Question

The centroid of the triangle formed by the lines x+y=1, 2x+3y=6 and 4x-y+4=0 lines in

• A. I quadrant
• B. II quadrant
• C. III quadrant
• D. IV quadrant

Answer 1

Answer: (B)

II quadrant

We have,

$x+y=1......\left(1\right)$

$2x+3y=6......\left(2\right)$

$4x-y=-4......\left(3\right)$

From equation (1) and (2) to and we get,

$(x+y=1)\times 2$

$2x+3y=6$

$2x+2y=2$

$2x+3y=6$

On subtracting and we get,

$y=4$ put in (1) and we get,

$2x+2y=2$

$2x+2\left(4\right)=2$

$2x+8=2$

$2x=-6$

$x=-3$

Then, $(x_1,y_1)=(-3,4)$

From equation (2) and (3) to, and we get

$(2x+3y=6)\times 2$

$4x-y=-4$

$4x+6y=12$

$4x-y=-4$

On subtracting and we get,

$7y=16$

$y=\frac{16}{7}$

Put the value of y in equation (2) and we get,

$2x+3y=6$

$2x+3\times \frac{16}{7}=6$

$2x+\frac{48}{7}=6$

$2x=6-\frac{48}{7}$

$2x=\frac{42-48}{7}$

$2x=\frac{-6}{7}$

$x=\frac{-3}{7}$

So, $(x_2,y_2)=(\frac{-3}{7},\frac{16}{7})$

From equation (3) and (1) to, and we get,

$4x-y=-4$

$x+y=1$

On adding and we get,

$5x=-3$

$x=\frac{-3}{5}$

Put the value of x in equation (4) and we get,

$4x-y=-4$

$4\times \frac{-3}{5}-y=-4$

$\frac{-12}{5}+4=y$

$y=\frac{-12+20}{5}$

$y=\frac{8}{5}$

So, $(x_3,y_3)=(\frac{-3}{5},\frac{8}{5})$

Then the orthocenter is

$(x,y)=(\frac{x_1+x_2+x_3}{3},\frac{y_1+y_2+y_3}{3})$

$(x,y)=\frac{(-3+\left(\frac{-3}{7}\right)+\left(\frac{-3}{5}\right))}{3},\frac{4+\frac{16}{7}+\frac{8}{5}}{3}$

$(x,y)=(\frac{-105-15-21}{105},\frac{140+80+56}{105})$

$(x,y)=(\frac{-141}{105},\frac{276}{105})$

$(x,y)=(-,+)IIndquadrent$

Hence, this is the answer.