The centroid of the triangle formed by the pa | vedclass

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

Question

(C) The pair of straight lines is given by $12x^2 - 20xy + 7y^2 = 0$. Factorizing this,we get: $12x^2 - 14xy - 6xy + 7y^2 = 0$ $2x(6x - 7y) - y(6x - 7

Vedclass · Accepted Answer

$8$

Vedclass · Answer

(C) The pair of straight lines is given by $12x^2 - 20xy + 7y^2 = 0$. Factorizing this,we get: $12x^2 - 14xy - 6xy + 7y^2 = 0$ $2x(6x - 7y) - y(6x - 7y) = 0$ $(2x - y)(6x - 7y) = 0$ So,the two lines are $2x - y = 0$ and $6x - 7y = 0$. The third line is $2x - 3y + 4 = 0$. To find the vertices of the triangle,we solve the lines in pairs: $1$. Intersection of $2x - y = 0$ and $6x - 7y = 0$: Solving these,we get $x = 0, y = 0$. So,vertex $A = (0, 0)$. $2$. Intersection of $2x - y = 0$ and $2x - 3y + 4 = 0$: Subtracting the equations: $(2x - y) - (2x - 3y + 4) = 0$ $\Rightarrow 2y - 4 = 0$ $\Rightarrow y = 2$. Substituting $y = 2$ in $2x - y = 0$,we get $x = 1$. So,vertex $B = (1, 2)$. $3$. Intersection of $6x - 7y = 0$ and $2x - 3y + 4 = 0$: From $6x - 7y = 0$,$x = \frac{7y}{6}$. Substituting in $2x - 3y + 4 = 0$: $2(\frac{7y}{6}) - 3y + 4 = 0$ $\Rightarrow \frac{7y}{3} - 3y + 4 = 0$ $\Rightarrow -\frac{2y}{3} = -4$ $\Rightarrow y = 6$. Then $x = \frac{7(6)}{6} = 7$. So,vertex $C = (7, 6)$. The centroid $(\alpha, \beta)$ is given by $(\frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3})$: $\alpha = \frac{0 + 1 + 7}{3} = \frac{8}{3}$ $\beta = \frac{0 + 2 + 6}{3} = \frac{8}{3}$ Thus,$\alpha + 2\beta = \frac{8}{3} + 2(\frac{8}{3}) = \frac{8 + 16}{3} = \frac{24}{3} = 8$.

The centroid of the triangle formed by the pair of straight lines $12x^2 - 20xy + 7y^2 = 0$ and the line $2x - 3y + 4 = 0$ is $(\alpha, \beta)$. Then,$\alpha + 2\beta =$

Similar Questions

The combined equation of the diagonals of the parallelogram formed by the lines $(7 x^2-4 x y+8 y^2)^2+(4 x-8 y-32)(7 x^2-4 x y+8 y^2)=0$ is

If the area of the triangle formed by the pair of lines $8x^2-6xy+y^2=0$ and the line $2x+3y=a$ is $7$,then $a$ is equal to

Suppose that the sides passing through the vertex $(\alpha, \beta)$ of a triangle are bisected at right angles by the lines $y^2-8xy-9x^2=0$. Then,the centroid of the triangle is

The triangle formed by the lines $2x^2 - 3xy - 2y^2 = 0$ and $3x - y = 7$ is...

If the lines $x^2-4xy+y^2=0$ and $x+y=10$ contain the sides of an equilateral triangle,then the area of the equilateral triangle is

Vedclass Test Series

Exam Paper Generator

Online Exam Module