Find the point of intersection of the lines r | vedclass

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

Question

(D) Let $\varphi(x, y) = 2x^2 - 7xy - 4y^2 - x + 22y - 10 = 0$. To find the point of intersection,we take the partial derivatives with respect to $x$

Vedclass · Accepted Answer

$(2, 1)$

Vedclass · Answer

(D) Let $\varphi(x, y) = 2x^2 - 7xy - 4y^2 - x + 22y - 10 = 0$. To find the point of intersection,we take the partial derivatives with respect to $x$ and $y$ and set them to zero. $\frac{\partial \varphi}{\partial x} = 4x - 7y - 1 = 0$  (Equation $1$) $\frac{\partial \varphi}{\partial y} = -7x - 8y + 22 = 0$  (Equation $2$) From Equation $1$,$4x = 7y + 1 \implies x = \frac{7y + 1}{4}$. Substituting this into Equation $2$: $-7(\frac{7y + 1}{4}) - 8y + 22 = 0$ $-49y - 7 - 32y + 88 = 0$ $-81y + 81 = 0 \implies y = 1$. Substituting $y = 1$ into $4x - 7(1) - 1 = 0$: $4x - 8 = 0 \implies x = 2$. Thus,the point of intersection is $(2, 1)$.

Find the point of intersection of the lines represented by $2x^2 - 7xy - 4y^2 - x + 22y - 10 = 0$.

Similar Questions

The joint equation of the bisectors of the angle between the lines represented by $3x^{2} + 2xy - y^{2} = 0$ is:

The joint equation of the bisectors of the angles between the lines $x=5$ and $y=3$ is

The area (in sq. units) of the triangle formed by the straight line $x+y=3$ and the angular bisectors of the pair of straight lines $x^2-y^2+2y=1$ is

If the combined equation of the angle bisectors of the pair of lines $x^2 - 2pxy - y^2 = 0$ is $x^2 - 2qxy - y^2 = 0$, then which of the following is true?

The combined equation of the bisectors of the angle between the lines represented by $({x^2} + {y^2})\sqrt{3} = 4xy$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module