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

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NCERT Class 10 Maths: Chapter 7 Coordinate Geometry Solutions

Example 1 :

Do the points $(3,2),(-2,-3)$ and $(2,3)$ form a triangle? If so, name the type of triangle formed.
Solution :

Let us apply the distance formula to find the distances $P Q, Q R$ and $P R$, where $P(3,2), Q(-2,-3)$ and $R(2,3)$ are the given points. We have
$
\begin{aligned}
& \mathrm{PQ}=\sqrt{(3+2)^2+(2+3)^2}=\sqrt{5^2+5^2}=\sqrt{50}=7.07 \text { (approx.) } \\
& \mathrm{QR}=\sqrt{(-2-2)^2+(-3-3)^2}=\sqrt{(-4)^2+(-6)^2}=\sqrt{52}=7.21 \text { (approx.) } \\
& \mathrm{PR}=\sqrt{(3-2)^2+(2-3)^2}=\sqrt{1^2+(-1)^2}=\sqrt{2}=1.41 \text { (approx.) }
\end{aligned}
$

Since the sum of any two of these distances is greater than the third distance, therefore, the points $\mathrm{P}, \mathrm{Q}$ and $\mathrm{R}$ form a triangle.

Also, $\mathrm{PQ}^2+\mathrm{PR}^2=\mathrm{QR}^2$, by the converse of Pythagoras theorem, we have $\angle \mathrm{P}=90^{\circ}$. Therefore, $\mathrm{PQR}$ is a right triangle.

Example 2 :

Show that the points $(1,7),(4,2),(-1,-1)$ and $(-4,4)$ are the vertices of a square.
Solution :

Let $A(1,7), B(4,2), C(-1,-1)$ and $D(-4,4)$ be the given points. One way of showing that $A B C D$ is a square is to use the property that all its sides should be equal and both its digonals should also be equal. Now,
$
\begin{aligned}
& \mathrm{AB}=\sqrt{(1-4)^2+(7-2)^2}=\sqrt{9+25}=\sqrt{34} \\
& \mathrm{BC}=\sqrt{(4+1)^2+(2+1)^2}=\sqrt{25+9}=\sqrt{34} \\
& \mathrm{CD}=\sqrt{(-1+4)^2+(-1-4)^2}=\sqrt{9+25}=\sqrt{34} \\
& \mathrm{DA}=\sqrt{(1+4)^2+(7-4)^2}=\sqrt{25+9}=\sqrt{34} \\
& \mathrm{AC}=\sqrt{(1+1)^2+(7+1)^2}=\sqrt{4+64}=\sqrt{68} \\
& \mathrm{BD}=\sqrt{(4+4)^2+(2-4)^2}=\sqrt{64+4}=\sqrt{68}
\end{aligned}
$

Since, $A B=B C=C D=D A$ and $A C=B D$, all the four sides of the quadrilateral $\mathrm{ABCD}$ are equal and its diagonals $\mathrm{AC}$ and $\mathrm{BD}$ are also equal. Thereore, $\mathrm{ABCD}$ is a square.

Alternative Solution :

We find the four sides and one diagonal, say, $\mathrm{AC}$ as above. Here $\mathrm{AD}^2+\mathrm{DC}^2=$ $34+34=68=\mathrm{AC}^2$. Therefore, by the converse of Pythagoras theorem, $\angle \mathrm{D}=90^{\circ}$. A quadrilateral with all four sides equal and one angle $90^{\circ}$ is a square. So, $\mathrm{ABCD}$ is a square.

Example 3 :

Fig. 7.6 shows the arrangement of desks in a classroom. Ashima, Bharti and Camella are seated at $\mathrm{A}(3,1)$, $B(6,4)$ and $C(8,6)$ respectively. Do you think they are seated in a line? Give reasons for your answer.

