What is the position of the point (8, -9) wit | vedclass

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

Question

(A) Let the lines be $L_1: 2x + 3y - 4 = 0$ and $L_2: 6x + 9y + 8 = 0$. First,observe that $L_2$ can be written as $3(2x + 3y) + 8 = 0$. Substitute th

Vedclass · Accepted Answer

The point lies outside the lines.

Vedclass · Answer

(A) Let the lines be $L_1: 2x + 3y - 4 = 0$ and $L_2: 6x + 9y + 8 = 0$. First,observe that $L_2$ can be written as $3(2x + 3y) + 8 = 0$. Substitute the point $(8, -9)$ into the expressions: For $L_1(8, -9) = 2(8) + 3(-9) - 4 = 16 - 27 - 4 = -15$. For $L_2(8, -9) = 6(8) + 9(-9) + 8 = 48 - 81 + 8 = -25$. Since both $L_1(8, -9)$ and $L_2(8, -9)$ have the same sign (both are negative),the point $(8, -9)$ lies on the same side of both lines. Therefore,the point lies outside the region between the two parallel lines.

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

Similar Questions

If a line $L$ is perpendicular to the line $5x - y = 1$,and the area of the triangle formed by the line $L$ and the coordinate axes is $5$,then the distance of line $L$ from the line $x + 5y = 0$ is

$A$ point equidistant from the lines $4x + 3y + 10 = 0$,$5x - 12y + 26 = 0$ and $7x + 24y - 50 = 0$ is

The distance between the lines given by $3x + 4y = 9$ and $6x + 8y = 15$ is (in $units$)

The equation of the line passing through $(1, 2)$ and having a distance equal to $7$ units from the point $(8, 9)$ is:

If $p$ and $p'$ are the distances of the origin from the lines $x \sec \alpha + y \csc \alpha = k$ and $x \cos \alpha - y \sin \alpha = k \cos 2\alpha$,then $4p^2 + p'^2$ equals:

Vedclass Test Series

Exam Paper Generator

Online Exam Module