Coordinate Geometry NCERT Exercise 7.4 part two Class Ten Mathematics

Coordinate Geometry

Exercise 7.4 Part 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. 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 figure. The students are to sow the seeds of flowering plants on the remaining area of the plot.

(a) Taking A as origin, find the coordinates of the vertices of the triangle.

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

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

Solution: P = (-12, -2), Q = (-13, -6), R = (-10, -3)


(c) Also calculate the areas of the triangles in these cases. What do you observe?

Solution: Area of triangle PQR in case of origin A:

`= ½ (x_1(y_2 – y_3) + x_2 (y_3 – y_1) + x_3 (y_1 – y_2))`

`= ½ (4(2 – 5) + 3 (5 – 6) + 6 (6 – 2))`

`= ½ ( - 12 – 3 + 24)`

`= 9/2` sq unit

Area of triangle PQR in case of origin C:

`= ½ (x_1(y_2 – y_3) + x_2 (y_3 – y_1) + x_3 (y_1 – y_2))`

`= ½ ( -12( - 6 + 3) + - 13 (- 3 + 2) + - 10( - 2 + 6))`

`= ½ ( 36 + 13 – 40)`

`= 9/2` sq unit

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


Question 6: The vertices of triangle 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 = ¼. Calculate the area of triangle ADE and compare it with area of triangle ABC.

Solution: Point D divides AB in ratio 1 : 3

Point E divides AC in the same ratio.

Hence, `m_1 = 1` and `m_2 = 3`

Coordinates of D can be calculated as follows:

`x=(m_1x_2+m_2x_1)/(m_1+m_2)`

`=(3xx4+1xx1)/(4)=13/4`

`y=(m_1y_2+m_2y_1)/(m_1+m_2)`

`=(3xx6+1xx5)/(4)=23/4`

Coordinates of E can be calculated as follows:

`x=(1xx7+3xx4)/(4)=19/4`

`y=(1xx2+3xx6)/(4)=5`

Area of triangle ABC can be calculated as follows:

`= ½ (x_1(y_2 – y_3) + x_2 (y_3 – y_1) + x_3 (y_1 – y_2))`

`= ½ (4(5 – 2) + 1( 2 – 6) + 7( 6 – 5))`

`= ½ (12 – 4 + 7) = 15/2` sq unit

Area of triangle ADE can be calculated as follows:

`= ½ (x_1(y_2 – y_3) + x_2 (y_3 – y_1) + x_3 (y_1 – y_2))`

`= ½ (4(23/4 – 5) + 13/4 (5 – 6) + 19/4 (6 – 23/4))`

`= ½ (3 – 13/4 + 19/16)`

`= ½ x 15/16 = 15/32` sq unit

Hence, ratio of area of triangle ADE to area of triangle ABC = 1 : 16

Question 7: Let A (4, 2), B (6, 5) and C (1, 4) be the vertices of triangle ABC.

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

Solution: Coordinates of D can be calculated as follows:

`x=(1+6)/(2)=7/2`

`y=(4+5)/(2)=9/2`

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

Solution: Coordinates of P can be calculated as follows:

`x=(1xx4+2xx7/2)/(2)`

`=(4+7)/(2)=11/2`

`y=(1xx2+2xx9/2)/(2)`

`=(2+9)/(2)=11/2`


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

Solution: Coordinates of E can be calculated as follows:

`x=(1+4)/(2)=5/2`

`y=(4+2)/(2)=3`

Point Q and P would be coincident because medians of a triangle intersect each other at a common point called centroid. Coordinate of Q can be given as follows:

`x=(1xx6+2xx5/2)/(2)=11/2`

`y=(1xx5+2xx3)/(2)=11/2`

(d) What do you observe?

Solution: Coordinates of P, Q and R are same which shows that medians intersect each other at a common point.

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

Solution: Looking at the coordinates of P in earlier sections of this question, coordinates of centroid can be given as follows:

`x=(x_1+x_2+x_3)/(3)`

`y=(y_1+y_2+y_3)/(3)`

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 midpoints of AB, BC, CD and DA respectively. Is the quadrilateral PQRS a square? a rectangle? or a rhombus? Justify your answer.

Solution: Coordinates of P can be calculated as follows:

`((-1-1)/(2), (4-1)/(2))=(-1, 3/2)`

Coordinates of Q can be calculated as follows:

`((5-1)/(2), (4+4)/(2))=(2. 4)`

Coordinates of R can be calculated as follows:

`((5+5)/(2), (4-1)/(2))=(5, 3/2)`

Coordinates of S can be calculated as follows:

`((5-1)/(2), (-1-1)/(2))=(2, -1)`

Length of PQ can be calculated as follows:

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

`=sqrt(3^2+(5/2)^2)=(sqrt61)/(2)`

Similarly, QR can be calculated as follows:

`OR=sqrt((5-2)^2+(3/2-4)^2)`

`=sqrt(3^2+(-5/2)^2)=(sqrt61)/(2)`

The above values show that RS = PS, i.e. all sides are equal.

Now let us calculate the diagonals

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

`=sqrt6^2=6`

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

`=sqrt-5^2=5`

Hence, it is clear that while all sides are equal, diagonals are not equal. So, the given figure is a rhombus.



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