The equation of the line perpendicular to the | vedclass

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

Question

(D) The given line is $\frac{x}{a} - \frac{y}{b} = 1$,which can be written as $bx - ay = ab$.The line cuts the $x$-axis where $y = 0$. Substituting $y

Vedclass · Accepted Answer

$\frac{x}{b} + \frac{y}{a} = \frac{a}{b}$

Vedclass · Answer

(D) The given line is $\frac{x}{a} - \frac{y}{b} = 1$,which can be written as $bx - ay = ab$.The line cuts the $x$-axis where $y = 0$. Substituting $y = 0$ in the equation,we get $bx = ab$,so $x = a$. Thus,the point of intersection is $(a, 0)$.The slope of the given line $bx - ay = ab$ is $m_1 = \frac{b}{a}$.The slope of the line perpendicular to it is $m_2 = -\frac{1}{m_1} = -\frac{a}{b}$.The equation of the line with slope $m_2 = -\frac{a}{b}$ passing through $(a, 0)$ is $y - 0 = -\frac{a}{b}(x - a)$.Multiplying by $b$,we get $by = -ax + a^2$,which simplifies to $ax + by = a^2$.Dividing both sides by $ab$,we get $\frac{ax}{ab} + \frac{by}{ab} = \frac{a^2}{ab}$,which results in $\frac{x}{b} + \frac{y}{a} = \frac{a}{b}$.

The equation of the line perpendicular to the line $\frac{x}{a} - \frac{y}{b} = 1$ and passing through the point at which it cuts the $x$-axis,is

Similar Questions

The vertex $A$ of a triangle lies on the lines $x+y=1$ and $2x+3y=6$. If the orthocentre of the triangle is $O\left(\frac{3}{7}, \frac{22}{7}\right)$,then the equation of $OA$ in the normal form is

The equations of the diagonals of the square formed by the lines $x = 0,$ $y = 0,$ $x = 1,$ and $y = 1$ are

Find the equation of a line which is bisected at the point $(x_1, y_1)$ by the axes.

Reduce the following equation into intercept form and find its intercepts on the axes: $4x - 3y = 6$.

The number of integral values of $m$ such that the abscissa of the point of intersection of the lines $3x + 4y = 9$ and $y = mx + 1$ is also an integer,is

Vedclass Test Series

Exam Paper Generator

Online Exam Module