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Sharlin Thomas 7 years, 2 months ago
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Sia ? 6 years, 4 months ago
Let A → (6, –6), B → (3, –7) and C → (3, 3).
Let the centre of the circle be I(x, y)
Then, IA = IB = IC [By definition of a circle]
{tex}\Rightarrow{/tex} IA2 = IB2 = IC2
{tex}\Rightarrow{/tex} (x - 6)2 + (y + 6)2 = (x - 3)2 + (y + 7)2 = (x - 3)2 + (y - 3)2
Taking first two, we get
(x - 6)2 + (y + 6)2 = (x - 3)2 + (y + 7)2
{tex}\Rightarrow{/tex} x2 - 12x + 36 + y2 + 12y + 36 = x2 - 6x + 9 + y2 + 14y + 49
{tex}\Rightarrow{/tex} 6x + 2y = 14
{tex}\Rightarrow{/tex} 3x + y = 7 ......(1) ....[Dividing throughout by 2]
Taking last two, we get
(x - 3)2 + (y + 7)2 = (x - 3)2 + (y - 3)2
{tex}\Rightarrow{/tex} (y + 7)2 = (y - 3)2
{tex}\Rightarrow{/tex} (y + 7) = {tex}\pm{/tex}(y-3)
taking +e sign, we get
y + 7 = y - 3
{tex}\Rightarrow{/tex} 7 = -3
which is impossible
Taking -ve sign, we get
y + 7 = -(y - 3)
{tex}\Rightarrow{/tex} y + 7 = -y + 3
{tex}\Rightarrow{/tex} 2y = -4
{tex}\Rightarrow y = \frac{{ - 4}}{2} = - 2{/tex}
Putting y = -2 in equation (1), we get
{tex}\Rightarrow{/tex} 3x - 2 = 7
{tex}\Rightarrow{/tex} 3x = 9
{tex}\Rightarrow{/tex} x = 3
Thus, I {tex}\rightarrow{/tex} (3, -2)
Hence, the centre of the circle is (3, -2).
Posted by Aniket Dalmia 7 years, 2 months ago
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Posted by Md Sameer Sam 6 years, 4 months ago
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Sia ? 6 years, 4 months ago
{tex}{/tex}We know that if three points A, B, C are collinear, then area of {tex}\Delta ABC =0{/tex}
Given that points A(k + 1, 1), B(2k + 1, 3) and C(2k + 2, 2k) are collinear.
Here, x1= k+1 , x2 =2k+1, x3= 2k+2, y1= 1, y2= 3, y3= 2k
{tex}\Rightarrow {/tex} Area of {tex}\Delta ABC =0{/tex}
{tex}\Rightarrow {/tex} {tex}\frac{1}{2}|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)|=0{/tex}
{tex}\Rightarrow {/tex} {tex}\frac{1}{2}|(k+1)(3-2k)+(2k+1)(2k-1)+(2k+2)(1-3)|=0{/tex}
{tex}\Rightarrow{/tex} {tex}|k(3-2k)+1(3-2k)+(2k)^2-(1)^2+(2k+2)(-2)|=0{/tex}
{tex}\Rightarrow{/tex} |3k - 2k2 + 3 - 2k + 4k2 - 1 - 4k - 4 |= 0
{tex}\Rightarrow{/tex} |2k2 - 3k - 2 |= 0
{tex}\Rightarrow{/tex} 2k2 - 4k + k - 2 = 0
{tex}\Rightarrow{/tex} 2k(k - 2) + 1(k - 2) = 0
{tex}\Rightarrow{/tex} (2k + 1) (k - 2) = 0
{tex}\Rightarrow{/tex} k = - {tex}\frac { 1 } { 2 }{/tex}, k = 2
Posted by Rahul Kumar 7 years, 2 months ago
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Manvi Munpariya 7 years, 2 months ago
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Posted by Ashwani Thakur 6 years, 4 months ago
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Sia ? 6 years, 4 months ago
Let us assume that point P divides the line segment AB in the ratio (k : 1)
Using section formula,
P(4, m) = {tex}\frac{6k+2}{k+1} , {/tex}{tex}\frac{-3k+3}{k+1}{/tex}
{tex}\therefore{/tex} {tex}\frac{6 \mathrm{k}+2}{\mathrm{k}+1}{/tex} = 4 {tex}\Rightarrow 6k+2=4k+4 \Rightarrow 2k=2{/tex}
{tex}\Rightarrow{/tex} k = 1
{tex}\therefore{/tex} The ratio is 1 : 1
Now, m = {tex}\frac{-3k+3}{k+1}{/tex} = {tex}\frac{-3+3}{2}{/tex} = 0
Posted by Ramanpreet Kaur Deol 6 years, 4 months ago
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Sia ? 6 years, 4 months ago
Given, {tex}\frac { \cos \theta } { 1 - \sin \theta } + \frac { \cos \theta } { 1 + \sin \theta } = 4{/tex}
Taking LCM
{tex}\frac { \cos \theta ( 1 + \sin \theta ) + \cos \theta ( 1 - \sin \theta ) } { ( 1 - \sin \theta ) ( 1 + \sin \theta ) } = 4{/tex}
{tex}\frac { \cos \theta [ 1 + \sin \theta + 1 - \sin \theta ] } { 1 - \sin ^ { 2 } \theta } = 4{/tex}
