NCERT Math Solutions for Class 10 Maths Chapter 7 Coordinate Geometry(English Medium) || CBSE Class 10 Maths Chapter 7 Coordinate Geometry Ex 7.4

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NCERT Math Solutions for Class 10 Maths Chapter 7 Coordinate Geometry(English Medium) || CBSE Class 10 Maths Chapter 7 Coordinate Geometry Ex 7.4
Exercise - 7.4

NCERT Math Solutions for Class 10 Maths Chapter 7 Coordinate Geometry(English Medium) || CBSE Class 10 Maths Chapter 7 Coordinate Geometry Ex 7.4

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1. Determine the ratio in which the line $ 2 x+y-4=0 $ divides the line segment joining the points $ \mathrm{A}(2,-2) $ and $ \mathrm{B}(3,7) $.

Answer

Let the given line divide the line segment joining the points $ \mathrm{A}(2,-2) $ and $ \mathrm{B}(3,7) $ in a ratio $ k: 1 $
Coordinates of the point of division $ =\left(\frac{3 k+2}{k+1}, \frac{7 k-2}{k+1}\right) $
This point also lies on $ 2 x+y-4=0 $
Therefore,
$
2\left(\frac{3 k+2}{k+1}\right)+\left(\frac{7 k-2}{k+1}\right)-4=0$
$\frac{6 k+4+7 k-2-4 k-4}{k+1}=0$
$9 \mathrm{k}-2=0$
$k=\frac{2}{9}
$
Hence, the ratio in which the line $ 2 x+y-4=0 $ divides the line segment joining the points $ \mathrm{A}(2,-2) $ and $ \mathrm{B}(3,7) $ is $ \mathbf{2 : 9} $.

2. Find a relation between $ x $ and $ y $ if the points $(x, y), (1,2)$ and $(7, 0 )$ are collinear

Answer

If the given points are collinear, then the area of triangle formed by these points will be 0
$
\text { Area of the triangle }=\frac{1}{2}\left[x_{1}\left(y_{2}-y_{3}\right)+x_{2}\left(y_{3}-y_{1}\right)+x_{3}\left(y_{1}-y_{2}\right)\right.$
$=\frac{1}{2}[x(2-0)+1(0-y)+7(y-2)$
$0=\frac{1}{2}(2 x-y+7 y-14)$
$2 x+6 y-14=0
$
$
x+3 y-7=0
$
This is the required relation between $ x $ and $ y $.

NCERT Math Solutions for Class 10 Maths Chapter 7 Coordinate Geometry(English Medium) || CBSE Class 10 Maths Chapter 7 Coordinate Geometry Ex 7.4

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3. Find the centre of a circle passing through the points $ (6,-6),(3 $, $ -7) $ and $ (3,3) $
The figure for the question is:

Answer

Let $ \mathrm{O}(x, y) $ be the centre of the circle. And let the points $ (6,-6),(3,- 7 )$, and $ (3,3) $ be representing the points $ \mathrm{A}, \mathrm{B} $, and C on the circumference of the circle
By distance formula, Distance between two points $ \mathrm{A}\left(x_{1}, y_{1}\right) $ and $ \mathrm{B}\left(x_{2}\right. $, $ y_{2} $ ) is
$
\mathrm{AB}=\sqrt{ }\left(\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}\right)
$
Then,
$
\mathrm{OA}=\sqrt{ }\left[(x-6)^{2}+(y+6)^{2}\right]$
$\mathrm{OB}=\sqrt{ }\left[(x-3)^{2}+(y+7)^{2}\right]$
$\mathrm{OC}=\sqrt{ }\left[(x-3)^{2}+(y-3)^{2}\right]
$
As $ \mathrm{OA}, \mathrm{OB} $ and OC are radii of same circle, $ \mathrm{OA}=\mathrm{OB} $
$
\left[(x-6)^{2}+(y+6)^{2}\right]^{\frac{1}{2}}=\left[(x-3)^{2}+(y+7)^{2}\right]^{\frac{1}{2}}
$
$
-6 x-2 y+14=0$
$3 x+y=7\ldots(i)$
Similarly, $ \mathrm{OA}=\mathrm{OC} $
$
{\left[(x-6)^{2}+(y+6)^{2}\right]^{\frac{1}{2}}=\left[(x-3)^{2}+(y-3)^{2}\right]^{\frac{1}{2}}}$
$-6 x+18 y+54=0$
$-3 x+9 y=-27\ldots(ii)$
On adding equation (i) and (ii), we obtain
$
10 y=-20$
$y=-2
$
From equation (i), we obtain
$
3 x-2=7$
$3 x=9$
$x=3
$
Therefore, the centre of the circle is $ (3,-2) $.

