Question:
The equation of straight line through the intersection of the lines $x - 2y = 1 $ and $x + 3y = 2$ and parallel $3x + 4y = 0$ is
The equation of straight line through the intersection of the lines $x - 2y = 1 $ and $x + 3y = 2$ and parallel $3x + 4y = 0$ is
Updated On: Jun 18, 2022
- 3x + 4y + 5 = 0
- 3x + 4y - 10 = 0
- 3x + 4y - 5 = 0
- 3x + 4y + 6 = 0
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The Correct Option is C
Solution and Explanation
The intersection point of lines $x - 2y = 1$ and $x +3y = 2$ is $\left( \frac{7}{5}, \frac{1}{5}\right)$
Since, required line is parallel to $3x + 4y = 0.$
Therefore, the slope of required line is $ - \frac{3}{4}$
$\therefore$ Equation of required line which passes throught $\left( \frac{7}{5}, \frac{1}{5}\right)$ is given by
$y - \frac{1}{5} = - \frac{3}{4}\left(x - \frac{7}{5}\right)$
$ \Rightarrow \frac{3x}{4} + y = \frac{21}{20} + \frac{1}{5}$
$\Rightarrow 3x+ 4y - 5 = 0 $
Since, required line is parallel to $3x + 4y = 0.$
Therefore, the slope of required line is $ - \frac{3}{4}$
$\therefore$ Equation of required line which passes throught $\left( \frac{7}{5}, \frac{1}{5}\right)$ is given by
$y - \frac{1}{5} = - \frac{3}{4}\left(x - \frac{7}{5}\right)$
$ \Rightarrow \frac{3x}{4} + y = \frac{21}{20} + \frac{1}{5}$
$\Rightarrow 3x+ 4y - 5 = 0 $
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Concepts Used:
General Equation of a Line
Equation of Straight Line Formula:
A straight line is a figure created when two points A (x1, y1) and B (x2, y2) are connected with a minimum distance between them, and both the ends are extended to immensity (infinity). With variables x and y, the standard form of a linear equation is: ax + by = c, where a, b, and c are constants and x, and y are variables.
Point Slope Form:
The equation of a straight line whose slope is m and passes through a point (x1, y1) is formed or created using the point-slope form. The equation of the point-slope form is:
y - y1 = m (x - x1),
where (x, y) = an arbitrary point on the line.