NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry

Q1 ) Find the coordinates of the point which divides the join of (- 1, 7) and (4, – 3) in the ratio 2:3.

Answer :

Let the points be P(x,y), Using the section formula

\(P(x, y) = [\frac{mx_2 + nx_1}{m + n} , \frac{my_2 + ny_1}{m + n}]\)

\(x = {{2(4) + 3(-1)} \over {2 + 3}} = {{8 - 3} \over {5}} = 1\)
\(y = {{2(-3) + 3(7)} \over {2 + 3}} = {{-6 + 21} \over {5}} = 3\)
Therefore the point is(1,3)

Q2 ) Find the coordinates of the points of trisection of the line segment joining (4, -1) and (-2, -3).

Answer :

Let P (x₁, y₁) and Q (x₂, y₂) are the points of trisection of the line segment joining the given points i.e., AP = PQ = QB

Therefore, point P divides AB internally in the ratio 1:2.

x₁ = \(\frac{(1×(-2)+2×4)}{1+2} = \frac{(-2+8)}{3} =\frac{6}{3} = 2 \)

y₁ = \(\frac{(1×(-3)+2×-1)}{1+2} = \frac{(-3-2)}{3} =\frac{-5}{3} \)

\(\therefore \)P = (x₁, y₁) = P(2, \(\frac{-5}{3} \))

Point Q divides AB internally in the ratio 2:1.

x₂ = \(\frac{(2×(-2)+1×4)}{2+1} = \frac{(-4+4)}{3} = 0\)

y₂ = \(\frac{(2×(-3)+1×-1)}{2+1} = \frac{(-6-1)}{3} =\frac{-7}{3} \)

The coordinates of the point Q(x₂ ,y₂ ) = (0, \(\frac{-7}{3} \) )

Q3 ) To conduct Sports Day activities, in your rectangular shaped school ground ABCD, lines have been drawn with chalk powder at a distance of 1 m each. 100 flower pots have been placed at a distance of 1 m from each other along AD, as shown in the following figure. Niharika runs \(\frac{1}{4} \) th the distance AD on the 2nd line and posts a green flag. Preet runs \(\frac{1}{5} \)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 segment joining the two flags, where should she post her flag?

Answer :

The green flag is \(\frac{1}{4} th \) of total distance =
\(\frac{1}{4} × 100 = 25 \) m in second line

\(\therefore \) the coordinate of green flag
= (2 , 25 )
Similarly coordinate of red flag
= (8 , 20)

Distance between red and green flag
= \( \sqrt{(8-2)^2 + (20-25)^2} \)
\( = \sqrt{6^2+(-5)^2} \)
\( = \sqrt{36+25} \)
\( = \sqrt{61} m \)

Now blue flag is posted at the mid point of two flag
Then let, the coordinate of blue flag = (x , y )

\(\therefore( x , y ) =( \frac{(8+2)}{2} , \frac{20+25}{2} ) \)
\( =( 5 , \frac{45}{2} )= ( 5, 22.5 ) \)

\(\therefore \) the blue flag is posted in fifth line at a distance of 22.5 m

Q4 ) Find the ratio in which the line segment joining the points (-3, 10) and (6, – 8) is divided by (-1, 6).

Answer :

Let the required ratio be k : 1
By section formula ,

\( x = \frac{m1x2 + m2x1}{m1+m2} \)
\( \Rightarrow -1 = \frac{k×6 + 1× -3 )}{k+1} \)
\( \Rightarrow -k-1 = 6k -3 \)
\( \Rightarrow 7k = 2 \)
\( \Rightarrow k = \frac{2}{7} \)

The required ratio = k : 1
= \(\frac{2}{7} \) : 1 = \(\frac{2}{7} ×7 : 1×7 \) = 2 : 7

Q5 ) 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.

