Students can Download Class 10 Maths Chapter 7 Coordinate Geometry Ex 7.2 Questions and Answers, Notes Pdf, KSEEB Solutions for Class 10 Maths helps you to revise the complete Karnataka State Board Syllabus and to clear all their doubts, score well in final exams.
Karnataka State Syllabus Class 10 Maths Chapter 7 Coordinate Geometry Ex 7.2
Question 1.
Find the coordinates of the point which divides the join of (- 1, 7) and (4, – 3) in the ratio 2 : 3.
Answer:
Let P(x, y) is the required point.
P(x, y) divides the points A(- 1, 7),
B (4, – 3) in the ratio m1 : m2 = 2 : 3.
As per Section Formula, coordinates of P(x, y) are
So, the coordinates of P are (1, 3)
Question 2.
Find the coordinates of the points of trisection of the line segment joining (4, -1) and (- 2, – 3)
Answer:
The given points be A(4, – 1) and B(- 2, – 3) and points of trisection be P and Q
AP = PQ = QB = K
PB = PQ ± QB = K + K = 2K
AP: PB = K : 2K = 1: 2
∴ m : n = 1 : 2
Coordinates of P are
Now AQ = AP + PQ = K + K = 2K
AQ : QB = 2K : K = 2 : 1
m : n = 2 : 1
Coordinates of Q are
Question 3.
To conduct Sports Day activities, in your rectangular shaped school ground ABCD, lines have been drawn with chalk powder at a distance of 1 m each. 100 flower pots have been placed at a distance of 1m from each other along with AD, as shown in the following figure.
Answer:
Distance AD = Number of Flowerpots in the x-axis, the distance between two flower pots is 1 m.
∴ take A as origin, AB as the x-axis, and AD as the y-axis
∴ Niharika runs \(\frac{1}{4}\) of distance AD on the second line Posts green Flag (2, 100/4) (or) (2, 25)
∴ Preet runs \(\frac{1}{5}\) th of distance AD on the eight-line posts red flag (8, 100/5) (or (8,20)
Distance between both Flags P(2, 25) & Q (8. 20)
Rashmi post a blue flag exactly middle b/w two flag posts
Rashmi posts her blue flag on the 5th line at a distance of 22.5 m from AB.
Question 4.
Find the ratio in which the line segment joining the points (-3, 10) and (6, – 8) is divided by (-1, 6).
Answer:
x1 = – 3, x2 = 10, m1 = ? m2 = ? x = – 1
– 1(m1 +m2) = 6m1 – 3m2,
– m1 – m2 = 6m1 – 3m2
– m2 + 3m2 = 6m1 + m1.
2m2 = 7m1
\(\frac{m_{1}}{m_{2}}=\frac{2}{7}\)
∴ m1 : m2 = 2 : 7
Question 5.
Find the ratio in which the line segment joining A(1, -5) and B(- 4, 5) is divided by the x-axis. Also, find the coordinates of the point of division.
Answer:
Let the x-axis divides line segment AB unit
m1 : m2 ratio
Section Formula
y – coordinate
0(m1 + m2 )= – 5m1 + 5m2
0 = – 5m1 +5m2
5m1 =5m2
\(\frac{\mathrm{m}_{1}}{\mathrm{m}_{2}}=\frac{5}{5}=\frac{1}{1}\)
m1 : m2 = 1 : 1
X – Coordinates x is mid point of AB
The coordinates of p are (- 3/2, 0)
Question 6.
If (1, 2), (4, y), (x, 6) and (3, 5) are the vertices of a parallelogram taken in order, find x and y.
Answer:
AC and BD are the diagonals of a parallelogram ABCD and diagonals bisect at O’.
∴ Coordinate of the midpoint of AC = Coordinate of midpoint BD.
As per the Mid-point formula,
Question 7.
Find the coordinates of point A, where AB is the diameter of a circle whose centre is (2, -3) and B is (1, 4).
Answer:
Let C be the centre of the circle then C → (2, – 3)
The coordinates point A → (x, y) Point C is the midpoint of diameter AB Mid Point Formula
Hence, coordinates of a point A are (3, -10)
Question 8.
If A and B are (-2, -2) and (2, -4), respectively, find the coordinates of P such that AP = \(=\frac{3}{7}\) AB and P lies on the line segment AB.
Answer:
Let P(x, y) be the point which divides the join of A(- 2 – 2) & B(2, – 4) in the ratio 3 : 4.
Section formula
Question 9.
Find the coordinates of the points which divide the line segment joining A(-2, 2) and B(2, 8) into four equal parts.
Answer:
Let P, Q, R be the points that divide the line segment joining A(- 2, 2) and B(2, 8) into four equal parts. Since Q divides the line segment AB into two equal parts i.e Q is the midpoint of AB.
Now, P divides AQ into two equal parts i.e, P is the midpoint of Q.
∴ Co-ordinates of P are
Question 10.
Find the area of a rhombus if its vertices are (3, 0), (4, 5), (-1, 4) and (-2, -1) taken in order. [Hint: Area of a rhombus = \(\frac{1}{2}\) (product of its diagonals]
Answer:
LetA → (3, 0), B → (4, 5), C → (-1, 4) and D → (- 2, – 1) be the vertices of a rhombus.
∴ AC and BD are diagonal
