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Posted by Mayank Dhandh 5 years, 6 months ago
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Yogita Ingle 5 years, 6 months ago
Let A(4,-1) and B(-2,-3) be the given points.
Let P(x,y) and Q(a,b) be the points of trisection of AB so that AP = PQ = QB
Hence P divides AB internally in the ratio 1 : 2 and Q divides AB internally in the ratio 2 : 1
By the section formula, the required points are
AP = 1
PQ = 1
QB = 1
Section formula internally = (lx₂ + mx₁)/(l + m) , (ly₂ + my₁)/(l + m)
P divides the line segment in the ratio 1:2
l = 1 m = 2
A(4,-1) and B(-2,-3)
= [(1(-2) + 2(4)]/(1+2) , [(1(-3) + 2(-1)]/(1+2)
= (-2+8)/3 , (-3-2)/3
= 6/3 , -5/3
= P (2 , -5/3)
Q divides the line segment in the ratio 2:1
l = 2 m = 1
= [(2(-2) + 1(4)]/(2+1) , [(2(-3) + 1(-1)]/(2+1)
= (-4+4)/3 , (-6-1)/3
= 0/3 , -7/3
= Q (0 , -7/3)
Posted by Sujal Gundaniya 5 years, 6 months ago
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Yogita Ingle 5 years, 6 months ago
Given: 2x +3y = 9 _______(1),
3x + 4y = 5 ______(2)
From (1) 2x = 9 – 3y
{tex} x=\frac{9-3 y}{2}{/tex} ______ (3)
Substitute (3) in (2)
{tex}3\left(\frac{9-3 y}{2}\right)+4 y=5\\\\\left(\frac{27-9 y}{2}\right)+4 y=5{/tex}
27 – 9y + 8y = 10
27 – 10 = y
y = 17 ______(4)
Substitute (4) in (1)
2x +3(17) = 9
2x = 9 – 51
2x = -42
x = -21 and y = 17
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Yogita Ingle 5 years, 6 months ago
Let the unit place digit be x
And tens place digit be y
No. Formed by x and y = 10y+x (I have multiplied y by ten because it is on tens place)
Now, no. Obtained by reversing digits = 10x+y
Here sum of digits = 8 (I)
And difference on reversing digits is 18
A/q
(10y+x) - (10x+y) = 18
Or, 10y+x-10x-y= 18
Or, 9y-9x = 18
Or, 9(y-x) = 18
Or y-x = 18/9= 2(ii)
On adding both the given equation
Now, y+x = 8
y-x = 2
------------------
2y. = 10
Or, y =5
Now putting the value of y in equation 1st
y+x=8
5+x=8
Or, x= 3
Now the formed no. = 53
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Bhavya Gupta 5 years, 6 months ago
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