Coordinate Geometry Ex 7.4 Solutions
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)2 + (y + 7)2 =(x – 3)2 + (y – 3)2
(y + 7)2 = (y – 3)2
y2 + 14y + 49 = y2 –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 : - 
Question - 7 : - Let A (4, 2), B (6, 5) and C (1, 4) be the vertices of ∆ ABC.
Answer - 7 : -
(i) The median fromA meets BC at D. Find the coordinates of point D.
(ii) Find thecoordinates of the point P on AD such that AP : PD = 2 : 1.
(iii) Find thecoordinates 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 youobserve?
[Note : The pointwhich is common to all the three medians is called the centroid
and this pointdivides 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 thetriangle.
Solution
Question - 8 : - ABCD is a rectangleformed by the points A (-1, – 1), B (-1, 4), C (5, 4) and D (5, -1). P, Q, Rand S are the midpoints of AB, BC, CD and DA respectively. Is the quadrilateralPQRS a square? a rectangle? or a rhombus? Justify your answer.
Answer - 8 : - 