The points on the x-axis whose perpendicular | vedclass

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

Question

(A) Let the point on the $x$-axis be $P(h, 0)$.The equation of the line is $\frac{x}{a} + \frac{y}{b} = 1$,which can be written as $bx + ay - ab = 0$.

Vedclass · Accepted Answer

$\left( \frac{a}{b}(b \pm \sqrt{a^2 + b^2}), 0 \right)$

Vedclass · Answer

(A) Let the point on the $x$-axis be $P(h, 0)$.The equation of the line is $\frac{x}{a} + \frac{y}{b} = 1$,which can be written as $bx + ay - ab = 0$.The perpendicular distance from $P(h, 0)$ to the line is given as $a$.Using the distance formula $d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}$,we have:$a = \frac{|bh + a(0) - ab|}{\sqrt{b^2 + a^2}}$$a\sqrt{a^2 + b^2} = |bh - ab|$$bh - ab = \pm a\sqrt{a^2 + b^2}$$bh = ab \pm a\sqrt{a^2 + b^2}$$h = \frac{a}{b}(b \pm \sqrt{a^2 + b^2})$Thus,the points are $\left( \frac{a}{b}(b \pm \sqrt{a^2 + b^2}), 0 \right)$.

The points on the $x$-axis whose perpendicular distance from the line $\frac{x}{a} + \frac{y}{b} = 1$ is $a$ are

Similar Questions

Let $R$ be the point $(3,7)$ and let $P$ and $Q$ be two points on the line $x+y=5$ such that $\triangle PQR$ is an equilateral triangle. Then the area of $\triangle PQR$ is

The position of the point $(8, -9)$ with respect to the lines $2x + 3y - 4 = 0$ and $6x + 9y + 8 = 0$ is

The vertices of $\Delta OBC$ are $(0, 0)$,$(-3, 1)$,and $(-1, -3)$ respectively. $A$ line parallel to $BC$ intersects $OB$ and $OC$ at a distance of $1/2$ from $O$. Find the equation of this line.

On which side of the line $7x + 5y - 9 = 0$ do the points $(3, 4)$ and $(-9, 6)$ lie?

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module