Answer :

Let p(x , 0) be the point which devides the line segment joining A(1, -5 ) and B(-4,5 ) in the ratio m : 1

Then the using section formula

(x , 0 ) = \(( \frac{m×(-4) + 1 × 1 }{m+1} , \frac{m×5+1×(-5)}{m+1} ) \)
\(\Rightarrow 0 = \frac{m×5+1×(-5)}{m+1} \)
\(\Rightarrow 0 = 5m - 5 \)
\(\Rightarrow m = 5 /5 = 1 \)

Hence the required ratio is 1 : 1

Since the ratio is 1 : 1 , so P is the mid point

\(\therefore \) x = \(\frac{m×(-4) + 1 × 1 }{m+1} \)
x = \(\frac{1×(-4) + 1 × 1 }{1+1} = \frac{-4+1}{2} = \frac{-3}{2} \)

\(\therefore ( \frac{-3}{2} , 0 )\) is the required point

Q6 ) If (1, 2), (4, y), (x, 6) and (3, 5) are the vertices of a parallelogram taken in order, find x and y.

Answer :

Mid point of AC = Mid point BD
\(\Rightarrow \frac{x+1}{2} , \frac{6+2}{2} = \frac{4+3}{2} , \frac{y+5}{2} \)
\(\Rightarrow \frac{x+1}{2} = \frac{7}{2} \)
\(\Rightarrow x +1 = 7 \)
\(\Rightarrow x = 7-1= 6 \)
\(\Rightarrow \frac{y+5}{2} = \frac{8}{2} \)
\(\Rightarrow y+5 = 8 \)
\(\Rightarrow y = 8 -5 = 3 \)

Q7 ) Find the coordinates of a point A, where AB is the diameter of circle whose centre is (2, – 3) and B is (1,4).

Answer :

Let the coordinates of point A be (x, y).

Mid-point of AB is (2, – 3), which is the centre of the circle.

Coordinate of B = (1, 4)
(2, -3) =\( (\frac{x+1}{2} , \frac{y+4}{2} )\)
\(\frac{x+1}{2} = 2\)
\( \Rightarrow x+1= 4 \)
\( \Rightarrow x = 4-1 = 3 \)
and\( \frac{y+4}{2} = -3 \)
\( \Rightarrow y+4 = -6 \)
\(\Rightarrow y = -6-4 = -10 \)
\(\therefore \) x = 3 and y = -10
The coordinates of A(3,-10).

Q8 ) If A and B are (-2, -2) and (2, -4), respectively, find the coordinates of P such that AP = \(\frac{3}{7} \) AB and P lies on the line segment AB.

Answer :

AP = \(\frac{3}{7} \) AB
7 AP = 3 AB
AB : AP = 7: 3
Let AB = 7x
\(\therefore \) AP = 3x
AB = AP + BP
7x = 3x + BP
BP = 7x - 3x = 4x

\(\therefore \frac{AP}{BP} = \frac{3x}{4x} \)
\(\therefore \) AP : BP = 3:4
by section formula ,

(x, y ) = \( (\frac{m1x2 + m2x1}{m1+m2} , \frac{m1y2 + m2y1}{m1+m2} ) \)
(x, y) = \( (\frac{3×2 +4×(-2) }{3+4} , \frac{3×(-4)+ 4(-2)}{3+4} ) \)
(x ,y ) = \( (\frac{6 -8 }{7} , \frac{-12+ -8}{7} ) \)
(x ,y ) = \( (\frac{-2}{7} , \frac{-20}{7} ) \)

Hence the coordinates of P = \( (\frac{-2}{7} , \frac{-20}{7} ) \)

Q9 ) Find the coordinates of the points which divide the line segment joining A (- 2, 2) and B (2, 8) into four equal parts.

Answer :

From the figure, it can be observed that points X, Y, Z are dividing the line segment in a ratio 1:3, 1:1, 3:1 respectively.

