If the area of the triangle formed by the pai | vedclass

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

Question

(D) The given pair of lines is $8x^2 - 6xy + y^2 = 0$.Factoring the quadratic equation: $8x^2 - 4xy - 2xy + y^2 = 0$ $\Rightarrow 4x(2x - y) - y(2x -

Vedclass · Accepted Answer

$28$

Vedclass · Answer

(D) The given pair of lines is $8x^2 - 6xy + y^2 = 0$.Factoring the quadratic equation: $8x^2 - 4xy - 2xy + y^2 = 0$ $\Rightarrow 4x(2x - y) - y(2x - y) = 0$ $\Rightarrow (4x - y)(2x - y) = 0$.Thus,the two lines are $y = 4x$ and $y = 2x$.The third line is $2x + 3y = a$.The intersection points are:$1$. Intersection of $y = 4x$ and $y = 2x$ is $O(0, 0)$.$2$. Intersection of $y = 4x$ and $2x + 3y = a$: $2x + 3(4x) = a$ $\Rightarrow 14x = a$ $\Rightarrow x = a/14, y = 4a/14 = 2a/7$. Point $A(a/14, 2a/7)$.$3$. Intersection of $y = 2x$ and $2x + 3y = a$: $2x + 3(2x) = a$ $\Rightarrow 8x = a$ $\Rightarrow x = a/8, y = 2a/8 = a/4$. Point $B(a/8, a/4)$.The area of the triangle with vertices $(0, 0), (x_1, y_1), (x_2, y_2)$ is $\frac{1}{2} |x_1y_2 - x_2y_1|$.Area $= \frac{1}{2} |(a/14)(a/4) - (a/8)(2a/7)| = \frac{1}{2} |a^2/56 - 2a^2/56| = \frac{1}{2} |-a^2/56| = a^2/112$.Given area is $7$,so $a^2/112 = 7$ $\Rightarrow a^2 = 784$ $\Rightarrow a = 28$ (assuming $a > 0$ for area).

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

Similar Questions

$\int \left( \frac{2 - \sin 2x}{1 - \cos 2x} \right) e^x \, dx$ is equal to

Data Entry - Physics

The number of ways that $8$ beads of different colours can be strung as a necklace is

The minimum value of $27 \tan ^2 \theta + 3 \cot ^2 \theta$ is

$A$ body,thrown upwards with some velocity,reaches a maximum height of $50 \ m$. Another body with double the mass thrown up with double the initial velocity will reach a maximum height of $.......... \ m$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module