Suppose a triangle of area 27 sq. units is fo | vedclass

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

Question

(A) The given equation is $18x^2 - 9xy + y^2 = 0$.This can be factored as $(y - 3x)(y - 6x) = 0$.Thus,the two lines are $y = 3x$ and $y = 6x$.The inte

Vedclass · Accepted Answer

$(3, 12)$

Vedclass · Answer

(A) The given equation is $18x^2 - 9xy + y^2 = 0$.This can be factored as $(y - 3x)(y - 6x) = 0$.Thus,the two lines are $y = 3x$ and $y = 6x$.The intersection point of these two lines is the origin $(0, 0)$.The intersection points of these lines with the line $y = c$ are $(\frac{c}{3}, c)$ and $(\frac{c}{6}, c)$.The area of the triangle formed by these three points $(0, 0)$,$(\frac{c}{3}, c)$,and $(\frac{c}{6}, c)$ is given by $\frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)| = 27$.$\frac{1}{2} |0(c - c) + \frac{c}{3}(c - 0) + \frac{c}{6}(0 - c)| = 27$.$\frac{1}{2} |\frac{c^2}{3} - \frac{c^2}{6}| = 27$ $\Rightarrow \frac{1}{2} |\frac{c^2}{6}| = 27$ $\Rightarrow c^2 = 324$.Taking $c = 18$ (assuming positive area),the vertices are $(0, 0)$,$(6, 18)$,and $(3, 18)$.The centroid is $(\frac{0 + 6 + 3}{3}, \frac{0 + 18 + 18}{3}) = (3, 12)$.

Suppose a triangle of area $27$ sq. units is formed by $18x^2 - 9xy + y^2 = 0$ and $y = c$. Then the centroid of the triangle is

Similar Questions

The equations to a pair of opposite sides of a parallelogram are $x^2 - 5x + 6 = 0$ and $y^2 - 6y + 5 = 0$. The equations to its diagonals are

If $9x^2-24xy+16y^2+\alpha x+\beta y+6=0$ represents a pair of parallel lines $1$ unit apart and one of those lines passes through $(1,1)$,then $\frac{\alpha}{\beta} = $

The normal to the curve $x^2 + 2xy - 3y^2 = 0$ at the point $(1,1)$ intersects the curve again in which quadrant?

If $2x^2-5xy+2y^2=0$ represents two sides of a triangle whose centroid is $(1,1)$,then the equation of the third side is

If the pair of straight lines given by $Ax^2+2Hxy+By^2=0$ $(H^2>AB)$ forms an equilateral triangle with the line $ax+by+c=0$,then $(A+3B)(3A+B)$ is equal to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module