{aligned}
& \mathrm{AC}=\sqrt{[4-4]^2+[3-5]^2}=\sqrt{(0)^2+(-2)^2}=\sqrt{0+4}=\sqrt{4}=2 \\
& \mathrm{BD}=\sqrt{[1-7]^2+[2-6]^2}=\sqrt{(-6)^2+(-4)^2}=\sqrt{36+16}=\sqrt{52}=2 \sqrt{13}
\end{aligned}
$

Here diagonals of $A B C D$ are not equal. ... (2)

From (1) and (2), we can say that $\mathrm{ABCD}$ is not a rectangle therefore it is a parallelogram.

Ex 7.1 Question 7.

Find the point on the $x$-axis which is equidistant from $(2,-5)$ and $(-2,9)$.

Answer.

Let the point be $(\mathrm{x}, 0)$ on $\mathrm{x}$-axis which is equidistant from $(2,-5)$ and $(-2,9)$.
Using Distance Formula and according to given conditions we have:
$
\begin{aligned}
& \sqrt{[x-2]^2+[0-(-5)]^2}=\sqrt{[x-(-2)]^2+[(0-9)]^2} \\
& \Rightarrow \sqrt{x^2+4-4 x+25}=\sqrt{x^2+4+4 x+81}
\end{aligned}
$

Squaring both sides, we get
$
\Rightarrow x^2+4-4 x+25=x^2+4+4 x+81
$

$
\begin{aligned}
& \Rightarrow-4 x+29=4 x+85 \\
& \Rightarrow 8 x=-56 \\
& \Rightarrow x=-7
\end{aligned}
$

Therefore, point on the $x$-axis which is equidistant from $(2,-5)$ and $(-2,9)$ is $(-7,0)$
Ex 7.1 Question 8.

Find the values of $y$ for which the distance between the points $P(2,-3)$ and $Q(10, y)$ is 10 units.

Answer.

Using Distance formula, we have
$
\begin{aligned}
& 10=\sqrt{(2-10)^2+(-3-y)^2} \\
& \Rightarrow 10=\sqrt{(-8)^2+9+y^2+6 y} \\
& \Rightarrow 10=\sqrt{64+9+y^2+6 y}
\end{aligned}
$

Squaring both sides, we get
$
\begin{aligned}
& 100=73+y^2+6 y \\
& \Rightarrow y^2+6 y-27=0
\end{aligned}
$

Solving this Quadratic equation by factorization, we can write
$
\begin{aligned}
& \Rightarrow y^2+9 y-3 y-27=0 \\
& \Rightarrow y(y+9)-3(y+9)=0 \\
& \Rightarrow(y+9)(y-3)=0 \\
& \Rightarrow y=3,-9
\end{aligned}
$

Ex 7.1 Question 9.

If, $Q(0,1)$ is equidistant from $P(5,-3)$ and $R(x, 6)$, find the values of $x$. Also, find the distances $Q R$ and PR.

Answer.

It is given that $\mathrm{Q}$ is equidistant from $\mathrm{P}$ and $\mathrm{R}$. Using Distance Formula, we get
$
\begin{aligned}
& P Q=R Q \\
& \Rightarrow \sqrt{(0-5)^2+[1-(-3)]^2}=\sqrt{(0-x)^2+(1-6)^2} \\
& \Rightarrow \sqrt{(-5)^2+[4]^2}=\sqrt{(-x)^2+(-5)^2} \\
& \Rightarrow \sqrt{25+16}=\sqrt{x^2+25}
\end{aligned}
$

Squaring both sides, we get
$
\begin{aligned}
& \Rightarrow 25+16=x^2+25 \\
& \Rightarrow x^2=16 \\
& \Rightarrow \mathbf{x}=4,-4
\end{aligned}
$

Thus, $Q$ is $(4,6)$ or $(-4,6)$.
Using Distance Formula to find QR, we get
Using value of $\mathrm{x}=4 \mathrm{QR}=\sqrt{(4-0)^2+(6-1)^2}=\sqrt{16+25}=\sqrt{41}$
Using value of $\mathrm{x}=-4 \mathrm{QR}=\sqrt{(-4-0)^2+(6-1)^2}=\sqrt{16+25}=\sqrt{41}$
Therefore, $\mathrm{QR}=\sqrt{41}$
Using Distance Formula to find PR, we get
Using value of $x=4 P R=\sqrt{(4-5)^2+[6-(-3)]^2}=\sqrt{1+81}=\sqrt{82}$
Using value of $\mathrm{x}=-4 \mathrm{PR}=\sqrt{(-4-5)^2+[6-(-3)]^2}=\sqrt{81+81}=\sqrt{162}=9 \sqrt{2}$
Therefore, $x=4,-4$

$\mathrm{QR}=\sqrt{41}, \mathrm{PR}=\sqrt{82}, 9 \sqrt{2}$

Ex 7.1 Question 10.

Find a relation between $x$ and $y$ such that the point ( $x, y)$ is equidistant from the point $(3,6)$ and $(-3,4)$.

Answer.

It is given that $(x, y)$ is equidistant from $(3,6)$ and $(-3,4)$.
Using Distance formula, we can write
$
\begin{aligned}
& \sqrt{(x-3)^2+(y-6)^2}=\sqrt{[x-(-3)]^2+(y-4)^2} \\
& \Rightarrow \sqrt{x^2+9-6 x+y^2+36-12 y}=\sqrt{x^2+9+6 x+y^2+16-8 y}
\end{aligned}
$

Squaring both sides, we get
$
\begin{aligned}
& \Rightarrow x^2+9-6 x+y^2+36-12 y=x^2+9+6 x+y^2+16-8 y \\
& \Rightarrow-6 x-12 y+45=6 x-8 y+25 \\
& \Rightarrow 12 x+4 y=20 \\
& \Rightarrow 3 x+y=5
\end{aligned}
$