Coordinate Geometry

NCERT Exercise 7.4(optional):2

Question (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 figure. 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 PQR if C is the origin?

Also calculate the areas of the triangles in these cases. What do you observer?

Solution

(i) Taking A as origin, finding the coordinates of the vertices of the triangle.

It can be clearly observed that, when taking A as origin, the coordinates of

P = (4, 6)

And, Q = (3, 2)

And, R = (6, 5)

Calculation of Area of Triangle PQR

We know that, in Coordinate Geometry when coordinate of vertices are given, then Area of Triangle =1/2[x₁(y₂–y₃) + x₂(y₃–y₁) + x₃(y₁–y₂)]

Here, we have, x₁ = 4, y₁ = 6

x₂ = 3, y₂ = 2

And, x₃ = 6, y₃ = 5

Thus, area of triangle PQR

= 1/2 [{4 (2 - 5)} + {3 (5 - 6)} + {6 (6 - 2)}]

= 1/2 [{4 (- 3)} + {3 × (- 1)} + {6 × 4}]

= 1/2 [- 12 - 3 + 24]

`=1/2xx9 = 9/2`

Thus, Area of triangle = `9/2` square unit

(ii) What will be the coordinates of the vertices of triangle PQR if C is the origin?

It can be clearly observed that, when taking C as origin, the coordinates of

P = (12, 2)

And, Q = (13, 6)

And, R = (10, 3)

Calculation of Area of Triangle PQR

We know that, in Coordinate Geometry when coordinate of vertices are given, then Area of Triangle =1/2[x₁(y₂–y₃) + x₂(y₃–y₁) + x₃(y₁–y₂)]

Here, we have, x₁ = 12, y₁ = 2

x₂ = 13, y₂ = 6

And, x₃ = 10, y₃ = 3

Thus, area of triangle PQR

= 1/2 [{12 (6 - 3)} + {13 (3 - 2)} + {10 (2 - 6)}]

= 1/2 [{12 × 3)} + {13 × 1} + {10 × (- 4)}]

= 1/2 [36 + 13 - 40]

`=1/2xx9 = 9/2`

Thus, Area of triangle = `9/2` square unit

Thus, (i) When A is taken as origin the coordinates are P(4, 6), Q (3, 2) and R (6, 5)

(ii) When C is taken as origin then coordinates are P(12, 2), Q(13, 6) and R(10, 3)

And Area of triangle `=9/2` Answer

Question (6) The vertices of a `Delta` 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 `(AD)/(AB) = (AE)/(AC) = 1/4`. Calculate the area of the `Delta` ADE and compare it with area of `Delta` ABC. (Recall Theorem 6.2 and Theorem 6.6)

Solution

Given, the vertices of `Delta` ABC are A(4, 6), B(1, 5) and C(7, 2)

And, `(AD)/(AB) = (AE)/(AC) = 1/4`

Then, Area of the `Delta` ADE : `Delta` ABC =?

Since, as given, AD = 1 and AB = 4

Thus, DB = AB – AD = 4 – 1 = 3

Similarly, as given AE = 1 and AC = 4

Thus, EC = AC – AE = 4 – 1 = 3

Thus, point D divides the AB in the ratio of 1:3

And, point E divides the AC in the ratio of 1:3

Now, We know that,

According to section formula, if the coordinates of the point P(x, y) divides the line segment joining the points A(x₁, y₁) and B(x₂, y₂), internally, in the ratio m₁:m₂ , then