{tex}\frac { \cos \theta ( 2 ) } { \cos ^ { 2 } \theta } = 4{/tex}
{tex}\frac { 2 } { \cos \theta } = 4{/tex}
{tex}\cos \theta = \frac { 2 } { 4 } = \frac { 1 } { 2 }{/tex}
{tex}\cos \theta = \cos 60 ^ { \circ }{/tex}
{tex}\therefore \theta = 60 ^ { \circ }{/tex}
Sia ? 6 years, 4 months ago
Given, {tex}\frac { \cos \theta } { 1 - \sin \theta } + \frac { \cos \theta } { 1 + \sin \theta } = 4{/tex}
Taking LCM
{tex}\frac { \cos \theta ( 1 + \sin \theta ) + \cos \theta ( 1 - \sin \theta ) } { ( 1 - \sin \theta ) ( 1 + \sin \theta ) } = 4{/tex}
{tex}\frac { \cos \theta [ 1 + \sin \theta + 1 - \sin \theta ] } { 1 - \sin ^ { 2 } \theta } = 4{/tex}
{tex}\frac { \cos \theta ( 2 ) } { \cos ^ { 2 } \theta } = 4{/tex}
{tex}\frac { 2 } { \cos \theta } = 4{/tex}
{tex}\cos \theta = \frac { 2 } { 4 } = \frac { 1 } { 2 }{/tex}
{tex}\cos \theta = \cos 60 ^ { \circ }{/tex}
{tex}\therefore \theta = 60 ^ { \circ }{/tex}
Posted by Asish Panigrahi Asish Panigrahi 7 years, 2 months ago
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Yash Kumar 7 years, 2 months ago
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Posted by Raju Hosur 6 years, 6 months ago
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Sia ? 6 years, 6 months ago
Given,
{tex}\frac { 1 } { ( a + b + x ) } = \frac { 1 } { a } + \frac { 1 } { b } + \frac { 1 } { x }{/tex}
{tex}\Rightarrow \quad \frac { 1 } { ( a + b + x ) } - \frac { 1 } { x } = \frac { 1 } { a } + \frac { 1 } { b } \Rightarrow \frac { x - ( a + b + x ) } { x ( a + b + x ) } = \frac { b + a } { a b }{/tex}
{tex}\Rightarrow \quad \frac { - ( a + b ) } { x ( a + b + x ) } = \frac { ( a + b ) } { a b }{/tex}
On dividing both sides by (a+b)
{tex}\Rightarrow \quad \frac { - 1 } { x ( a + b + x ) } = \frac { 1 } { a b }{/tex}
Now cross multiply
{tex}\Rightarrow{/tex} x(a + b + x) = -ab
{tex}\Rightarrow{/tex} x2 + ax + bx + ab = 0
{tex}\Rightarrow{/tex} x(x +a) + b(x +a) = 0
{tex}\Rightarrow{/tex} (x + a) (x + b) = 0
{tex}\Rightarrow{/tex} x + a = 0 or x + b = 0
{tex}\Rightarrow{/tex} x = -a or x = -b.
Therefore, -a and -b are the roots of the equation.
Posted by Saif Khan 6 years, 6 months ago
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Sia ? 6 years, 6 months ago
L.H.S
= (1 + cotA + tanA) (sinA - cosA)
= sinA - cosA + cotA sinA - cotA cosA + sinA tanA - tanA cosA
{tex}= \sin A - \cos A + \frac{{\cos A}}{{\sin A}} \times \sin A - \cot A\cos A + \sin A\;\tan A - \frac{{\sin A}}{{\cos A}} \times \cos A{/tex}
= sinA - cosA + cosA - cotA cosA + sinA tanA - sinA
= sinA tanA - cotA cosA........(1)
Now taking ;
{tex}\quad \frac{{\sec A}}{{\cos e{c^2}A}} - \frac{{\cos ecA}}{{{{\sec }^2}A}}{/tex}
{tex} = \frac{{\frac{1}{{\cos A}}}}{{\frac{1}{{{{\sin }^2}A}}}} - \frac{{\frac{1}{{\sin A}}}}{{\frac{1}{{{{\cos }^2}A}}}}{/tex}
{tex} = \frac{{{{\sin }^2}A}}{{\cos A}} - \frac{{{{\cos }^2}A}}{{\sin A}}{/tex}
{tex} = \sin A \times \frac{{\sin A}}{{\cos A}} - \cos A \times \frac{{\cos A}}{{\sin A}}{/tex}
{tex} = \sin A \times \tan A - \cos A \times \cot A{/tex}.......(2)
From (1) & (2),
(1 + cotA + tanA) (sinA - cosA) = {tex}\frac { \sec A } { cosec ^ { 2 } A } - \frac { cosec A } { \sec ^ { 2 } A }{/tex} = sinA.tanA - cosA.cotA
Hence, Proved.
Posted by Lakshika Sood 7 years, 2 months ago
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Mp Bvc 7 years, 2 months ago
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Nisha Pawar 7 years, 2 months ago
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Mohit Kumar 5 years, 8 months ago
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