Solution :

Using the distance formula, we have
$
\begin{aligned}
& \mathrm{AB}=\sqrt{(6-3)^2+(4-1)^2}=\sqrt{9+9}=\sqrt{18}=3 \sqrt{2} \\
& \mathrm{BC}=\sqrt{(8-6)^2+(6-4)^2}=\sqrt{4+4}=\sqrt{8}=2 \sqrt{2} \\
& \mathrm{AC}=\sqrt{(8-3)^2+(6-1)^2}=\sqrt{25+25}=\sqrt{50}=5 \sqrt{2}
\end{aligned}
$

Since, $A B+B C=3 \sqrt{2}+2 \sqrt{2}=5 \sqrt{2}=A C$, we can say that the points $A, B$ and $C$ are collinear. Therefore, they are seated in a line.

Example 4 :

Find a relation between $x$ and $y$ such that the point $(x, y)$ is equidistant from the points $(7,1)$ and $(3,5)$.
Solution :

Let $\mathrm{P}(x, y)$ be equidistant from the points $\mathrm{A}(7,1)$ and $\mathrm{B}(3,5)$.
We are given that $\mathrm{AP}=\mathrm{BP}$. So, $\mathrm{AP}^2=\mathrm{BP}^2$
i.e., $(x-7)^2+(y-1)^2=(x-3)^2+(y-5)^2$
i.e., $x^2-14 x+49+y^2-2 y+1=x^2-6 x+9+y^2-10 y+25$
i.e., $x-y=2$

which is the required relation. 

Remark: Note that the graph of the equation $x-y=2$ is a line. From your earlier studies, you know that a point which is equidistant from $A$ and $B$ lies on the perpendicular bisector of $A B$. Therefore, the graph of $x-y=2$ is the perpendicular bisector of $\mathrm{AB}$ (see Fig. 7.7).

Example 5 :

Find a point on the $y$-axis which is equidistant from the points $A(6,5)$ and $B(-4,3)$.

Solution:

We know that a point on the $y$-axis is of the form $(0, y)$. So, let the point $\mathrm{P}(0, y)$ be equidistant from $\mathrm{A}$ and $\mathrm{B}$. Then 

i.e.,
$
\begin{aligned}
(6-0)^2+(5-y)^2 & =(-4-0)^2+(3-y)^2 \\
36+25+y^2-10 y & =16+9+y^2-6 y \\
4 y & =36 \\
y & =9
\end{aligned}
$

So, the required point is $(0,9)$.
Let us check our solution
$
\begin{aligned}
& \mathrm{AP}=\sqrt{(6-0)^2+(5-9)^2}=\sqrt{36+16}=\sqrt{52} \\
& \mathrm{BP}=\sqrt{(-4-0)^2+(3-9)^2}=\sqrt{16+36}=\sqrt{52}
\end{aligned}
$

Note: Using the remark above, we see that $(0,9)$ is the intersection of the $y$-axis and the perpendicular bisector of $A B$.

Example 6 :

Find the coordinates of the point which divides the line segment joining the points $(4,-3)$ and $(8,5)$ in the ratio $3: 1$ internally.
Solution :

Let $\mathrm{P}(x, y)$ be the required point. Using the section formula, we get
$
x=\frac{3(8)+1(4)}{3+1}=7, y=\frac{3(5)+1(-3)}{3+1}=3
$

Therefore, $(7,3)$ is the required point.

Example 7 :

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

Let $(-4,6)$ divide $\mathrm{AB}$ internally in the ratio $m_1: m_2$. Using the section formula, we get
$
(-4,6)=\left(\frac{3 m_1-6 m_2}{m_1+m_2}, \frac{-8 m_1+10 m_2}{m_1+m_2}\right)
$

Recall that if $(x, y)=(a, b)$ then $x=a$ and $y=b$.

So, $-4=\frac{3 m_1-6 m_2}{m_1+m_2}$ and $6=\frac{-8 m_1+10 m_2}{m_1+m_2}$

Now, $-4=\frac{3 m_1-6 m_2}{m_1+m_2} \quad$ gives us
$
\begin{aligned}
-4 m_1-4 m_2 & =3 m_1-6 m_2 \\
7 m_1 & =2 m_2 \\
m_1: m_2 & =2: 7
\end{aligned}
$

You should verify that the ratio satisfies the $y$-coordinate also.