`x=(m_1x_2+m_2x_1)/(m_1+m_2)` and `y=(m_1y_2+m_2y_1)/(m_1+m_2)`

Calculation of coordinates of point D

Here, we have, x₁ = 4, y₁= 6

And, x₂ = 1, y₂ = 5

And, m₁ = 1, m₂ =3

Thus, Using Section Formula, we have the coordinates of point D

`=((1xx1+3xx4)/(1+3), (1xx5+3xx6)/(1+3))`

`=((1+12)/4, (5+18)/4)`

Thus, coordinates of point D`=(13/4, 23/4)`

Calculation of coordinates of point E

Here, we have, x₁ = 4, y₁= 6

And, x₂ = 7, y₂ = 2

And, m₁ = 1, m₂ =3

Thus, Using Section Formula, we have the coordinates of point E

`=((1xx7+3xx4)/(1+3), (1xx2+3xx6)/(1+3))`

`=((7+12)/4, (2+18)/4)`

`=(19/4, 20/4)`

Thus, coordinates of point E `=(19/4, 5)`

We know that, in Coordinate Geometry when coordinate of vertices are given, then Area of Triangle =1/2[x₁(y₂–y₃) + x₂(y₃–y₁) + x₃(y₁–y₂)]

Calculation of Area of Triangle ADE

Here, we have, x₁ = 4, y₁ = 6

`x_2 = 13/4`, and `y_2=23/4`

And, `x_3=19/4` and `y_3 = 5`

Thus, area of triangle ADE

`=1/2[3-13/4+19/16]`

`=1/2[(48-52+19)/16]`

`=1/2xx15/16 =15/32`

⇒ Area of triangle ADE `=15/32` square unit

Calculation of Area of triangle ABC

Here, we have, x₁ = 4, y₁ = 6

x₂ = 1, y₂ = 5

And, x₃ = 7, y₃ = 2

Thus, Area of triangle ABC

= 1/2 [{4 (5 – 2)} + {1 (2 – 6)} + {7 (6 – 5)}

= 1/2 [{4 × 3} + {1 ( –4 )} + {7 × 1}]

= 1/2 [12 – 4 + 7]

`=1/2xx15 = 15/2`

Thus, Area of triangle ABC `=15/2` square unit

Now, Area of triangle ADE : Area of triangle ABC

`=15/32:15/2`

`=15/32xx2/15`

`=1/16` = 1:16

Thus, ratio of area of triangle ADE to the Area of triangle ABC = 1:16 Answer

Question (7) Let A (4, 2), B(6, 5) and C(1, 4) be the vertices of `Delta`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 AP : PD = 2:1

(iii) Find the coordinates of points 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`(x_1, y_1)`, B`(x_2, y_2)` and C`(x_3, y_3)` are the vertices of `Delta`ABC, find the coordinates of the centroid of the triangle.

Solution

Given, coordinates triangle are A(4, 2), B(6, 5) and C(1, 4)

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

Since, AD divides the BC at mid-point, thus, BD: DC = 1:1

Now, we know that, in coordinate geometry, the coordinates of mid-point are

`((x_1+x_2)/2, (y_1+y_2)/2)`

Calculation of coordinates of mid-point D of B(6, 5) and C(1, 4)

Here we have, x₁ = 6, y₁ = 5

And, x₂ = 1, y₂ = 4

Thus, coordinates of D `=((6+1)/2, (5+4)/2)`

Thus, coordinates of D `=(7/2, 9/2)` Answer

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

Here, AP: PD = 2:1

Now, We know that,

According to section formula, if the coordinates of the point P(x, y) divides the line segment joining the points A(x₁, y₁) and B(x₂, y₂), internally, in the ratio m₁:m₂ , then

`x=(m_1x_2+m_2x_1)/(m_1+m_2)` and `y=(m_1y_2+m_2y_1)/(m_1+m_2)`

Calculation of coordinates of point P

Here, we have, x₁ = 4, y₁ = 2

And, `x_2=7/2, y_2 = 9/2`

And, m₁ = 2, m₂ = 1

Thus, coordinates of point P

`=((2xx7/2+1xx4)/(2+1), (2xx9/2+1xx2)/(2+1))`

`=((7+4)/3, (9+2)/3)`

⇒ coordinates of point P `=(11/4, 11/4)` Answer

(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

Here, given, BQ : QE = 2:1 and CR: RF= 2:1

Since, CF and BE are medians of AB and BC respectively,

Thus, F and E are mid-points of AB and AC respectively.