Coordinate of X = \(( \frac{1×2+3×(-2)}{1+3} , \frac{1×8+3×2}{1+3} ) \)
\( =( \frac{2-6}{4} , \frac{8+6}{4} ) \)
\( = ( \frac{-4}{4} , \frac{14}{4} ) \)
\( = ( -2 , \frac{7}{2} )\)
Coordinate of Y = \(( \frac{1×2+1×(-2)}{1+1} , \frac{1×8+1×2}{1+1} ) \)
\( =( \frac{2-2}{2} , \frac{8+2}{2} ) \)
\( = ( \frac{0}{2} , \frac{10}{2} ) \)
\( = ( 0 , 5) \)
Coordinate of Y = \(( \frac{3×2+1×(-2)}{3+1} , \frac{3×8+1×2}{3+1} ) \)
\( =( \frac{6-2}{4} , \frac{24+2}{4} ) \)
\( = ( \frac{4}{4} , \frac{26}{4} ) \)
\( = ( 1 , \frac{13}{2} ) \)

Q10 ) 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 = \(\frac{1}{2} \) (product of its diagonals)

Answer :

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

Length of diagonal AC
\(=\sqrt{[3-(-1)]^2 + (0-4)^2} \)
\( = \sqrt{16+16} \)
\( = \sqrt{32} \)
\( = 4\sqrt{2} \)
Length of diagonal BD
\(=\sqrt{[(-2)-4]^2+ (-1-5)^2} \)
\( = \sqrt{(-6)^2+(-6)^2} \)
\( = \sqrt{36+36} \)
\( = \sqrt{72} \)
\( = 6\sqrt{2}\)

Area of rhombus = \(\frac{1}{2} \) × product of diagonals
= \(\frac{1}{2} × 4\sqrt{2}×6\sqrt{2} = 24 \)sq unit

Q1 ) Find the area of the triangle whose vertices are:
(i) (2, 3), (-1, 0), (2, -4)
(ii) (-5, -1), (3, -5), (5, 2)

Answer :

We know that formula for area of a triangle whose vertices are \( (x_1,y_1) , (x_2,y_2) , (x_3,y_3) \) is,
= \( \frac{1}{2} \ | \ [x_1(y_2 - y_3) \ + \ x_2(y_3 - y_1) \ + \ x_3(y_1 - y_2)] \ | \)

(i) So, here \(x_1 \ = \ 2 \ , \ y_1 \ = \ 3 \ , \ x_2 \ = \ -1 \ , \ y_2 \ = \ 0 \ x_3 \ = \ 2 \ , \ y_3 \ = \ -4 \)

So, area of triangle = \( \frac{1}{2} \ | \ [ 2( 0 + 4 ) \ - \ 1(-4 - 3) \ + \ 2(3 - 0)] \ | \ = \ \frac{1}{2} \ | \ [ 8 + 7 + 6 ] \ | \ = \ \frac{21}{2} \)

∴ Area of triangle is \( \frac{21}{2} \) sq. units.

(ii) Similarly, here \(x_1 \ = \ -5 \ , \ y_1 \ = \ -1 \ , \ x_2 \ = \ 3 \ , \ y_2 \ = \ -5 \ x_3 \ = \ 5 \ , \ y_3 \ = \ 2 \)
So, area of triangle = \( \frac{1}{2} \ | \ [ (-5)( -5 - 2 ) \ - \ 3(2 + 1) \ + \ 5(-1 + 5)] \ | \ = \ \frac{1}{2} \ | \ [ 35 + 9 + 20 ] \ | \ = \ 32 \)

∴ Area of triangle is \( 32 \) sq. units.