4. The two opposite vertices of a square are $ (-1,2) $ and $ (3,2) $. Find the coordinates of the other two vertices

Answer

Let ABCD be a square having $ (-1,2) $ and $ (3,2) $ as vertices A and C respectively. Let $ (x, y),\left(x_{1}, y_{1}\right) $ be the coordinate of vertex B and D respectively
We know that the sides of a square are equal to each other

From the figure we can see that the sides of square $ A B $ and $ B C $ are equal
And also we know that for two points $ \mathrm{P}\left(x_{1}, y_{1}\right) $ and $ \mathrm{Q}\left(x_{2}, y_{2}\right) $, The distance between points P and Q is given by the formula
$
P Q=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}
$
From the figure,
$
\mathrm{AB}=\mathrm{BCTherefore}, \sqrt{ }\left[(x+1)^{2}+(y-2)^{2}\right]=\sqrt{ }\left[(x-3)^{2}+(y-2)^{2}\right]
$
Squaring both side, and evaluating
$
x^{2}+2 x+1+y^{2}-4 y+4=x^{2}+9-6 x+y^{2}+4-4 y$
$8 x=8$
$x=1
$
We know that in a square, all interior angles are of $ 90^{\circ} $.
Pythagoras theorem: the square of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the other two sides.
In $ \triangle \mathrm{ABC} $,
Applying pythagoras theorem we get,
$
\mathrm{AB}^{2}+\mathrm{BC}^{2}=A C^{2}$
$\sqrt{ }\left[(x+1)^{2}+(y-2)^{2}\right]^{2}+\sqrt{ }\left[(x-3)^{2}+(y-2)^{2}\right]^{2}=\sqrt{ }\left[(3+1)^{2}+(2-2)^{2}\right]^{2}
$
$
4+y^{2}+4-4 y+4+y^{2}-4 y+4=16$
$2 y^{2}+16-8 y=16$
$2 y^{2}-8 y=0$
$y(y-4)=0$
$y=0 \text { or } 4
$
Now we have obtained the $ B(x, y) $ point and we are left to find point $ \mathrm{C}\left(x_{1}, y_{1}\right) $
We know that in a square, the diagonals are equal in length and bisect each other at $ 90^{\circ} $.
Let O be the mid-point of AC. Therefore, it will also be the mid-point of BD
If $ O $ is mid-point of $ A C $, then
we know, mid-point of ( $ x_{1}, y_{1} $ ) and ( $ x_{2}, y_{2} $ ) is
$
\left(\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2}\right)
$
i.e. $ O=\left(\frac{-1+3}{2}, \frac{2+2}{2}\right) $
therefore, $ \mathrm{O}=(1,2) $ Also, if we take O as midpoint of BD then
$
O=\left(\frac{x+x_{1}}{2}, \frac{y+y_{1}}{2}\right)
$
If $ y=0 $
$
\Rightarrow(1,2)=\left(\frac{1+x_{1}}{2}, \frac{0+y_{1}}{2}\right)$
$\Rightarrow \frac{1+x_{1}}{2}=1 \text { and } \frac{0+y_{1}}{2}=2$
$x_{1}=1 \text { and } y_{1}=4
$
If $ y=4 $
$
\Rightarrow(1,2)=\left(\frac{1+x_{1}}{2}, \frac{4+y_{1}}{2}\right)
$
$
\Rightarrow \frac{1+x_{1}}{2}=1 \text { and } \frac{4+y_{1}}{2}=2$
$x_{1}=1 \text { and } y_{1}=0
$
Therefore, The sides of square are $ \mathrm{A}(-1,2), \mathrm{B}(1,4), \mathrm{C}(1,0) $ and $ \mathrm{D}(3 $, 2) or $ \mathrm{A}(-1,2), \mathrm{B}(1,0), \mathrm{C}(1,4) $ and $ \mathrm{D}(3,2) $.

5. The Class X students of a secondary school in Krishinagar have been allotted a rectangular plot of land for their gardening activity.
Sapling of Gulmohar are planted on the boundary at a distance of 1 m from each other. There is a triangular grassy lawn in the plot as shown in the Fig. 7.14. The students are to sow seeds of flowering plants on the remaining area of the plot
(i) Taking A as origin, find the coordinates of the vertices of the triangle.
(ii) What will be the coordinates of the vertices of $ \triangle \mathrm{PQR} $ if C is the origin?
Also calculate the areas of the triangles in these cases. What do you observe?

