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) To find the vertices of the triangle,we solve the equations of the lines in pairs:$1$. Intersection of $x+y-1=0$ and $x-y-1=0$:Adding the two equa

Vedclass · Accepted Answer

$\left(\frac{4}{3}, 1\right)$

Vedclass · Answer

(A) To find the vertices of the triangle,we solve the equations of the lines in pairs:$1$. Intersection of $x+y-1=0$ and $x-y-1=0$:Adding the two equations gives $2x-2=0$,so $x=1$. Substituting $x=1$ into $x+y-1=0$ gives $y=0$. Thus,vertex $A$ is $(1, 0)$.$2$. Intersection of $x-y-1=0$ and $x-3y+3=0$:Subtracting the second from the first gives $2y-4=0$,so $y=2$. Substituting $y=2$ into $x-y-1=0$ gives $x=3$. Thus,vertex $B$ is $(3, 2)$.$3$. Intersection of $x-3y+3=0$ and $x+y-1=0$:Subtracting the second from the first gives $-4y+4=0$,so $y=1$. Substituting $y=1$ into $x+y-1=0$ gives $x=0$. Thus,vertex $C$ is $(0, 1)$.The centroid $(G)$ of a triangle with vertices $(x_1, y_1)$,$(x_2, y_2)$,and $(x_3, y_3)$ is given by $\left(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}\right)$.Substituting the coordinates of $A, B, C$:$G = \left(\frac{1+3+0}{3}, \frac{0+2+1}{3}\right) = \left(\frac{4}{3}, \frac{3}{3}\right) = \left(\frac{4}{3}, 1\right)$.

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

Similar Questions

If the points $(1, 1)$,$(-1, -1)$ and $(-\sqrt{3}, k)$ are vertices of an equilateral triangle,then the value of $k$ will be:

Let $(\alpha, \beta)$ be the centroid of the triangle formed by the lines $15x - y = 82$,$6x - 5y = -4$,and $9x + 4y = 17$. Then $\alpha + 2\beta$ and $2\alpha - \beta$ are the roots of the equation $...........$.

In a rectangle $ABCD$,points $X$ and $Y$ are the mid-points of $AD$ and $DC$,respectively. Lines $BX$ and $CD$ when extended intersect at $E$,lines $BY$ and $AD$ when extended intersect at $F$. If the area of $ABCD$ is $60$,then the area of $\triangle BEF$ is

The area of the triangle bounded by the straight line $ax + by + c = 0$ $(a, b, c \neq 0)$ and the coordinate axes is

The set of values that $\beta$ can assume so that the point $(0, \beta)$ lies on or inside the triangle formed by the lines $3x+y+2=0$,$2x-3y+5=0$,and $x+4y-14=0$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module