Q2 ) In each of the following find the value of ‘k’, for which the points are collinear.
(i) (7, -2), (5, 1), (3, -k)
(ii) (8, 1), (k, -4), (2, -5)

Answer :

We know that for collinear points area of triangle = 0 ,i.e.,
\( 0 \ = \frac{1}{2}[x_1(y_2 - y_3) \ + \ x_2(y_3 - y_1) \ + \ x_3(y_1 - y_2)] \)

(i) \( 7(1-k) + 5(k+2) + 3(-2-1) \ = \ 0 \)
=> \( 7 - 7k + 5k + 10 - 9 \ = \ 0 \)
=> \( 2k \ = \ 8 \)
=> \( k \ = \ 4 \)

(ii) \( 8(-4+5) + k(-5-1) + 2(1+4) \ = \ 0 \)
=> \( 8 - 6k + 10 \ = \ 0 \)
=> \( 6k \ = \ 18 \)
=> \( k \ = \ 3 \)

Q3 ) Find the area of the triangle formed by joining the mid-points of the sides of the triangle whose vertices are (0, -1), (2, 1) and (0, 3). Find the ratio of this area to the area of the given triangle.

Answer :

The vertices of the triangle ABC are A (0, -1), B (2, 1), C (0, 3).
Let D, E, F be the mid-points of the sides AB,BC,CA of this triangle ABC respectively.
Coordinates of D, E, and F are given by :

D = \( (\frac{0+2}{2} \ , \ \frac{-1+1}{2}) \) = (1,0)
E = ( \( \frac{0+2}{2} \ , \ \frac{1+3}{2} \) ) = (1,2)
F = ( \( \frac{0+0}{2} \ , \ \frac{3-1}{2} \) ) = (0,1)

So, Area of \( ∆ \ DEF \ = \ \frac{1}{2} \ | \ [ 1(2-1) + 1(1-0) + 0(0-2)] \ | \ = \ 1 \)

∴ Area of ∆ DEF = 1 sq. units

Area of ∆ ABC = \( \frac{1}{2} \ | \ [0(1-3) \ + \ 2(3+1) \ + \ 0(-1-1) ] \ | \ = \ 4 \)

\(\therefore \) Area of \(\triangle \) ABC = 4 sq. units

So, \( ∆ \ DEF \ : \ ∆ \ ABC \ = \ 1 \ : \ 4 \).

Q4 ) Find the area of the quadrilateral whose vertices, taken in order, are (-4, -2), (-3, -5), (3, -2) and (2, 3).

Answer :

Draw a line from B to D



Now we have,
Area of quad. ABCD = | Area of ∆ DAB | + | Area of ∆ BCD |

So,
Area of ∆ DAB = \( | \ \frac{1}{2} [2(-2+5) \ - \ 4(-5-3) \ - \ 3(3+2)] \ | \) = \( \ | \frac{1}{2} [6 + 32 - 15 ] \ | \) = \( \frac{23}{2} \) sq. units

Similarly, Area of ∆ BCD = \( | \ \frac{1}{2} [-3(-2-3) \ + \ 3(3+5) \ + \ 2(-5+2)] \ | \ = \ | \ \frac{1}{2} [15 + 24 - 6 ] \ | \) = \( \frac{33}{2} \) sq. units

\(\therefore \) Area of quad. ABCD = \( \frac{23}{2} \ + \ \frac{33}{2} \) = \( \frac{56}{2} \ = \ 28 \) sq. units

Q5 ) You have studied in Class IX that a median of a triangle divides it into two triangles of equal areas. Verify this result for \(\triangle \) ABC whose vertices are A (4, – 6), B (3, – 2) and C (5, 2).

Answer :

The vertices of triangle ABC are A(4,-6), B (3, – 2) and C (5, 2).

Let D be the mid-point of side BC of ∆ ABC. Therefore, AD is the median in ∆ ABC.



So, coordinates of D are ( \( \frac{3+5}{2} \ , \ \frac{-2+2}{2} \) )
Therefore, coordinates of D = (4,0)

Now,
Area of ∆ ABD = \( \frac{1}{2} \ | \ [4(-2-0) \ + \ 3(0+6) \ + \ 4(-6+2)] | \)
= \( \frac{1}{2} \ | \ [ -8 + 18 - 16] \ | \)
= \( |-3| \ = \ 3 \) sq. units

Similarly,
Area of ∆ ACD = \( \frac{1}{2} \ | \ [4(0-2) \ + \ 4(2+6) \ + \ 5(-6-0)] \ | \)
= \( \frac{1}{2} \ | \ [ -8 + 32 - 30] \ | \)
= \( |-3| \ = \ 3 \) sq. units

\(\therefore \) Clearly, Area of ∆ ABD = Area of ∆ ACD = 3 sq. units.

