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

If a point $(a, a)$ lies between the lines $|x+y|=4$,then

If $x + 2y = 3$ is a line and $A(-1, 3)$,$B(2, -3)$,and $C(4, 9)$ are three points,then:

Two points $P(a, 2)$ and $Q(1, b)$ lie on either side of the line $2x - 3y + 1 = 0$. If $P$ is the point of intersection of the lines $4x + 3y + k = 0$ and $3x + 4y + k = 0$,then the range of $b$ is

If $O$ is the origin and $P, Q$ are points on the line $3x + 4y + 15 = 0$ such that $OP = OQ = 9$, then the area of $\triangle OPQ$ is (in $\sqrt{2}$)

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module