A straight line x/a - y/b = 1 passes through | vedclass

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

Question

(B) The line $x/a - y/b = 1$ passes through $(8, 6)$,so $8/a - 6/b = 1$ . . . $(i)$.The intercepts are $(a, 0)$ and $(0, -b)$. The area of the triangl

Vedclass · Accepted Answer

$3 x - 2 y - 12 = 0$ and $3 x - 8 y + 24 = 0$

Vedclass · Answer

(B) The line $x/a - y/b = 1$ passes through $(8, 6)$,so $8/a - 6/b = 1$ . . . $(i)$.The intercepts are $(a, 0)$ and $(0, -b)$. The area of the triangle formed with the axes is $1/2 	imes |a| 	imes |-b| = 12$,so $|ab| = 24$,which means $ab = 24$ or $ab = -24$.Case $1$: $b = 24/a$. Substituting into $(i)$: $8/a - 6/(24/a) = 1 \implies 8/a - a/4 = 1 \implies 32 - a^2 = 4a \implies a^2 + 4a - 32 = 0 \implies (a+8)(a-4) = 0$.If $a = 4$,$b = 6$. The line is $x/4 - y/6 = 1 \implies 3x - 2y = 12 \implies 3x - 2y - 12 = 0$.If $a = -8$,$b = -3$. The line is $x/(-8) - y/(-3) = 1 \implies -x/8 + y/3 = 1 \implies -3x + 8y = 24 \implies 3x - 8y + 24 = 0$.Thus,the equations are $3x - 2y - 12 = 0$ and $3x - 8y + 24 = 0$.

$A$ straight line $x/a - y/b = 1$ passes through the point $(8, 6)$ and cuts a triangle of area $12 \text{ sq units}$ from the axes of coordinates. The equations of the straight lines are

Similar Questions

$A$ line passes through the origin and is perpendicular to two given lines $2x + y + 6 = 0$ and $4x + 2y - 9 = 0$. What is the ratio in which the origin divides the segment formed by the intersection points of this line with the two given lines?

Consider the family of lines $x(a + b) + y = 1$,where $a, b$ and $c$ are the roots of the equation $x^3 - 3x^2 + x + \lambda = 0$ such that $c \in [1, 2]$. If the given family of lines makes a triangle of area $A$ with the coordinate axes,then the maximum value of $A$ (in sq. units) will be:

If $a, b, c$ are in $A.P.$,then the straight line $ax + by + c = 0$ will always pass through the point

$A$ straight line $x-2y-4=0$ is shifted parallel to it by $3$ units away from the origin and then rotated by an angle of $30^{\circ}$ in the anti-clockwise direction. If the slope of the new line formed is $m$,then the integral part of $m$ is

If the two lines $x + (a - 1)y = 1$ and $2x + a^2y = 1$ $(a \in R - \{0, 1\})$ are perpendicular,then the distance of their point of intersection from the origin is

Vedclass Test Series

Exam Paper Generator

Online Exam Module