Question

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Answer 1

Check the attachment.

Answer 2

$\large{\underline{\underline{\mathfrak{\green{\sf{QUESTION:-}}}}}}$.

• If the equation S = $\:(ax^2+2hxy+by^2+2gx+2fy+c)\:=\:0$ represent a pair of parallel straight line , then show that ,

$\implies\:(1)\:h\:=ab$

$\implies\:(2)\:af^2\:=\:bg^2$

$\large{\underline{\underline{\mathfrak{\bf{SOLUTION:-}}}}}$.

$\large{\underline{\underline{\mathfrak{\sf{Given\:Here:-}}}}}$.

• Combine Equation of two parallel straight line S = $\:(ax^2+2hxy+2gx+2fy+c)\:=\:0$

$\large{\underline{\underline{\mathfrak{\sf{Show\:that:-}}}}}$.

• $\:h\:=\:ab$

• $\:af^2\:=\:bg^2$

• Distance between two parallel straight line will be $\:2\sqrt{\frac{(g^2-ac)}{a(a+b)}}\:=\:2\sqrt{\frac{(f^2-bc)}{a(a+b)}}$

$\large{\underline{\underline{\mathfrak{\sf{Explanation:-}}}}}$.

Let,

• L1 and L2 are two parallel straight

So, equation will be straight line

• $\:(a1x+b1y+c1)\:=\:0$....(1)

• $\:(a1x+b1y+c2)\:=\:0$....(2)

So ,Combination of two straight line will be

$\:(a1x+b1y+c1)(a1x+b1y+c2)\:=\:(ax^2+by^2+2hxy+2gx+2fy+c)$

$\leadsto\:x^2a1^2\:+\:2(a1b1)xy\:+\:b1^2y^2\:+\:a1(c1+c2)x\:+\:b1(c1+c2)y\:+\:c1c2\:=\:(ax^2+by^2+2hxy+2gx+2fy+c)$

Compare Coefficient of x^2, y^2, x, y, Constant part

we find here ,

• $\:a\:=\:a1^2$....(3)

• $\:b\:=\:b1^2$.....(4)

• $\:a1b1\:=\:h$....(5)

• $\:2g\:=\:a1(c1+c2)$....(6)

• $\:2f\:=\:b1(c1+c2)$...(7)

• $\:c1c2\:=\:c$....(8)

now, we show that ,

Case (1):-

• $\:h\:=\:ab$

We take , by equation (5)

$\leadsto\:h\:=\:a1b1$

by equation (3) and (4) , we get there

• a1 = a
• b1 = b

So,

$\leadsto\boxed{\boxed{\:h\:=\:ab}}$

Case(2):-

• $\:af^2\:=\:bg^2$

we take , by equation (3) and (7)

$\leadsto\:af^2\:=\:a1^2\times(\frac{b1(c1+c2)}{2})^2$

$\leadsto\:af^2\:=\:b1^2\times\:(\frac{a1(c1+c2)}{2})^2$

but, by equation(4) and (7) ,

$\leadsto\boxed{\boxed{\:af^2\:=\:bg^2}}$

Case(3):-

• Distance between two parallel straight line will be $\:2\sqrt{\frac{(g^2-ac)}{a(a+b)}}\:=\:2\sqrt{\frac{(f^2-bc)}{a(a+b)}}$

we know here ,

[ Distance between two parallel straight line ]

$\large\boxed{\boxed{\:Distance\:=\frac{(|c1-c2|)}{\sqrt{(a1^2+b1^2)}}}}$

$\leadsto\:Distance\:=\frac{\sqrt{(c1+c2)^2-4c1c2)}}{\sqrt{a+b}}$....(9)

Keep value by equation (6) ,

$\leadsto\:Distance\:=\frac{(\frac{2g}{a1})^2-4c}{\sqrt{(a+b)}}$

$\leadsto\:Distance\:=\frac{(\frac{4g^2}{a}-4c}{\sqrt{(a+b)}}$

$\leadsto\:Distance\:=\frac{(4g^2-4ac}{\sqrt{a(a+b)}}$

$\leadsto\boxed{\boxed{\:Distance\:=\:2\frac{(4g^2-4ac)}{\sqrt{a(a+b)}}}}$....(10)

Again, we take by equation (7) keep in (9)

we get ,

$\leadsto\:Distance\:=\frac{(\frac{2f}{b1})^2-4c}{\sqrt{(a+b)}}$

$\leadsto\:Distance\:=\frac{\frac{(4f^2)}{b}-4c}{\sqrt{(a+b)}}$

$\leadsto\:Distance\:=\frac{(4f^2-4bc)}{\sqrt{b(a+b)}}$

$\leadsto\boxed{\boxed{\:Distance\:=\:2\frac{(f^2-bc)}{\sqrt{b(a+b)}}}}$..(11)

by equation (10) and (11) ,

we find here

$\large\boxed{\boxed{\:2\frac{(g^2-ac)}{\sqrt{a(a+b)}}\:=\:2\frac{(f^2-bc)}{\sqrt{b(a+b)}}}}$

here three cases are proved

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