Answer

(i) Taking $ A $ as origin,
We will take AD as $ x $-axis and AB as $ y $-axis.
It can be observed that the coordinates of point $ P, Q $, and $ R $ are $ (4,6) $, $ (3,2) $, and $ (6,5) $ respectively
Area of the triangle $PQR$
$ =\frac{1}{2}\left[x_{1}\left(y_{2}-y_{3}\right)+x_{2}\left(y_{3}-y_{1}\right)+x_{3}\left(y_{1}-y_{2}\right)\right. $
$
=\frac{1}{2}[4(2-5)+3(5-6)+6(6-2)$
$=\frac{1}{2}(-12-3+24)$
$=\frac{9}{2} \text { Square units }
$
(ii) Taking C as origin, CB as $ x $-axis, and CD as $ y $-axis, the coordinates of vertices $ \mathrm{P}, \mathrm{Q} $, and R are $ (12,2),(13,6) $, and $ (10,3) $ respectively
Area of the triangle $ P Q R$
$=\frac{1}{2}\left[x_{1}\left(y_{2}-y_{3}\right)+x_{2}\left(y_{3}-y_{1}\right)+x_{3}\left(y_{1}-y_{2}\right)\right. $
$
=\frac{1}{2}[12(6-3)+13(3-2)+10(2-6)$
$=\frac{1}{2}(36+13-40)$
$=\frac{9}{2} \text { Square units }
$
It can be observed that the area of the triangle is same in both the cases.

NCERT Math Solutions for Class 10 Maths Chapter 7 Coordinate Geometry(English Medium) || CBSE Class 10 Maths Chapter 7 Coordinate Geometry Ex 7.4

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6. The vertices of a $ \triangle A B C $ are $ A(4,6), B(1,5) $ and $ C(7,2) $. $ A $ line is drawn to intersect sides AB and AC at D and E respectively, such that $ \frac{A D}{A B}=\frac{A E}{A C}=\frac{1}{4} $ Calculate the area of the $ \triangle \mathrm{ADE} $ and compare it with the area of $ \triangle \mathrm{ABC} $ (Recall Theorem 6.2 and Theorem 6.6).

Answer

Given that

$\frac{A D}{D B}=\frac{A E}{A C}=\frac{1}{4}$
$\frac{A D}{A D+D B}=\frac{A E}{A E+E C}=\frac{1}{4}$
$\frac{A D}{D B}=\frac{A E}{E C}=\frac{1}{3}$
Therefore, D and E are two points on side AB and AC respectively such that they divide side $ A B $ and $ A C $ in a ratio of $ 1: 3 $
Coordinates of point $ \mathrm{D}=\left(\frac{1 * 1+3 * 4}{1+3}, \frac{1 * 5+3 * 6}{1+3}\right) $
$
=\left(\frac{13}{4}, \frac{23}{4}\right)
$
Coordinates of point $ \mathrm{D}=\left(\frac{1 * 7+3 * 4}{1+3}, \frac{1 * 2+3 * 6}{1+3}\right) $
$
=\left(\frac{19}{4}, \frac{20}{4}\right)
$
Area of the triangle
$ =^{\frac{1}{2}}\left[x_{1}\left(y_{2}-y_{3}\right)+x_{2}\left(y_{3}-y_{1}\right)+x_{3}\left(y_{1}-y_{2}\right)\right. $
$ =\frac{1}{2}\left[4\left(\frac{23}{4}-\frac{20}{4}\right)+\frac{13}{4}\left(\frac{20}{4}-6\right)+\frac{19}{4}\left(6-\frac{23}{4}\right)\right]$
$=\frac{1}{2}\left[3-\frac{13}{4}+\frac{19}{16}\right]$
$=\frac{1}{2}\left[\frac{48-52+19}{16}\right]$
$=\frac{ 15 }{ 32 } \text { square units } $
Area of triangle $ \mathrm{ABC} =\frac{[4(5-2)+1(2-6)+7(6-5)] }{ 2 }$
$=\frac{ (12-4+7) }{ 2 }$
$=\frac{ 15 }{ 2 } \text { square units }
$
Clearly, the ratio between the areas of $ \triangle \mathrm{ADE} $ and $ \triangle \mathrm{ABC} $ is $ 1: 16 $
We know that if a line segment in a triangle divides its two sides in the same ratio, then the line segment is parallel to the third side of the triangle.
These two triangles so formed (here $ \triangle \mathrm{ADE} $ and $ \triangle \mathrm{ABC} $ ) will be similar to each other.
Hence, the ratio between the areas of these two triangles will be the square of the ratio between the sides of these two triangles
And, ratio between the areas of $ \triangle \mathrm{ADE} $ and $ \Delta \mathrm{ABC}=(\frac{ 1 }{ 4 }) ^{2}$ $ =\frac{ 1 }{ 16 } $

