Find the equation of a line perpendicular to | vedclass

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

Question

(A) The given line is $x-2y+3=0$,which can be rewritten as $2y = x+3$,or $y = \frac{1}{2}x + \frac{3}{2}$.The slope of this line is $m_1 = \frac{1}{2}

Vedclass · Accepted Answer

$2x+y=0$

Vedclass · Answer

(A) The given line is $x-2y+3=0$,which can be rewritten as $2y = x+3$,or $y = \frac{1}{2}x + \frac{3}{2}$.The slope of this line is $m_1 = \frac{1}{2}$.The slope of a line perpendicular to this line is $m_2 = -\frac{1}{m_1} = -\frac{1}{1/2} = -2$.The equation of the line passing through the point $(1, -2)$ with slope $m_2 = -2$ is given by the point-slope form: $y - y_1 = m(x - x_1)$.Substituting the values: $y - (-2) = -2(x - 1)$.$y + 2 = -2x + 2$.$2x + y = 0$.

Find the equation of a line perpendicular to the line $x-2y+3=0$ and passing through the point $(1, -2).$

Similar Questions

What is the equation of the line that makes an angle of $45^{\circ}$ with the $x$-axis and passes through the point $(3, -4)$?

The equation of the straight line which is perpendicular to $y = x$ and passes through $(3, 2)$ is

The equation of the line parallel to the line $3x - 4y + 2 = 0$ and passing through $(-2, 3)$ 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 .....$^o$.

$A$ line passes through the point $(3, 4)$ and cuts off intercepts from the coordinate axes such that their sum is $14$. The equation of the line is

Vedclass Test Series

Exam Paper Generator

Online Exam Module