These NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry Ex 7.4 Questions and Answers are prepared by our highly skilled subject experts.
NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry Exercise 7.4
Question 1.
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).
Solution:
Given line is 2x + y – 4 = 0 … (i)
Given point A (2, – 2)
Putting x = 2, and y = 2 in equation (i)
4 – 2 – 4 = 0 = 0
Given point B (3, 7)
Putting x = 3, y = 7 in equation (i)
6 + 7 – 4 = 9 = 9
Ratio = \(\frac { 2 }{ 9 }\) = 2 : 9
Question 2.
Find a relation between x and y, if the points (x, y), (1, 2) and (7, 0) are collinear.
Solution:
We know that
which is the required relation between x and y.
Question 3.
Find the centre of a circle passing through the points (6, -6), (3, -7) and (3, 3).
Solution:
We know that equation of circle is
x² + y² + 2gx + 2fy + c = 0
where (- g, – f) are the centre of circle.
(6, – 6) passes through the circle
36 + 36 + 12g + 2f(- 6) + r = 0
72 + 12g – 12f + c = 0 … (1)
(3, – 7) passes through the circle
(3)² + (- 7)² + 2g (3) + 2f (- 7) + c = 0
9 + 49 + 6g – 14f+ c = 0
58 + 6g – 14f + c = 0 …. (2)
(3, 3) passes through the circle
(3)² + (3)² + 2g (3) + 2f(3) + c = 0
18 + 6g + 6f + c = 0 … (3)
From (2) and (3),
Question 4.
The two opposite vertices of a square are (-1, 2) and (3, 2). Find the coordinates of the other two vertices.
Solution:
ABCD is a square
AB = BC
(AB)² = (BC)²
Hence the coordinates are (1. 0) and (1, 4)
Question 5.
The class X students school in krishnanagar 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 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 ∆PQR, if C is the origin?
Also, calculate the areas of the triangles in these cases. What do you observe?
Solution:
(i) Coordinates are
P (4, 6), Q (3, 2), R (6, 5) taking AD and AB as coordinate axis.
Area of ∆PQR
(ii) Coordinates are (12, 2) (13, 6) (10,3) taking CB and CD as coordinate axis.
Area of ∆PQR
Area of ∆PQR is same in both regions.
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 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 the area of ∆ABC.
Solution:
Area of ∆ADE
Question 7.
Let A(4, 2), B(6,5) and C(1, 4) be the vertices of ∆ABC.
(i) The median from A meters 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 points which is common to all the three medians is called 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 ∆ABC, find the coordinates of the centroid of the triangles.
Solution:
(i) D is median from A
(ii) Let the coordinate of the point P be (x1, y1)
(iii) Coordinate of E
(iv) The coordinate of the points are
P(\(\frac { 11 }{ 3 }\), \(\frac { 11 }{ 3 }\)), Q(\(\frac { 11 }{ 3 }\), \(\frac { 11 }{ 3 }\)), R(\(\frac { 11 }{ 3 }\), \(\frac { 11 }{ 3 }\))
P, Q and R the same point
(v) Given A(x1, y1), B(x2, y2) and C(x3, y3) are the coordinates of the vertices of ∆ ABC
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:
P is the mind point of AB.
In quadrilateral PQRS each side are \(\frac{\sqrt{61}}{2}\) and the diagonal is distinct. Hence is a rhombus.