Calculation of mid-point F of AB

Here, we have, x₁ = 4, y₁ = 2

And, x₂ = 6, y₂ = 5

Thus, coordinates of mid-point F of AB

`=((4+6)/2, (2+5)/2)`

`=(10/2, 7/2)`

Thus, coordinates of mid-point F of AB `=(5, 7/2)`

Calculation of coordinates of point R

Here, given, CR:RF = 2:1

Thus, here we have x₁=1, y₁=4 (coordinates of C)

And, `x_2 = 5, y_2 = 7/2` (coordinates of F as calculated above)

And, m₁ = 2, m₂ = 1

Thus, coordinates of R `=((2xx5+1xx1)/(2+1), (2xx7/2+1xx4)/(2+1))`

`=((10+1)/3, (7+4)/3)`

Thus, coordinates of R `=(11/3, 11/3)`

Calculation of mid-point E of AC

Here we have, x₁ = 4, y₁ = 2

And, x₂ = 1, y₂ = 4

Thus, coordinates of mid-point E of AC

`=((4+1)/2, (2+4)/2)`

`=(5/2, 6/2)`

Thus, coordinates of mid-point E of AC `=(5/2, 3)`

Calculation of coordinates of Q

Here, we have x₁= 6, y₁ = 5 (Coordinates of B)

And, `x_2=5/2, y_2 =3` (Coordinates of E)

And, m₁ = 2, m₂ = 1

Thus, coordinates of Q `=((2xx5/2+1xx6)/(2+1), (2xx3+1xx5)/(2+1))`

`=((5+6)/3, (6+5)/3)`

`=(11/3, 11/3)`

Thus, coordinates of Q and R `=(11/3, 11/3)` Answer

(iv) What do you observe

It has been observed that, the coordinates of P, Q and R are same.

This means are these points P, Q and R are same point and are the centroid of the given triangle.

(v) If A`(x_1, y_1)`, B`(x_2, y_2)` and C`(x_3, y_3)` are the vertices of `Delta`ABC, find the coordinates of the centroid of the triangle.

Let, median AD divides the BC into two equal parts at point D.

And P is the centroid of the triangle.

Calculation of coordinates of mid-point D of BC

Thus, coordinates of point D

`=((x_2+x_3)/2, (y_2+y_3)/2)`

Now, P is the centroid, thus, AP : PD = 2:1

Calculation of coordinates of point P

Here we have, `x_1=x_1, y_1 = y_1` (coordinates of A)

And, `x_2=(x_2+x_3)/2`, and, `y_2 = (y_2+Y_3)/2`

And, m₁ = 2, m₂ = 1

Thus, According to section formula of coordinate geometry, the coordinates of point P

`=((x_2+x_3+x_1)/3, (y_2+y_3+Y_1)/3)`

Thus, coordinates of point P `=((x_1+x_2+x_3)/3, (y_1+y_2 + y_3)/3)` Answer

Question (8) 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 mid points of AB, BC, CD and DA respectively. Is the quadrilateral PQRS a square? a rectangle? or a rhombus? Justify your answer.

Solution

Let ABCD is the given rectangle.

And as per question coordinates of ABCD are A(–1, –1), B(–1, 4), C(5, 4) and D(5, –1)

Again given, P, Q, R and S are the mid-points of AB, BC, CD and DA respectively.

Thus, find PQRS is which type of rectangle.

We know that, Now, we know that, in coordinate geometry, the coordinates of mid-point of line joining two points are

`((x_1+x_2)/2, (y_1+y_2)/2)`

Calculation of mid-point P of AB

Here, we have x₁ = –1, y₁ = –1

And, x₂ = –1, y₂ = 4

Thus, coordinates of mid-point P of AB

`=((-1+(-1))/2, (-1+4)/2)`

`=((-2)/2, 3/2)`

Thus, coordinates of P `=(-2/2, 3/2)`

Calculation of mid-point Q of BC

Here, we have x₁ = –1, y₁ = 4

And, x₂ = 5, y₂ = 4

Thus, coordinate of mid-point Q

`=((-1+5)/2, (4+4)/2)`

`=(4/2, 8/2)`

Thus, coordinate of Q =(2, 4)