Hence, median of a triangle divides it into two triangles of equal areas.

Q1 ) Determine the ratio in which the line 2x + y – 4 = 0 divides the line segment joining the points A(2, –2) and B(3, 7).

Answer :

Let the line 2x + y - 4 = 0 divides AB in 1 : k ratio, then the coordinates of the point of division is

\( x \ = \ \frac{2+3k}{k+1} \ and \ y \ = \ \frac{-2+7k}{k+1} \)

The point of intersection lie on both lines

\(\therefore x \ = \ \frac{2+3k}{k+1} \ and \ y \ = \ \frac{-2+7k}{k+1} \) should satisfy 2x + y - 4 = 0
\( \Rightarrow 2( \frac{2+3k}{k+1}) \ + \ (\frac{-2+7k}{k+1}) \ - \ 4 \ = \ 0 \)  
\( \Rightarrow 4 \ + \ 6k \ - \ 2 \ + \ 7k \ = \ 4k \ + \ 4 \)
\(\Rightarrow 9k - 2 = 0\)
\( \Rightarrow k \ = \ \frac{2}{9} \)

\( \therefore \) The line 2x + y – 4 = 0 divides the line segment joining the points A(2, –2) and B(3, 7) in ratio 2 : 9 internally.

Q2 ) Find the relation between x and y if the points (x, y), (1, 2) and (7, 0) are collinear.

Answer :

If given points are collinear then area of triangle formed by them must be zero.

Let (x, y), (1, 2) and (7, 0) are vertices of a triangle,

Area of triangle = \( \frac{1}{2} \ | \ [ (x_1(y_2 – y_3 ) + x_2(y_3 – y_1) + x_3(y_1 – y_2) ) ] \ | \ = \ 0 \)

\( \Rightarrow \frac{1}{2} \ | \ [x(2 – 0) + 1 (0 – y) + 7( y – 2)] \ | \ = \ 0 \)

\( \Rightarrow 2x \ – \ y \ + \ 7y \ – \ 14 \ = \ 0 \)

\(\Rightarrow 2x \ + \ 6y \ – \ 14 \ = \ 0 \)

\(\Rightarrow x \ + \ 3y \ – \ 7 \ = \ 0 \) .

Hence this is the required relation between x and y.

Q3 ) Find the centre of a circle passing through points (6, -6), (3, -7) and (3, 3).

Answer :

Let the coordinate of centre O be (x,y)

Then, OA = OB = OC {\(\because \) radius is equal}

Now,
OA = \( \sqrt{ (x-6)^2+(y+6)^2 } \)
OB = \( \sqrt{(x-3)^2 + (y+7)^2 } \)
OC = \( \sqrt{(x-3)^2 + (y-3)^2 } \)

Now, OA = OB
\( \Rightarrow \sqrt{(x-6)^2 + (y+6)^2 } = \sqrt{(x-3)^2 + (y+7)^2 } \)
\(\Rightarrow x^2 \ + \ 36 \ - \ 12x \ + \ y^2 \ + \ 36 \ - \ 12 \ = \ x^2 \ + \ 9 \ - \ 6x \ + \ y^2 \ + \ 9 \ - \ 6y \)
\( \Rightarrow 6x \ + 2y \ - \ 14 \ = \ 0 \) ..............(i)

Similarly, OB = OC
\( \Rightarrow \sqrt{(x-3)^2 + (y+7)^2 } = \sqrt{(x-3)^2 + (y-3)^2 } \)
\( \Rightarrow y^2 \ + \ 14y \ + \ 49 \ = \ y^2 \ - \ 6y \ + \ 9 \)
\(\Rightarrow y \ = \ -2 \)

