Coordinate Geometry

Jai Ganesh · 2020-06-19 12:43:02

Solution CG#98 is correct. Good work!

CG#99. If the vertices of ΔABC are A(5,-1), B(-3,-2), and C(-1,8), find the length of the median through A.

666 bro · 2020-06-20 01:10:34

The length of the median through A = √65 units.

Jai Ganesh · 2020-06-20 16:05:22

Hi,

Neat work!

CG#100. The point (4,2) divides the line segment joining (5,-1) and (2,y) in the ratio 1:2.

666 bro · 2020-06-21 02:01:15

CG#100 : Let the points of the line segment will be AB, where
A = (5, -1)
B = ( 2 , y)
Let the point divides line segment will be C.
According to given conditions:
Point c divides line segment AB in the ratio 1:2
Coordinates of c (x, y) = (4,2)
Using section formula =
((m₁x₂+m₂x₁)/(m₁+m₂) , (m₁y₂+m₂y₁)/(m₁+m₂))
where, x=4, y=2 , x₁ = 5 , x₂ = 2 , y₁= -1, y₂ =y , m₁= 1 , m₂ =2

(4,2) = ( ( 2 +10) / (1+2) , (y-2)/(1+2) )

(4,2) = ( 4, ( y-2)/3)
comparing corresponding y coordinates ,we get
2 = ( y-2) /3
6 = y-2
y= 8
∴ the value of y = 8.

Jai Ganesh · 2020-06-21 13:38:23

The solution CG#100 is correct. Good work!

CG#101. Find the mid-point of side BC of ΔABC, with A(1,-4) and the mid-points of the sides through A being (2,-1) and (0,1).

666 bro · 2020-06-21 22:40:20

Given:∆ABC with midpoints of sides through A being (2,-1) and (0,1)
A = (1,-4)
let the midpoints are E & F, where
E = (2, -1)
F = (0 ,1)
Let B = (x₂ , y₂)
Let C = (x₃, y₃)
Let midpoint of BC will be = G
According to the given conditions:
Midpoint of AB = E
Using midpoint formula (m) = ( (x₁+x₂)/2 , (y₁+y₂)/2 ) , we get
(2,-1) = ( (1+x₂)/2 , (-4+y₂)/2 )
Compairing corresponding Coefficients , we get
(1+x₂)/2 = 2 and (-4+y₂)/2 =-1
x₂+1=4 and y₂-4 =-2
x₂ = 3 and y₂ = 2
∴ B = (3,2) = (x₂,y₂)
similarly midpoint of AC = F
( (1+x₃)/2 , (-4+y₃) ) = ( 0,1)
( 1+x₃)/2 = 0 and (-4+y₃)/2 = 1
x₃ =-1 and y₃ = 6
∴ c = ( -1,6) = (x₃,y₃)
now, midpoint of BC = G
( (x₃+x₂)/2 , (y₂+y₃)/2 ) = ( (-1+3)/2 , (6+2)/2 )
(1,4) =G
∴ the midpoint of BC(g) = (1,4)

Jai Ganesh · 2020-06-22 00:05:07

Hi,

The solution CG#101 is correct. Neat work!

CG # 102. Three consecutive vertices of a parallelogram are A(1,2), B(2,3), and C(8,5). Find the fourth vertex.

666 bro · 2020-06-22 22:01:28

CG #102 the fourth vertex(D) = (7, 4)

Jai Ganesh · 2020-06-23 00:14:08

Hi,

CG # 103. Find the area of the quadrilateral whose vertices, taken in order, are (-4,-2), (-3,-5), (3,-2), and (2,3).

666 bro · 2020-06-23 17:16:22

CG#103: the area of quadrilateral is = 28 sq units.

Jai Ganesh · 2020-06-23 20:07:06

Hi,

Good work!

CG # 104. Find the center of a circle passing through the points (6,-6), (3,-7), and (3,3).

666 bro · 2020-06-24 18:41:13

CG#104 the center of the circle (h, k) = (3, -2)
The radius of the circle = 5 units
Equation of the circle
(x-3)^2 + ( y+2)^2 = 25.

Jai Ganesh · 2020-06-24 19:59:15

Neat work!

CG # 105. The two opposite vertices of a square are (-1,2), and (3,2). Find the coordinates of the other two vertices.

666 bro · 2020-06-25 20:15:28

The other two vertices are = (1, 2) and (1, 2) respectively.

Jai Ganesh · 2020-06-26 00:07:36

CG # 106. Determine the ratio in which the line 2x + y - 4 = 0 divides the line segment joining the point A(2,-2), and B(3,7).

666 bro · 2020-06-26 01:24:33

CG#106: The ratio(k:1) is = 2:9.

Jai Ganesh · 2020-06-26 13:56:59

Good work!

CG # 107. Find the coordinates of the points where a circle of radius √2, centered on the point with coordinates (1,1), cut the axes.

666 bro · 2020-06-26 23:08:24

CG#107: the coordinates are (2, 0) (0, 2) .

Jai Ganesh · 2020-06-26 23:35:16

Good work!

CG # 108. Calculate the length of the sides and diagonals of the quadrilateral A(-3,-2), B(-1,-2), C(0,0), and D(-3,1).

666 bro · 2020-06-27 17:02:22

CG#108: ABCD is quadrilateral
where , A = (-3,-2)
B = (-1,-2)
C = (0,0)
D = (-3,1)
AC & BD are diagonals
Now, using distance formula = √( (x₂-x₁)^2 + (y₂-y₁)^2 ), we get
Distance of AB = √( (-1+3)^2 +(-2+2) ^2)
AB = 2 units.
Distance of BC = √( (0+1)^2 + (0+2)^2)
BC = √(5) units.
distance of CD = √( (-3+0)^2 +(1-0)^2)
CD = √(10) units.
distance of AD = √( (-3+3)^2 + (1+2) ^2)
AD = 3 units.
Distance of diagonal AC = √( (0+3)^2 + (0+2)^2)
AC = √(13) units.
distance of diagonal BD = √( (-3+1)^2+(1+2)^2)
BD = √(13) units.

Jai Ganesh · 2020-06-27 18:27:26

A transposing error!

CG # 109. Find the coordinates of the vertices of an equilateral triangle of side 2a.

666 bro · 2020-06-28 02:49:11

CG#109: the vertices are (0, ±√(3) ) , (-a,0) , (a,0) respectively.

Jai Ganesh · 2020-06-28 13:31:01

Good attempt!

CG # 110. If the vertices of right angle triangle are (5,2), (2,-2), and (-2,a) and right angle between point (5,2) , (2,-2), and (2,-2), (-2,a), find the value of 'a'.

666 bro · 2020-06-28 22:31:43

CG#110: the value of a = 1.

Last edited by 666 bro (2020-06-28 22:56:12)

Jai Ganesh · 2020-06-30 14:55:55

Neat work!

(When questions on specific topics are posted, like coordinate geometry, algebra, mensuration etc., it may take some time to check and reply, say a fortnight or a month.)

CG # 111. If (-3,0) and (3,0) are two vertices of an equilateral triangle, find the coordinates of the third vertex.

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