The centroid of the triangle formed by the li | vedclass

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

Question

(A) Let the intersection of lines $x-3y+3=0$ and $x+3y+3=0$ be $A$. Adding the equations: $2x+6=0 \implies x=-3$. Substituting $x=-3$ in $x-3y+3=0$,we

Vedclass · Accepted Answer

$\left(0, -\frac{1}{3}\right)$

Vedclass · Answer

(A) Let the intersection of lines $x-3y+3=0$ and $x+3y+3=0$ be $A$. Adding the equations: $2x+6=0 \implies x=-3$. Substituting $x=-3$ in $x-3y+3=0$,we get $y=0$. So,$A = (-3, 0)$.Let the intersection of lines $x+3y+3=0$ and $x+y-1=0$ be $B$. Subtracting the equations: $(x+3y+3) - (x+y-1) = 0 \implies 2y+4=0 \implies y=-2$. Substituting $y=-2$ in $x+y-1=0$,we get $x-2-1=0 \implies x=3$. So,$B = (3, -2)$.Let the intersection of lines $x-3y+3=0$ and $x+y-1=0$ be $C$. Subtracting the equations: $(x-3y+3) - (x+y-1) = 0 \implies -4y+4=0 \implies y=1$. Substituting $y=1$ in $x+y-1=0$,we get $x+1-1=0 \implies x=0$. So,$C = (0, 1)$.The centroid $G$ of a triangle with vertices $(x_1, y_1)$,$(x_2, y_2)$,and $(x_3, y_3)$ is given by $\left(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}\right)$.$G = \left(\frac{-3+3+0}{3}, \frac{0-2+1}{3}\right) = \left(0, -\frac{1}{3}\right)$.

The centroid of the triangle formed by the lines $x-3y+3=0$,$x+3y+3=0$,and $x+y-1=0$ is

Similar Questions

The two vertices of a triangle are $(-1, 4)$ and $(5, 2)$. If its centroid is $(0, -3)$,find the third vertex.

Let $A(0,0), B(3,0), C(0,-4)$ be the vertices of $\triangle ABC$. Then the coordinates of the incentre of $\triangle ABC$ are:

If the coordinates of the vertices of the triangle $ABC$ are $A(-1, 6)$,$B(-3, -9)$,and $C(5, -8)$,then the equation of the median through $C$ is

The coordinates of the orthocentre of the triangle bounded by the lines $4x - 7y + 10 = 0$,$x + y = 5$,and $7x + 4y = 15$ are:

Let $ABC$ be an equilateral triangle with orthocenter at the origin and the side $BC$ on the line $x+2\sqrt{2}y=4$. If the coordinates of the vertex $A$ are $(\alpha, \beta)$, then the greatest integer less than or equal to $|\alpha+\sqrt{2}\beta|$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module