7. Let $ A(4,2), B(6,5) $ and $ C(1,4) $ be the vertices of $ \Delta \mathrm{ABC} $
(i) The median from A meets BC at D . Find the coordinates of the point $ D $
(ii) Find the coordinates of the point P on AD such that $ \mathrm{AP}: \mathrm{PD}=2: 1 $
(iii) Find the coordinates of points Q and R on medians BE and CF respectively such that $ \mathrm{BQ}: \mathrm{QE}=2: 1 $ and $ \mathrm{CR}: \mathrm{RF}=2: 1 $
(iv) What do you observe?
(v) If $ A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right) $ and $ B\left(x_{3}, y_{3}\right) $ are the vertices of $ \Delta A B C $, find the coordinates of the centroid of the triangle.

Answer

(i) Median AD of the triangle will divide the side BC in two equal parts
Therefore, D is the mid-point of side BC

And according to midpoint formula, midpoints of $ \left(x_{1}, y_{1}\right),\left(x_{2}, y_{2}\right) $ is given by
$
(x, y)=\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2}
$
Coordinate of $ \mathrm{D}=\left(\frac{6+1}{2}, \frac{5+4}{2}\right) $
$
=\left(\frac{7}{2}, \frac{9}{2}\right)
$
(ii) Point P divides the side AD in a ratio $ 2: 1 $
By section formula if $ (x, y) $ divides line joining points $ \mathrm{A}\left(x_{1}, y_{1}\right) $, $ \mathrm{B}\left(x_{2}, y_{2}\right) $ in ratio m:n
Then, $ x, y=\frac{m x_{2}+n x_{1}}{m+n}+\frac{m y_{2}+n y_{1}}{m+n} $
Coordinates of $ \mathrm{p}=\left(\frac{2 \times \frac{7}{2}+1 \times 4}{2+1}, \frac{2 \times \frac{9}{2}+1 \times 2}{2+1}\right) $
$
=\left(\frac{11}{3}, \frac{11}{3}\right)
$
(iii) Median BE of the triangle will divide the side AC in two equal parts.
Therefore, E is the mid-point of side AC
Coordinates of $ \mathrm{E}=\left(\frac{4+1}{2}, \frac{2+4}{2}\right) $
$
=\left(\frac{5}{2,3}\right)
$
Point Q divides the side BE in a ratio 2:1
By section formula if $ (x, y) $ divides line joining points $ \mathrm{A}\left(x_{1}, y_{1}\right) $, $ \mathrm{B}\left(x_{2}, y_{2}\right) $ in ratio m:n
Then, $ x, y=\frac{m x_{2}+n x_{1}}{m+n}+\frac{m y_{2}+n y_{1}}{m+n} $
Coordinates of $ \mathrm{p}=\left(\frac{2 \times \frac{5}{2}+1 \times 6}{2+1}, \frac{2 \times 3+1 \times 5}{2+1}\right) $
$
=\left(\frac{11}{3}, \frac{11}{3}\right)
$
Median CF of the triangle will divide the side AB in two equal parts. Therefore, $ F $ is the mid-point of side $ A B $
Coordinates of $ \mathrm{F}=\left(\frac{4+6}{2}, \frac{2+5}{2}\right) $
$
=\left(5, \frac{7}{2}\right)
$
Point R divides the side CF in a ratio $ 2: 1 $
By section formula if $ (x, y) $ divides line joining points $ \mathrm{A}\left(x_{1}, y_{1}\right) $, $ \mathrm{B}\left(x_{2}, y_{2}\right) $ in ratio m:n
Then, $ x, y=\frac{m x_{2}+n x_{1}}{m+n}+\frac{m y_{2}+n y_{1}}{m+n} $
Coordinates of $ \mathrm{p}=\left(\frac{2 \times 5+1 \times 6}{2+1}, \frac{2 \times \frac{7}{2}+1 \times 4}{2+1}\right) $
$
=\left(\frac{11}{3}, \frac{11}{3}\right)
$
(iv) It can be observed that the coordinates of point $ P, Q, R $ are the same. Therefore, all these are representing the same point on the plane i.e., the centroid of the triangle.
(v)

