The length of the perpendicular from the poin | vedclass

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

Question

(B) The given line is $\frac{x}{a} + \frac{y}{b} = 1$,which can be rewritten as $bx + ay - ab = 0$. The length of the perpendicular from a point $(x_{

Vedclass · Accepted Answer

$\left| \frac{ab}{\sqrt{a^{2} + b^{2}}} \right|$ units

Vedclass · Answer

(B) The given line is $\frac{x}{a} + \frac{y}{b} = 1$,which can be rewritten as $bx + ay - ab = 0$. The length of the perpendicular from a point $(x_{1}, y_{1})$ to the line $Ax + By + C = 0$ is given by $d = \left| \frac{Ax_{1} + By_{1} + C}{\sqrt{A^{2} + B^{2}}} \right|$. Substituting $A = b$,$B = a$,$C = -ab$,$x_{1} = a$,and $y_{1} = b$: $d = \left| \frac{b(a) + a(b) - ab}{\sqrt{b^{2} + a^{2}}} \right| = \left| \frac{ab + ab - ab}{\sqrt{a^{2} + b^{2}}} \right| = \left| \frac{ab}{\sqrt{a^{2} + b^{2}}} \right|$ units.

The length of the perpendicular from the point $P(a, b)$ to the line $\frac{x}{a} + \frac{y}{b} = 1$ is

Similar Questions

If $O$ is the origin and $A$ and $B$ are points on the line $3x - 4y + 25 = 0$ such that $OA = OB = 13$,then the area of $\triangle OAB$ (in sq units) is

The distance between the lines $5x + 3y - 7 = 0$ and $15x + 9y + 14 = 0$ is

The equation of the base of an equilateral triangle is $x + y = 2$ and the vertex is $(2, -1)$. The length of the side of the triangle is

The positions of the points $(3, 4)$ and $(2, -6)$ with respect to the line $3x - 4y = 8$ are:

Let $d_{1}$ and $d_{2}$ be the lengths of the perpendiculars drawn from any point on the line $7x - 9y + 10 = 0$ to the lines $3x + 4y = 5$ and $12x + 5y = 7$, respectively. Then,

Vedclass Test Series

Exam Paper Generator

Online Exam Module