Find the equation of the line passing through the points (2, 3) and the point of intersection of the lines $4 x - 3 y = 7$ and $3 x + 4 y + 1 = 0$ .

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Solution

$4 x - 3 y = 7 \dots \dots (i) 3 x + 4 y = - 1 \dots \dots (i i)$
On multiplying eq (i) by 4 and eq (ii) by 3, we get
$16 x - 12 y = 28 \dots (i i i) 9 x + 12 y = - 3 - -------------- - \dots (i v) a d d i n g 25 x = 25 \Rightarrow x = 1$
Putting $x = 1$ , in equation $4 x - 3 y = 7$ , we get
$\Rightarrow 4 \times 1 - 3 y = 7 \Rightarrow - 3 y = 7 - 4 \Rightarrow - 3 y = 3 \Rightarrow y = - 1$
$∴$ Point of intersection = (1, -1)
Equation of line passing through (2, 3) and (1, -1) is
$y - y_{1} = \frac{y_{2} - y_{1}}{x_{2} - x_{1}} (x - x_{1}) \Rightarrow y - 3 = \frac{- 1 - 3}{1 - 2} (x - 2) \Rightarrow y - 3 = 4 (x - 2) \Rightarrow y - 3 = 4 x - 8 \Rightarrow 4 x - y - 5 = 0$

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Find the equation of the line passing through the points (2, 3) and the point of intersection of the lines 4x−3y=7 and 3x+4y+1=0.

Find the equation of the line passing through the points (2, 3) and the point of intersection of the lines $4 x - 3 y = 7$ and $3 x + 4 y + 1 = 0$ .