Calculation of mid-point R of DC

Here, we have x₁ = 5, y₁ = –1

And, x₂ = 5, y₂ = 4

Thus, coordinate of mid-point R

`=((5+5)/2, (-1+4)/2)`

`=(10/2, 3/2)`

Thus, coordinate of R `=(5, 3/2)`

Calculation of mid-point S of AD

Here, we have x₁ = –1, y₁ = –1

And, x₂ = 5, y₂ = –1

Thus, coordinate of mid-point S

`=((-1+5)/2, (-1+(-1))/2)`

`=(4/2, (-1-1)/2)`

`=(2, (-2)/2)`

Thus, coordinate of S = (2, –1)

Thus, coordinates of PQRS are `P((-2)/2, 3/2)`, Q(2, 4), `R(5, 3/2)` and S (2, –1)

Now, calculation of length of sides of PQRS

Now, we know that, in coordinate geometry, the distance between two points P (x₁, y₁) and Q (x₂, y₂), i.e

PQ `=sqrt((x_1-x_2)^2+(y_1-y_2)^2)`

Calculation of distance between P and Q, i.e. PQ

Here we have, x₁ = –1, y₁=3/2

x₂ = 2, y₂ = 4

Thus, according to distance formula,

PQ `=sqrt((-1-2)^2+(3/2-4)^2)`

`=sqrt((-3)^2+((3-8)/2)^2)`

`=sqrt(9+(-5/2)^2)`

`=sqrt(9+25/4)`

`=sqrt((36+25)/4)`

Thus PQ `=sqrt(61/4)`

Calculation of distance between Q and R, i.e. QR

Here we have, x₁ = 2 , y₁= 4

x₂ = 5, y₂ =3/2

Thus distance between QR `=sqrt((2-5)^2+(4-3/2)^2)`

`=sqrt((-3)^2+((8-3)/2)^2)`

`=sqrt(9 + (5/2)^2)`

`=sqrt(9+25/4)`

`=sqrt((36+25)/4)`

Thus, QR `=sqrt(61/4)`

Calculation of distance between R and S, i.e. RS

Here we have, x₁ = 5, y₁= 3/2

x₂ = 2, y₂ = –1

Thus, according to distance formula, the distance between R and S, i.e.

RS`=sqrt((5-2)^2+(3/2-(-1))^2)`

`=sqrt(3^2+(3/2+1)^2)`

`=sqrt(9+((3+2)/2)^2)`

`=sqrt(9+(5/2)^2)`

`=sqrt(9+25/4)`

`=sqrt((36+25)/4)`

Thus, RS `=sqrt(61/4)`

Calculation of distance between S and P, i.e. SP

Here we have, x₁ = 2, y₁= –1

x₂ = –1, y₂ = 3/2

Thus, according to distance formula,

SP `=sqrt((2+1)^2+(-1-3/2)^2)`

`=sqrt(3^2+((-2-3)/2)^2)`

`=sqrt(9+(-5/2)^2)`

`=sqrt(9+25/4)`

`=sqrt((36+25)/4)`

Thus, SP `=sqrt(61/4)`

Now, distance between P and R, i.e. PR

Here we have, x₁ = –1, y₁=3/2

x₂ = 5, y₂ =3/2

Thus, According to distance formula,

PR `=sqrt((-1-5)^2+(3/2-3/2)^2)`

`=sqrt((-6)^2+0)`

`=sqrt(36) =6`

Thus, PR = 6

Calculation of distance between S and Q, i.e. SQ

Here we have, x₁ = 2, y₁= –1

x₂ = 2, y₂ = 4

Thus, according to distance formula

SQ `=sqrt((2-2)^2+(4-(-1))^2)`

`=sqrt(0+5^2)`

`=sqrt(25) =5`

Thus, SQ = 5

Here, it can be observed that, all sides of PQRS are equal but their diagonals are not equal, thus PQRS is a Rhombus. Answer

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