Now, $\quad \frac{-8 m_1+10 m_2}{m_1+m_2}=\frac{-8 \frac{m_1}{m_2}+10}{\frac{m_1}{m_2}+1} \quad$ (Dividing throughout by $m_2$ )
$
=\frac{-8 \times \frac{2}{7}+10}{\frac{2}{7}+1}=6
$

Therefore, the point $(-4,6)$ divides the line segment joining the points $\mathrm{A}(-6,10)$ and $B(3,-8)$ in the ratio $2: 7$.

Alternatively :

 The ratio $m_1: m_2$ can also be written as $\frac{m_1}{m_2}: 1$, or $k: 1$. Let $(-4,6)$ divide $\mathrm{AB}$ internally in the ratio $k: 1$. Using the section formula, we get
$
(-4,6)=\left(\frac{3 k-6}{k+1}, \frac{-8 k+10}{k+1}\right)
$

So,
$
-4=\frac{3 k-6}{k+1}
$
i.e.,
$
-4 k-4=3 k-6
$
i.e.,
$
7 k=2
$
i.e.,
$
k: 1=2: 7
$

You can check for the $y$-coordinate also.

So, the point $(-4,6)$ divides the line segment joining the points $A(-6,10)$ and $\mathrm{B}(3,-8)$ in the ratio $2: 7$.
Note : You can also find this ratio by calculating the distances $\mathrm{PA}$ and $\mathrm{PB}$ and taking their ratios provided you know that A, P and B are collinear.

Example 8 :

Find the coordinates of the points of trisection (i.e., points dividing in three equal parts) of the line segment joining the points $\mathrm{A}(2,-2)$ and $\mathrm{B}(-7,4)$.

Solution :

Let $\mathrm{P}$ and $\mathrm{Q}$ be the points of trisection of $A B$ i.e., $A P=P Q=Q B$ (see Fig. 7.11).

Therefore, $\mathrm{P}$ divides $\mathrm{AB}$ internally in the ratio $1: 2$. Therefore, the coordinates of $\mathrm{P}$, by applying the section formula, are
$
\left(\frac{1(-7)+2(2)}{1+2}, \frac{1(4)+2(-2)}{1+2}\right) \text {, i.e., }(-1,0)
$

Now, $\mathrm{Q}$ also divides $\mathrm{AB}$ internally in the ratio $2: 1$. So, the coordinates of $\mathrm{Q}$ are
$
\left(\frac{2(-7)+1(2)}{2+1}, \frac{2(4)+1(-2)}{2+1}\right) \text {,i.e., }(-4,2)
$

Therefore, the coordinates of the points of trisection of the line segment joining A and $B$ are $(-1,0)$ and $(-4,2)$.
Note : We could also have obtained $Q$ by noting that it is the mid-point of $P B$. So, we could have obtained its coordinates using the mid-point formula.

Example 9 :

Find the ratio in which the $y$-axis divides the line segment joining the points $(5,-6)$ and $(-1,-4)$. Also find the point of intersection.
Solution :

Let the ratio be $k: 1$. Then by the section formula, the coordinates of the point which divides $\mathrm{AB}$ in the ratio $k: 1$ are $\left(\frac{-k+5}{k+1}, \frac{-4 k-6}{k+1}\right)$.
This point lies on the $y$-axis, and we know that on the $y$-axis the abscissa is 0 .
Therefore,
$
\frac{-k+5}{k+1}=0
$

So,
$
k=5
$

That is, the ratio is $5: 1$. Putting the value of $k=5$, we get the point of intersection as $\left(0, \frac{-13}{3}\right)$.

Example 10 :

If the points $A(6,1), B(8,2), C(9,4)$ and $D(p, 3)$ are the vertices of a parallelogram, taken in order, find the value of $p$.
Solution :

We know that diagonals of a parallelogram bisect each other.
So, the coordinates of the mid-point of $\mathrm{AC}=$ coordinates of the mid-point of $\mathrm{BD}$
i.e.,
$
\left(\frac{6+9}{2}, \frac{1+4}{2}\right)=\left(\frac{8+p}{2}, \frac{2+3}{2}\right)
$
i.e.,
$
\left(\frac{15}{2}, \frac{5}{2}\right)=\left(\frac{8+p}{2}, \frac{5}{2}\right)
$
so,
$
\frac{15}{2}=\frac{8+p}{2}
$
i.e.,
$
p=7
$