If p is the length of the perpendicular from | vedclass

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

Question

(D) The equation of the line with intercepts $a$ and $b$ on the axes is given by $\frac{x}{a} + \frac{y}{b} = 1$,which can be rewritten as $bx + ay -

Vedclass · Accepted Answer

$\frac{1}{p^2}$

Vedclass · Answer

(D) The equation of the line with intercepts $a$ and $b$ on the axes is given by $\frac{x}{a} + \frac{y}{b} = 1$,which can be rewritten as $bx + ay - ab = 0$.The length of the perpendicular $p$ from the origin $(0, 0)$ to this line is given by the formula $p = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}}$.Substituting the values,we get $p = \frac{|b(0) + a(0) - ab|}{\sqrt{b^2 + a^2}} = \frac{|-ab|}{\sqrt{a^2 + b^2}}$.Squaring both sides,we have $p^2 = \frac{a^2 b^2}{a^2 + b^2}$.Taking the reciprocal,we get $\frac{1}{p^2} = \frac{a^2 + b^2}{a^2 b^2} = \frac{a^2}{a^2 b^2} + \frac{b^2}{a^2 b^2} = \frac{1}{b^2} + \frac{1}{a^2}$.Thus,$\frac{1}{a^2} + \frac{1}{b^2} = \frac{1}{p^2}$.

If $p$ is the length of the perpendicular from the origin to the line whose intercepts on the axes are $a$ and $b$,then $\frac{1}{a^2}+\frac{1}{b^2}=$

Similar Questions

Let $A$ be the point of intersection of the lines $3x + 2y = 14$ and $5x - y = 6$. Let $B$ be the point of intersection of the lines $4x + 3y = 8$ and $6x + y = 5$. The distance of the point $P(5, -2)$ from the line $AB$ is:

The ratio in which the line $3x + 4y + 2 = 0$ divides the distance between $3x + 4y + 5 = 0$ and $3x + 4y - 5 = 0$ is:

The line $L_1$ given by $\frac{x}{p} + \frac{y}{2} = 1$ passes through the point $(5, 0)$. The line $L_2$ given by $\frac{x}{10} + \frac{y}{q} = 1$ is parallel to $L_1$. Find the distance between the lines $L_1$ and $L_2$.

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 distance between the lines $4x + 3y = 11$ and $8x + 6y = 15$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module