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**ganesh****Administrator**- Registered: 2005-06-28
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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.

It is no good to try to stop knowledge from going forward. Ignorance is never better than knowledge - Enrico Fermi.

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**666 bro****Member**- From: Flatland
- Registered: 2019-04-26
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The length of the median through A = √65 units.

"An equation for me has no meaning, unless it expresses a thought of God"- Srinivasa ramanujan

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**ganesh****Administrator**- Registered: 2005-06-28
- Posts: 30,587

Hi,

Neat work!

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

It is no good to try to stop knowledge from going forward. Ignorance is never better than knowledge - Enrico Fermi.

Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.

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**666 bro****Member**- From: Flatland
- Registered: 2019-04-26
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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.

"An equation for me has no meaning, unless it expresses a thought of God"- Srinivasa ramanujan

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**ganesh****Administrator**- Registered: 2005-06-28
- Posts: 30,587

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).

It is no good to try to stop knowledge from going forward. Ignorance is never better than knowledge - Enrico Fermi.

Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.

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**666 bro****Member**- From: Flatland
- Registered: 2019-04-26
- Posts: 244

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)

"An equation for me has no meaning, unless it expresses a thought of God"- Srinivasa ramanujan

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**ganesh****Administrator**- Registered: 2005-06-28
- Posts: 30,587

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.

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**666 bro****Member**- From: Flatland
- Registered: 2019-04-26
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CG #102 the fourth vertex(D) = (7, 4)

"An equation for me has no meaning, unless it expresses a thought of God"- Srinivasa ramanujan

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**ganesh****Administrator**- Registered: 2005-06-28
- Posts: 30,587

Hi,

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

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**666 bro****Member**- From: Flatland
- Registered: 2019-04-26
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CG#103: the area of quadrilateral is = 28 sq units.

"An equation for me has no meaning, unless it expresses a thought of God"- Srinivasa ramanujan

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**ganesh****Administrator**- Registered: 2005-06-28
- Posts: 30,587

Hi,

Good work!

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

Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.

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**666 bro****Member**- From: Flatland
- Registered: 2019-04-26
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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.

"An equation for me has no meaning, unless it expresses a thought of God"- Srinivasa ramanujan

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**ganesh****Administrator**- Registered: 2005-06-28
- Posts: 30,587

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.

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**666 bro****Member**- From: Flatland
- Registered: 2019-04-26
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The other two vertices are = (1, 2) and (1, 2) respectively.

"An equation for me has no meaning, unless it expresses a thought of God"- Srinivasa ramanujan

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**ganesh****Administrator**- Registered: 2005-06-28
- Posts: 30,587

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).

Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.

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**666 bro****Member**- From: Flatland
- Registered: 2019-04-26
- Posts: 244

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

"An equation for me has no meaning, unless it expresses a thought of God"- Srinivasa ramanujan

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**ganesh****Administrator**- Registered: 2005-06-28
- Posts: 30,587

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.

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**666 bro****Member**- From: Flatland
- Registered: 2019-04-26
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CG#107: the coordinates are (2, 0) (0, 2) .

"An equation for me has no meaning, unless it expresses a thought of God"- Srinivasa ramanujan

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**ganesh****Administrator**- Registered: 2005-06-28
- Posts: 30,587

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).

Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.

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**666 bro****Member**- From: Flatland
- Registered: 2019-04-26
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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.

"An equation for me has no meaning, unless it expresses a thought of God"- Srinivasa ramanujan

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**ganesh****Administrator**- Registered: 2005-06-28
- Posts: 30,587

A transposing error!

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

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**666 bro****Member**- From: Flatland
- Registered: 2019-04-26
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CG#109: the vertices are (0, ±√(3) ) , (-a,0) , (a,0) respectively.

"An equation for me has no meaning, unless it expresses a thought of God"- Srinivasa ramanujan

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**ganesh****Administrator**- Registered: 2005-06-28
- Posts: 30,587

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'.

Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.

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**666 bro****Member**- From: Flatland
- Registered: 2019-04-26
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CG#110: the value of a = 1.

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

"An equation for me has no meaning, unless it expresses a thought of God"- Srinivasa ramanujan

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**ganesh****Administrator**- Registered: 2005-06-28
- Posts: 30,587

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.

Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.

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