Putting this value of y in eq. (i) , we get

\(\Rightarrow 6x \ + \ 2(-2) \ - \ 14 \ = \ 0 \)
\( \Rightarrow x \ = \ 3 \)

\(\therefore \) The centre of circle is at (3,-2)

Q4 ) 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 with vertices A( \( x_1, y_1) , B(3 , 2) , C(x_2 , y_2) , D(-1 , 2)\)

Let each side of square be 'a'.
So applying Pythagoras theorem in ∆ABD, we get

\( BD^2 = AB^2 + DA^2 \)
\( \Rightarrow \sqrt{(3+1)^2 \ + \ (2-2)^2} \ = \ a^2 \ + \ a^2 \)
\( \Rightarrow 4 \ = \ 2 a^2 \)
\(\Rightarrow a \ = \ 2 \sqrt{2} \)

Now, DA = AB {\(\because \) ABCD is a square }

\( \Rightarrow \sqrt{(x_1 +1)^2 + (y_1 - 2)^2 } \ = \ \sqrt{(3-x_1)^2 + (2 - y_1)^2} \) {by distance formula}
\(\Rightarrow (x_1 +1)^2 \ = \ (3-x_1)^2 \)
\( \Rightarrow x_1^2 \ + \ 1 \ + \ 2x_1 \ = \ x_1^2 \ + \ 9 \ - \ 6x_1 \)
\(\Rightarrow x_1 \ = \ 1 \)

Now since each side is \( 2 \sqrt{2} \)

\(\therefore \) AB \( = \ 2 \sqrt{2} \)
\( \Rightarrow \sqrt{(3-x_1)^2 + (2 - y_1)^2} \ = \ 2 \sqrt{2} \)
\( \Rightarrow(2-y_1)^2 = \ 4 \)
\(\Rightarrow y \ = \ 4 \)

Now, point O is id point of BD . SO coordinates of O are

x = \( \frac{3-1}{2} \) , y = \( \frac{2+2}{2} \)
\(\Rightarrow \) x = 1 , y = 2

Now, since ABCD is a square \(\therefore \) O is mid point of AC also

\( \frac{1+x_2}{2} \ = \ 1 \) , \( \frac{4+ y_2}{2} \ = \ 2 \)
\( \Rightarrow x_2 \ = \ 1 \) , \( y_2 \ = \ 0 \)

\(\therefore \) Coordinates of vertices square ABCD are A(1, 4) , B(3,2) , C(1,0) , D(-1,2)

Q5 ) The class X students of a secondary school in Krishinagar have been allotted a rectangular plot of land for their gardening activity. Saplings of Gulmohar are planted on the boundary at a distance of 1 m from each other. There is a triangular lawn in the plot as shown in the fig. 7.14. The students are to sow the 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 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, coordinates of the vertices P, Q and R are,

From figure:
P = (4, 6)
Q = (3, 2)
R = (6, 5)

Here AD is the x-axis and AB is the y-axis.

Area of triangle PQR in case of origin A:

Area of a triangle = \( \frac{1}{2} \ | \ [4(2 – 5) + 3 (5 – 6) + 6 (6 – 2)] \ | \ \)
\( = \ \frac{1}{2} \ | \ [– 12 – 3 + 24] \ | \ \)
\( = \ \frac{9}{2} \) sq. units

(ii) Taking C as origin, coordinates of vertices P, Q and R are,

From figure:
P = (12, 2)
Q = (13, 6)
R = (10, 3)

Here CB is the x-axis and CD is the y-axis.

Area of triangle PQR in case of origin C:

Area of a triangle = \( \frac{1}{2} \ | \ [ 12(6 – 3) + 13 ( 3 – 2) + 10( 2 – 6)] \ | \ \)
\( = \frac{1}{2} \ | \ [36 + 13 – 40] \ | \ \)
\( = \ \frac{9}{2} \) sq unit

This implies, Area of triangle PQR at origin A = Area of triangle PQR at origin C

Area is same in both case because triangle remains the same no matter which point is considered as origin.

