Given: #160;9x #160;+ #160;5y #160;= #160;10 #160; #160; #160; #160; #160; #160; #160; #160; #160; #160; #160; #160; #160; ...

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Bisectors of Angle between Two Lines

9 x+5 y=103 y...

Question

9x + 5y = 10
3y − 2x = 8

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Solution

$Given : 9 x + 5 y = 10 3 y - 2 x = 8 Rearranging the second equation, the two equations can be written as 9 x + 5 y = 10 - 2 x + 3 y = 8 Now, D = |9 5 - 2 3| = 27 + 10 = 37 D_{1} = |10 5 8 3| = 30 - 40 = - 10 D_{2} = |9 10 - 2 8| = 72 + 20 = 92 Using Cramer' s rule, we get x = \frac{D_{1}}{D} = \frac{- 10}{37} y = \frac{D_{2}}{D} = \frac{92}{37} ∴ x = \frac{- 10}{37} and y = \frac{92}{37}$

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$\frac{F i r s t m o n o m i a l \to}{S e c o n d m o n o m i a l ↓}$	$2 x$	$- 5 y$	$3 x^{2}$	$- 4 x y$	$7 x^{2} y$	$- 9 x^{2} y^{2}$
$2 x$	$4 x^{2}$	.....	.....	.....	.....	.....
$- 5 y$	.....	.....	$- 15 x^{2} y$	.....	.....	.....
$3 x^{2}$	.....	.....	.....	.....	.....	.....
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\frac{3 x + 4 y}{2 x + 5 y} = \frac{4}{3}

9 x + 12 y = 8 x + 20 y

\frac{3 x + 4 y}{2 x + 5 y} = \frac{4}{3}

\frac{x}{y} = 8 \Rightarrow x : y = 8 : 1

3 (3 x + 4 y) = 4 (2 x + 5 y)

x = 8 y

x^{2} + 4 y^{2} - 15 x = 10 (3 y - 8), x y = 6

$F i r s t m o n o m i a l \to - --------------- \to ↓ S e c o n d m o n o m i a l$	2x	-5y	$3 x^{2}$	-4xy	$7 x^{2} y$	$- 9 x^{2} y^{2}$
2x	$4 x^{2}$	-	-	-	-	-
- 5y	-	-	$- 15 x^{2} y$	-	-	-
$3 x^{2}$	-	-	-	-	-	-
- 4xy	-	-	-	-	-	-
$7 x^{2} y$	-	-	-	-	-	-
$- 9 x^{2} y^{2}$	-	-	-	-	-	-

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