The equation of a line passing through (3, -4 | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) The given line is $3x + 4y = 5$,which can be written as $4y = -3x + 5$ or $y = -\frac{3}{4}x + \frac{5}{4}$.The slope of this line is $m_1 = -\fra

Vedclass · Accepted Answer

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

Vedclass · Answer

(D) The given line is $3x + 4y = 5$,which can be written as $4y = -3x + 5$ or $y = -\frac{3}{4}x + \frac{5}{4}$.The slope of this line is $m_1 = -\frac{3}{4}$.Since the required line is perpendicular to this line,its slope $m_2$ must satisfy $m_1 	imes m_2 = -1$.Thus,$m_2 = -\frac{1}{m_1} = -\frac{1}{-3/4} = \frac{4}{3}$.The equation of a line passing through $(x_1, y_1) = (3, -4)$ with slope $m = \frac{4}{3}$ is given by the point-slope form: $y - y_1 = m(x - x_1)$.Substituting the values,we get $y - (-4) = \frac{4}{3}(x - 3)$,which simplifies to $y + 4 = \frac{4}{3}(x - 3)$.

The equation of a line passing through $(3, -4)$ and perpendicular to the line $3x + 4y = 5$ is

Similar Questions

The point $P(a, b)$ lies on the straight line $3x + 2y = 13$ and the point $Q(b, a)$ lies on the straight line $4x - y = 5$. Then the equation of the line $PQ$ is

The equation of the line passing through $(c, d)$ and parallel to $ax + by + c = 0$ is

If the coordinates of the points $A$ and $B$ are $(1, 0)$ and $(2, \sqrt{3})$,then the angle made by the line $AB$ with the $x$-axis is .....$^o$.

If the straight line through the point $P(3, 4)$ makes an angle $\frac{\pi}{6}$ with the $x$-axis and meets the line $12x + 5y + 10 = 0$ at $Q$,then the length $PQ$ is

Reduce the following equation into slope-intercept form and find its slope and the $y$-intercept.
$x+7y=0$

Vedclass Test Series

Exam Paper Generator

Online Exam Module