Q6 ) The vertices of a \( ∆ \ ABC \) 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 \( \frac{AD}{AB} \ = \ \frac{AE}{AC} \ = \ \frac{1}{4} \) . Calculate the area of the \( ∆ \ ADE \) and compare it with area of \( ∆ \ ABC \) . (Recall Theorem 6.2 and Theorem 6.6)

Answer :

Given, the vertices of a \( ∆ \ ABC \) are A (4, 6), B (1, 5) and C (7, 2) and,
\( \frac{AD}{AB} \ = \ \frac{AE}{AC} \ = \ \frac{1}{4} \)
\( \Rightarrow \frac{AB}{AD} \ = \ \frac{AC}{AE} \ = \ 4 \)
\( \Rightarrow \frac{AD + DB}{AD} \ = \ \frac{AE + EC}{AE} \ = \ 4 \)
\( \Rightarrow 1 + \frac{DB}{AD} \ = \ 1 + \frac{EC}{AE} \ = \ 4 \)
\(\Rightarrow \frac{DB}{AD} \ = \ \frac{EC}{AE} \ = \ 3 \)

\(\therefore \) Point D and E divides AB and AC in ratio 1 : 3

Now coordinates of D calculated by section formula are,
\( x \ = \ \frac{3(4) + 1(1)}{4} \ = \ \frac{13}{4} \) and,
\( y \ = \ \frac{3(6) + 1(5)}{4} \ = \ \frac{23}{4} \)

Similarly, coordinates of E are,
\( x \ = \ \frac{1(7) + 3(4)}{4} \ = \ \frac{19}{4} \) and,
\( y \ = \ \frac{1(2) + 3(6)}{4} \ = \ \frac{20}{4} \ = \ 5 \)

Now area of triangle ADE is,

\( \frac{1}{2} \ | \ [4(5-2) \ + \ 1(2-6) \ + \ 7(6-5)] \ | \)
= \( \frac{1}{2} \ | \ [(12-4+7)] \ | \)
= \( \frac{15}{2} \) sq. units

Similarly, area of triangle ABC is,

\( \frac{1}{2} \ | \ [4( \frac{23}{4} -5) \ + \ \frac{13}{4}(5-6) \ + \ \frac{19}{4}(6- \frac{23}{4})] \ | \)
= \( \frac{1}{2} \ | \ [(3- \frac{13}{4} + \frac{19}{4})] \ | \)
= \( \frac{15}{32} \) sq. units

\(\therefore \) ratio of \(area \ of \ ∆ \ ADE \ to \ area \ of \ ∆ \ ABC \ = \ 1:16 \)

Q7 ) Let A (4, 2), B (6, 5) and C (1, 4) be the vertices of \( ∆ \ ABC \) .

(i) The median from A meets BC at D. Find the coordinates of point D.

(ii) Find the coordinates of the point P on AD such that AP : PD = 2 : 1.

(iii) Find the coordinates of point Q and R on medians BE and CF respectively such that BQ : QE = 2:1 and CR : RF = 2 : 1.

(iv) What do you observe?

[Note : The point which is common to all the three medians is called the centroid and this point divides each median in the ratio 2 : 1.]

(v) If A (x₁, y₁), B (x₂, y₂) and C (x₃, y₃) are the vertices of triangle ABC, find the coordinates of the centroid of the triangle.

