Find the equation of a line passing through the point $(2, 3)$ and parallel to the line $3 x - 4 y + 5 = 0.$

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Solution

Equation of line through $(2, 3)$ is

$y - 1 = m (x - x_{1}) . . . (1)$

$2, 3$ is $(x_{1} y_{1})$

Since the line is parallel to

$3 x - 4 y + 5 = 0$

$\Rightarrow$ The slope will be equal

Slope of $3 x - 4 y + 5 = 0$

$4 y = 3 x + 5$

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

$\Rightarrow m = \frac{3}{4}$

Substituting m and $(x_{1} y_{1})$ is (1)

$y - 3 = \frac{3}{4} (x - 2)$

$4 y - 12 = 3 x - 6$

$3 x - 4 y = - 12 + 6 = - 6$

$3 x - 4 y + 6 = 0$

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Find the equation of a line passing through the point (2, 3) and parallel to the line 3 x−4 y+5=0.

Equation of line through (2, 3) is y−1=m(x−x1) ...(1) 2, 3 is (x1y1) Since the line is parallel to 3x−4y+5=0 ⇒ The slope will be equal Slope of 3x−4y+5=0 4y=3x+5 y=34x+54 ⇒ m=34 Substituting m and (x1y1) is (1) y−3=34(x−2) 4y−12=3x−6 3x−4y=−12+6=−6 3x−4y+6=0

Find the equation of a line passing through the point $(2, 3)$ and parallel to the line $3 x - 4 y + 5 = 0.$