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Coordinate Geometry Ex-7.4 Questions Answers

Subjects

Question 1 :

Determine the ratioin which the line 2x + y – 4 = 0 divides the line segment joining the pointsA(2, –2) and B(3, 7).

Answer 1 :

Consider line 2x + y – 4 = 0 divides line ABjoined by the two points A(2, -2) and B(3, 7) in k : 1 ratio.

Coordinates of point of division can be givenas follows:

x = and y =

Substituting the values of x and y givenequation, i.e. 2x + y – 4 = 0, we have

2( + (– 4 = 0

+ ( = 4

4 + 6k – 2 + 7k = 4(k+1)

-2 + 9k = 0

Or k = 2/9

Hence, the ratio is 2:9.

Question 2 : Find the relation between x and y if thepoints (x, y), (1, 2) and (7, 0) are collinear.

Answer 2 :

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

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

Area of a triangle = = 0

[x(2 – 0) + 1 (0 – y) + 7( y– 2)] = 0

  • 2x – y + 7y – 14 = 0
  • 2x + 6y – 14 = 0
  • x + 3y – 7 = 0. Which is required result.

Question 3 :

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

Answer 3 :

Let A = (6, -6), B = (3, -7), C = (3, 3) arethe points on a circle.

If O is the centre, then OA = OB = OC (radiiare equal)

If O = (x, y) then

OA =

OB =

OC =

Choose: OA = OB, we have

After simplifying above, we get -6x = 2y – 14….(1)

Similarly: OB = OC

(x – 3)+ (y + 7)2 =(x – 3)2 + (y – 3)2

(y + 7)2 = (y – 3)2

y+ 14y + 49 = y–6y + 9

20y =-40

or y = -2

Substituting the value of y in equation (1),we get;

-6x = 2y – 14

-6x = -4 – 14 = -18

x = 3

Hence, centre of the circle located at point(3,-2).

Question 4 :

The two oppositevertices of a square are (-1, 2) and (3, 2). Find the coordinates of the othertwo vertices.

Answer 4 :

Question 5 :

The class X studentsof a secondary school in Krishinagar have been allotted a rectangular plot ofland for their gardening activity. 

Answer 5 :

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?

Solution:

(i) Taking A as origin, coordinates of thevertices 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.

(ii) Taking C as origin,

Coordinates of vertices P, Q and R are ( 12,2) , (13, 6) and (10, 3) respectively.

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

Find the area of triangles:

Area of triangle PQR in case of origin A:

Using formula: Area of a triangle =

= ½ [4(2 – 5) + 3 (5 – 6) + 6 (6 – 2)]

= ½ ( – 12 – 3 + 24 )

= 9/2 sq unit

(ii) Area of triangle PQR in case of origin C:

Area of a triangle =

= ½ [ 12(6 – 3) + 13 ( 3 – 2) + 10( 2 – 6)]

= ½ ( 36 + 13 – 40)

= 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 triangleremains the same no matter which point is considered as origin.


Question 6 :

The vertices of a ∆ABC are A (4, 6), B (1, 5) and C (7, 2). A line is drawn to intersect sides ABand AC at D and E respectively, such that . Calculate the area of the ∆ ADE andcompare it with area of ∆ ABC. (Recall Theorem 6.2 and Theorem 6.6)

Answer 6 :