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Andhra Pradesh SSC Class 10 Solutions For Maths – Coordinate Geometry (English Medium)
These Solutions are part of AP SSC Class 10 Solutions for Maths. In this article, we have provided Andhra Pradesh SSC Class 10 Solutions For Maths Chapter 7 Coordinate Geometry
Exercise 7.1:
Question 1:
Find the distance between the following pair of points
(i) (2, 3) and (4, 1)
(ii) (-5,7) and (-1,3)
(iii) (-2, -3) and (3,2)
(iv) (a,b) and (-a,-b)
Solution :
AP SSC 10th Class Textbook Solutions
Question 2:
Find the distance between the points (0,0) and (36,15).
Solution :
Question 3:
Verify whether the points (1,5)(2,3) and (-2,-1) are collinear or not.
Solution :
Question 4:
Check whether (5, -2), (6,4) and (7, -2) are the vertices of an isosceles triangle.
Solution :
Question 5:
In a class room, 4 friends are seated at the points A, B, C and D as shown in Figure. Jarina and Phani walk into the class and after observing for a few minutes Jarina asks Phani “Don’t you notice that ABCD is a square?” Phani disagrees. Using distance formula, find which of them is correct. Why?
Solution :
Question 6:
Show that the following points form an equilateral triangle A (a, 0), B(-a, 0), C(0, a√3 )
Solution :
Question 7:
Prove that the points (-7, -3), (5,10), (15,8) and (3, -5) taken in order are the comers of a parallelogram. And find its area.
Solution :
Question 8:
Show that the points (-4, -7), (-1,2), (8,5) and (5, -4) taken in order are the vertices of a rhombus.
(Hint: Area of rhombus = 1/2 x product of its diagonals).
Solution :
Question 9:
Name the type of quadrilateral formed, if any, by the following points, and give reasons for your answer. (i) (-1, -2), (1, 0), (-1,2), (-3, 0) (ii) (-3, 5), (3, 1), (1, -3), (-1, -4)
(ii) (-3, 5), (3, 1), (1, -3), (-1, -4)
(iii) (4, 5), (7, 6), (4, 3), (1,2)
Solution :
Question 10:
Find the point on the x-axis which is equidistant from (2, -5) and (-2,9).
Solution :
Question 11:
If the distance between two points (x, 7) and (1,15) is 10, find the value of x.
Solution :
Question 12:
Find the values of y for which the distance between the points P(2, -3) and Q( 10,>’) is 10 units.
Solution :
Question 13:
Find the radius of the circle whose centre is (3,2) and passes through (-5,6).
Solution :
Question 14:
Can you draw a triangle with vertices (1,5), (5, 8) and (13,14) ? Give reason.
Solution :
Question 15:
Find a relation between x and y such that the point (x,y) is equidistant from the points (-2, 8) and (-3, -5)
Solution :
Let P(x, y) be equidistance from the points A(-2, 8) and B(-3, -5)
Given that AP=BP.
Squaring both sides, we have, AP2=BP2
i.e.,
(x+2)2+(y-8)2=(x+3)2+(y+5)2
(x2+4x+4)+(y2-16y+64)=(x2+6x+9)+(y2+10y+25)
X2+4x+y2-16y+68=x2+6x+y2+10y+34
-2x-26y+34=0
x+13y-17=0
x+13y=17 which is the required equation.
Exercise 7.2:
Question 1:
Find the coordinates of the point which divides the line segment joining the points (-1,7) and (4,-3) in the ratio 2:3.
Solution :
Question 2:
Find the coordinates of the points of trisection of the line segment joining (4,-1) and (-2,-3).
Solution :
Question 3:
Find the ratio in which the line segment joining the points (-3,10) and (6,-8) is divided by (-1,6).
Solution :
Question 4:
If (1,2), (4, v), (x, 6) and (3, 5) are the vertices of a parallelogram taken in order, find x and y.
Solution :
Question 5:
Find the coordinates of a point A, where AB is the diameter of a circle whose centre is (2, -3) and B is (1,4).
Solution :
Question 6:
If A and B are (-2, -2) and (2, -4) respectively. Find the coordinates of P such that AP = 3/7 AB and P lies on the segment AB.
Solution :
Question 7:
Find the coordinates of points which divide the line segment joining A(-4,0) and B(0,6) into four equal parts.
Solution :
Question 8:
Find the coordinates of the points which divides the line segment joining A(-2,2) and B(2,8) into four equal parts.
Solution :
Question 9:
Find the coordinates of the point which divides the line segment joining the points (a + b, a – h) and (a-b,a + b) in the ratio 3 : 2 internally.
Solution :
Question 10:
Find the coordinates of centroid of the triangle with vertices following:
- (-1,3), (6,-3) and (-3, 6)
- (6, 2), (0, 0) and (4,-7)
- (1,-1), (0,6) and (-3,0)
Solution :
Exercise 7.3:
Question 1:
Find the area of the triangle whose vertices are
(i) (2, 3) (-1,0), (2, -4)
(ii) (-5, -1), (3, -5), (5,2)
(iii) (0,0), (3,0) and (0,2)
Solution :
Question 2:
Find the value of ‘K’ for which the points are collinear.
(i) (7,-2) (5,1) (3, K)
(ii) (8,l),(K,-4),(2,-5)
(iii) (K, K) (2, 3) and (4, -1).
Solution :
Question 3:
Find the area of the triangle formed by joining the mid-points of the sides of the triangle whose vertices are (0, -1), (2,1) and (0,3). Find the ratio of this area to the area of the given triangle.
Solution :
Question 4:
Find the area of the quadrilateral whose vertices, taken in order, are (-4, -2), (-3, -5), (3, -2) and (2, 3).
Solution :
Question 5:
Find the area of the triangle formed by the points (8, -5), (-2, -7) and (5,1) by using Heron’s formula.
Solution :
Exercise 7.4:
Question 1:
Find the slope of the line passing the two given points
Solution :
Hope given Andhra Pradesh SSC Class 10 Solutions For Maths chapter 7 Coordinate Geometry are helpful to complete your math homework.
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