Answer :

(i) Given, D is the median to BC. So, coordinates of D using distance formula are

\( x \ = \ \frac{6+1}{2} \ = \ \frac{7}{2} \) and \( y \ = \ \frac{5+4}{2} \ = \ \frac{9}{2} \)

(ii) Coordinates of P can be calculated using section formula

\( x \ = \ \frac{2( \frac{7}{2}) + 1(4)}{3} \ = \ \frac{11}{3} \) and \( y \ = \ \frac{2( \frac{9}{2}) + 1(2)}{3} \ = \ \frac{11}{3} \)

(iii) Coordinates of E
= ( \( \frac{4+1}{2} \ , \ \frac{2+4}{2} \) )
= ( \( \frac{5}{2} , 3 \) )

So, coordinates of Q are,
( \( \frac{2( \frac{5}{2}) + 1(6)}{3} \ , \ \frac{2(3) + 1(5)}{3} \) )
= ( \( \frac{11}{3} , \frac{11}{3} \) )

Coordinates of F
= ( \( \frac{4+6}{2} , \frac{2+5}{2} \) )
= ( \( 5, \frac{7}{2} \) )

So, coordinates of R are,
( \( \frac{2(5) + 1(1)}{3} , \frac{2 ( \frac{7}{2}) + 1(4)}{3} \) )
= ( \( \frac{11}{3} , \frac{11}{3} \) )

(iv) We observed that coordinates of P , Q, R are same i.e., ( \( \frac{11}{3} , \frac{11}{3} \) ) which shows that medians intersect each other at a common point which is called centroid of the triangle.

(v) If A (x₁, y₁), B (x₂, y₂) and C (x₃, y₃) are the vertices of triangle ABC, then the coordinates of the centroid of the triangle is

\( x \ = \ \frac{x_1 + x_2 + x_3}{3} \ and \ \frac{y_1 + y_2 + y_3}{3} \)

Q8 ) ABCD is a rectangle formed by the points A (-1, – 1), B (-1, 4), C (5, 4) and D (5, -1). P, Q, R and S are the midpoints of AB, BC, CD and DA respectively. Is the quadrilateral PQRS a square? a rectangle? or a rhombus? Justify your answer.

Answer :

P, Q, R and S are the midpoints of AB, BC, CD and DA respectively

So, coordinate of P
= ( \( \frac{-1-1}{2} , \frac{-1+4}{2} \) )
= ( \( -1 , \frac{3}{2} \) )

Similarly, coordinate of Q
= ( \( \frac{5-1}{2} , \frac{4+4}{2} \) )
= ( 2,4)

R = ( \( \frac{5+5}{2} , \frac{4-1}{2} \) ) = ( \( 5 , \frac{3}{2} \) )

S = ( \( \frac{5-1}{2} , \frac{-1-1}{2} \) ) = ( 2,-1)

Now, PQ = \( \sqrt{(-1-2)^2 + ( \frac{3}{2} - 4)^2} \ \)
\( = \ \sqrt{ \frac{61}{4} } \ \)
\( = \ \frac{ \sqrt{61}}{2} \)

SP = \( \sqrt{(2+1)^2 + (-1- \frac{3}{2})^2} \ \)
\( = \ \sqrt{ \frac{61}{4} } \ \)
\( = \ \frac{ \sqrt{61}}{2} \)

QR = \( \sqrt{(2-5)^2 + (4- \frac{3}{2})^2} \ \)
\( = \ \sqrt{ \frac{61}{4} } \ \)
\( = \ \frac{ \sqrt{61}}{2} \)

RS = \( \sqrt{(5-2)^2 + ( \frac{3}{2} +1)^2} \ \)
\(= \ \sqrt{ \frac{61}{4} } \ \)
\( = \ \frac{ \sqrt{61}}{2} \)

PR = \( \sqrt{(-1-5)^2 + ( \frac{3}{2} - \frac{3}{2})^2 } \ \)
\( = \ 6 \)

QS = \( \sqrt{(2-2)^2 + (4+1)^2} \ \)
\( = \ 5 \)

Clearly, PQ = QR = RS = SP

But, \( PR \ne QS \)


Hence the lengths of all the sides are equal, while the lengths of the diagonals are different.

\(\therefore \) The PQRS is a rhombus.