The equations of the sides AB,AC,and BC of a | vedclass

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

Question

(A) The equations of the sides are given as:$AB: x-3y=0$$AC: 3x-y=0$$BC: x+y+4=0$To find the coordinates of vertex $B$,we solve the equations of lines

Vedclass · Accepted Answer

$8$

Vedclass · Answer

(A) The equations of the sides are given as:$AB: x-3y=0$$AC: 3x-y=0$$BC: x+y+4=0$To find the coordinates of vertex $B$,we solve the equations of lines $AB$ and $BC$:$x-3y=0 \Rightarrow x=3y$Substitute $x=3y$ into $x+y+4=0$:$3y+y+4=0$ $\Rightarrow 4y=-4$ $\Rightarrow y=-1$Then $x=3(-1)=-3$.So,the vertex $B$ is $(-3, -1)$.Since the line $3x-y+k=0$ passes through $B(-3, -1)$,we substitute these coordinates into the line equation:$3(-3)-(-1)+k=0$$-9+1+k=0$$-8+k=0 \Rightarrow k=8$.

The equations of the sides $AB$,$AC$,and $BC$ of a $\triangle ABC$ are respectively $x-3y=0$,$3x-y=0$,and $x+y+4=0$. If $P$ and $Q$ are points on the line $3x-y+k=0$ passing through $B$ such that $PB:BQ=1:1$,then $k=$

Similar Questions

If the $x-$intercept of some line $L$ is double that of the line $3x + 4y = 12$ and the $y-$intercept of $L$ is half that of the same line,then the slope of $L$ is

The equation of the straight line which is perpendicular to $y = x$ and passes through $(3, 2)$ is

Find the angle between the $x$-axis and the line joining the points $(3, -1)$ and $(4, -2).$ (in $^{\circ}$)

The length of the perpendicular from the origin to a line is $9$ and the perpendicular makes an angle of $120^{\circ}$ with the positive direction of the $y$-axis. Find the equation of the line.

The $y$-intercept made by the line $x \cos \alpha + y \sin \alpha = a$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module