Consider a triangle, $ \triangle \mathrm{ABC} $, having its vertices as $ \mathrm{A}\left(x_{1}, y_{1}\right), \mathrm{B}\left(x_{2}, y_{2}\right) $, and $ C\left(x_{3}, y_{3}\right) $
Median AD of the triangle will divide the side BC in two equal parts. Therefore, D is the mid-point of side BC
Coordinates of $ \mathrm{D}=\left(\frac{x_{2}+x_{3}}{2}, \frac{y_{2}+y_{3}}{2}\right) $
Let the centroid of this triangle be O . Point O divides the side AD in a ratio $ 2: 1 $
By section formula if $ (x, y) $ divides line joining points $ \mathrm{A}\left(x_{1}, y_{1}\right) $, $ \mathrm{B}\left(x_{2}, y_{2}\right) $ in ratio m:n
then $ x, y=\frac{m x_{2}+n x_{1}}{m+n}+\frac{m y_{2}+n y_{1}}{m+n} $
Coordinates of $ \mathrm{O}=\left(\frac{2 \times \frac{x_{2}}{3}+\frac{x_{3}}{2}+1 \times x_{1}}{2+1}, \frac{2 \times \frac{y_{2}}{3}+\frac{y_{3}}{2}+1 \times y_{1}}{2+1}\right) $
Centroid of $ \mathrm{ABC}=\left(\frac{x_{1}+x_{2}+x_{3}}{3}, \frac{y_{1}+y_{2}+y_{3}}{3}\right) $

8. ABCD is a rectangle formed by the points $ \mathrm{A}(-1,-1), \mathrm{B}(-1,4) $, $ C(5,4) $ and $ D(5,-1) . P, Q, R $ and $ S $ are the mid-points of $ A B, B C, C D $ and DA respectively. Is the quadrilateral PQRS a square? a rectangle? or a rhombus? Justify your answer.

Answer

$ P $ is the mid-point of $ A B $

Coordinates of $ P=\left(-1, \frac{3}{2}\right) $
Similarly coordinates of $ \mathrm{Q}, \mathrm{R} $ and S are $ (2,4),\left(5, \frac{3}{2}\right) $ and $ (2,-1) $ respectively.
Length of $ \mathrm{PQ}=\left[(-1-2)^{2}+\left(\frac{3}{2}-4\right)^{2}\right]^{\frac{ 1 }{ 2 }} $
$
=\sqrt{\frac{61}{4}}
$
Length of $ \mathrm{QR}=\left[(2-5)^{2}+\left(4-\frac{3}{2}\right)^{2}\right]^{\frac{ 1 }{ 2 }} $
$
=\sqrt{\frac{61}{4}}
$
Length of $ \mathrm{RS}=\left[(5-2)^{2}+\left(\frac{3}{2}+1\right)^{2}\right]^{\frac{ 1 }{ 2 }} $
$
=\sqrt{\frac{61}{4}}
$
Length of SP $ =\left[(2+1)^{2}+\left(-1-\frac{3}{2}\right)^{2}\right]^{\frac{ 1 }{ 2 }} $
$
=\sqrt{\frac{61}{4}}
$
Length of $ \mathrm{SP}=\left[(-1-5)^{2}+\left(\frac{3}{2}-\frac{3}{2}\right)^{2}\right]^{\frac{ 1 }{ 2 }} $ $ =6 $
Length of $ \mathrm{QS}=\left[(2-2)^{2}+(4+1)^{2}\right]^{\frac{ 1 }{ 2 }}=5 $
It can be observed that all sides of the given quadrilateral are of the same measure. However, the diagonals are of different lengths.
Therefore, PQRS is a rhombus

NCERT Math Solutions for Class 10 Maths Chapter 7 Coordinate Geometry(English Medium) || CBSE Class 10 Maths Chapter 7 Coordinate Geometry Ex 7.4

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NCERT Math Solutions for Class 10 Maths Chapter 7 Coordinate Geometry(English Medium) || CBSE Class 10 Maths Chapter 7 Coordinate Geometry Ex 7.4

Download the Math Ninja App Now