The line 3x + 2y = 24 meets the y-axis at A a | vedclass

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

Question

(B) The coordinates of $A$ are found by setting $x=0$ in $3x + 2y = 24$,giving $A(0, 12)$.Setting $y=0$ gives $B(8, 0)$.The midpoint of $AB$ is $M\lef

Vedclass · Accepted Answer

$91$

Vedclass · Answer

(B) The coordinates of $A$ are found by setting $x=0$ in $3x + 2y = 24$,giving $A(0, 12)$.Setting $y=0$ gives $B(8, 0)$.The midpoint of $AB$ is $M\left(\frac{0+8}{2}, \frac{12+0}{2}ight) = (4, 6)$.The slope of $AB$ is $m = \frac{0-12}{8-0} = -\frac{3}{2}$.The slope of the perpendicular bisector is $m' = \frac{2}{3}$.The equation of the perpendicular bisector is $y - 6 = \frac{2}{3}(x - 4)$,which simplifies to $2x - 3y + 10 = 0$.The line through $(0, -1)$ parallel to the $x$-axis is $y = -1$.Substituting $y = -1$ into the bisector equation: $2x - 3(-1) + 10 = 0 \implies 2x + 13 = 0 \implies x = -\frac{13}{2}$.Thus,$C$ is $\left(-\frac{13}{2}, -1ight)$.The area of $	riangle ABC$ with vertices $(0, 12), (8, 0), \left(-\frac{13}{2}, -1ight)$ is:$	ext{Area} = \frac{1}{2} |0(0 - (-1)) + 8(-1 - 12) + (-\frac{13}{2})(12 - 0)|$$= \frac{1}{2} |0 + 8(-13) - \frac{13}{2}(12)|$$= \frac{1}{2} |-104 - 78| = \frac{1}{2} |-182| = 91 \, sq. \, units$.

The line $3x + 2y = 24$ meets the $y$-axis at $A$ and the $x$-axis at $B$. The perpendicular bisector of $AB$ meets the line through $(0, -1)$ parallel to the $x$-axis at $C$. The area of the triangle $ABC$ is ............... $sq. \, units$.

Similar Questions

If a straight line $L$ perpendicular to the line $3x - 4y = 6$ forms a triangle of area $6$ square units with the coordinate axes,then the minimum perpendicular distance from the point $(1, 1)$ to the line $L$ is

$A$ straight line $L \equiv 0$ passing through the point $A=(-5,-4)$ and having slope $\tan \theta$ meets the lines $x+3y+2=0$ and $2x+y+4=0$ respectively at the points $B$ and $C$. If $\frac{100}{AC^2}-\frac{225}{AB^2}=4 \cos 2\theta+\sin 2\theta$,then the slope of the line $L \equiv 0$ is

$A$ straight 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 $5$ square units. The equation of the line $L$ can be

Let a variable line of slope $m > 0$ passing through the point $(4, -9)$ intersect the coordinate axes at the points $A$ and $B$. The minimum value of the sum of the distances of $A$ and $B$ from the origin is

The points $(1, 1)$,$(0, \sec^2 \theta)$,and $(\csc^2 \theta, 0)$ are collinear for:

Vedclass Test Series

Exam Paper Generator

Online Exam Module