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 given pair of straight lines is $12x^2 - 20xy + 7y^2 = 0$. Factoring the quadratic expression: $12x^2 - 6xy - 14xy + 7y^2 = 0 \implies 6x(2x -

Vedclass · Accepted Answer

$\left(\frac{8}{3}, \frac{8}{3}\right)$

Vedclass · Answer

(C) The given pair of straight lines is $12x^2 - 20xy + 7y^2 = 0$. Factoring the quadratic expression: $12x^2 - 6xy - 14xy + 7y^2 = 0 \implies 6x(2x - y) - 7y(2x - y) = 0 \implies (6x - 7y)(2x - y) = 0$. The equations of the sides of the triangle are: $L_1: 6x - 7y = 0$ $L_2: 2x - y = 0$ $L_3: 2x - 3y + 4 = 0$ Solving for the vertices: Intersection of $L_1$ and $L_2$: $(0, 0)$. Intersection of $L_1$ and $L_3$: $6x - 7y = 0 \implies x = \frac{7y}{6}$. Substituting into $L_3$: $2(\frac{7y}{6}) - 3y + 4 = 0 \implies \frac{7y}{3} - 3y = -4 \implies -\frac{2y}{3} = -4 \implies y = 6, x = 7$. Vertex is $(7, 6)$. Intersection of $L_2$ and $L_3$: $2x - y = 0 \implies y = 2x$. Substituting into $L_3$: $2x - 3(2x) + 4 = 0 \implies -4x = -4 \implies x = 1, y = 2$. Vertex is $(1, 2)$. The vertices are $(0, 0), (7, 6), (1, 2)$. The centroid is $\left(\frac{0+7+1}{3}, \frac{0+6+2}{3}\right) = \left(\frac{8}{3}, \frac{8}{3}\right)$.

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:

Similar Questions

The binding energy per nucleon of ${O^{16}}$ is $7.97 \, MeV$ and that of ${O^{17}}$ is $7.75 \, MeV$. The energy (in $MeV$) required to remove a neutron from ${O^{17}}$ is

The length of the perpendicular from the point $P(2, -1, 4)$ to the straight line $\frac{x+3}{10} = \frac{y-2}{-7} = \frac{z}{1}$ is:

When two sound sources of the same amplitude but of slightly different frequencies $v_1$ and $v_2$ are sounded simultaneously,the sound one hears has a frequency equal to

When a wire of length $10 \ m$ is subjected to a force of $100 \ N$ along its length,the lateral strain produced is $0.01 \times 10^{-3} \ m$. The Poisson's ratio was found to be $0.4$. If the area of cross-section of the wire is $0.025 \ m^2$,its Young's modulus is

If a root of the equations $x^2+px+q=0$ and $x^2+\alpha x+\beta=0$ is common,then its value will be (where $p \neq \alpha$ and $q \neq \beta$).

Vedclass Test Series

Exam Paper Generator

Online Exam Module