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Bisectors of Angle between Two Lines

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Question

Solve:
(i) $\frac{16 x^{2} - 20 x + 9}{8 x^{2} + 12 x + 21} = \frac{4 x - 5}{2 x + 3}$
(ii) $\frac{5 y^{2} + 40 y - 12}{5 y + 10 y^{2} - 4} = \frac{y + 8}{1 + 2 y}$

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Solution

(i) $\frac{16 x^{2} - 20 x + 9}{8 x^{2} + 12 x + 21} = \frac{4 x - 5}{2 x + 3}$
Let, $k = \frac{16 x^{2} - 20 x + 9}{8 x^{2} + 12 x + 21} = \frac{4 x - 5}{2 x + 3}$
Multiplying numerator and denominator of $\frac{4 x - 5}{2 x + 3}$ , we get,
$\Rightarrow \frac{16 x^{2} - 20 x + 9}{8 x^{2} + 12 x + 21} = \frac{4 x (4 x - 5)}{4 x (2 x + 3)}$
$\Rightarrow \frac{16 x^{2} - 20 x + 9}{8 x^{2} + 12 x + 21} = \frac{16 x^{2} - 20 x}{8 x^{2} + 12 x}$
Using Theorems on Equal ratios ie. $\frac{a}{b} = \frac{c}{d} = \frac{a - c}{b - d}$ , we get,
$\Rightarrow \frac{16 x^{2} - 20 x + 9}{8 x^{2} + 12 x + 21} = \frac{16 x^{2} - 20 x}{8 x^{2} + 12 x} = \frac{16 x^{2} - 20 x + 9 - 16 x^{2} + 20 x}{8 x^{2} + 12 x + 21 - 8 x^{2} - 12 x}$
$\Rightarrow k = \frac{9}{21} = \frac{3}{7}$
Hence, $\frac{16 x^{2} - 20 x + 9}{8 x^{2} + 12 x + 21} = \frac{4 x - 5}{2 x + 3} = \frac{3}{7}$
(ii) $\frac{5 y^{2} + 40 y - 12}{5 y + 10 y^{2} - 4} = \frac{y + 8}{1 + 2 y}$
Let, $z = \frac{5 y^{2} + 40 y - 12}{5 y + 10 y^{2} - 4} = \frac{y + 8}{1 + 2 y}$
Multiplying numerator and denominator of $\frac{y + 8}{1 + 2 y}$ by $5 y$ , we get,
$\Rightarrow \frac{5 y^{2} + 40 y - 12}{5 y + 10 y^{2} - 4} = \frac{5 y (y + 8)}{5 y (1 + 2 y)}$
$\Rightarrow \frac{5 y^{2} + 40 y - 12}{5 y + 10 y^{2} - 4} = \frac{5 y^{2} + 40 y}{5 y + 10 y^{2}}$
Using Theorems on Equal ratios, we get,
$\Rightarrow \frac{5 y^{2} + 40 y - 12}{5 y + 10 y^{2} - 4} = \frac{5 y^{2} + 40 y}{5 y + 10 y^{2}} = \frac{5 y^{2} + 40 y - 12 - 5 y^{2} - 40 y}{5 y + 10 y^{2} - 4 - 5 y - 10 y^{2}}$
$\Rightarrow z = \frac{- 12}{- 4} = 3$
Hence, $\frac{5 y^{2} + 40 y - 12}{5 y + 10 y^{2} - 4} = \frac{y + 8}{1 + 2 y} = 3$

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