The point of intersection of the lines repres | vedclass

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

Question

(A) Given the equation of the pair of lines: $f(x, y) = 3x^2 - 11xy + 10y^2 - 7x + 13y + 4 = 0$. To find the point of intersection,we solve the partia

Vedclass · Accepted Answer

$(3, 1)$

Vedclass · Answer

(A) Given the equation of the pair of lines: $f(x, y) = 3x^2 - 11xy + 10y^2 - 7x + 13y + 4 = 0$. To find the point of intersection,we solve the partial derivatives $\frac{\partial f}{\partial x} = 0$ and $\frac{\partial f}{\partial y} = 0$. $\frac{\partial f}{\partial x} = 6x - 11y - 7 = 0$ $\frac{\partial f}{\partial y} = -11x + 20y + 13 = 0$ Solving these simultaneous equations using the cross-multiplication method: $\frac{x}{(-11)(13) - (20)(-7)} = \frac{-y}{(6)(13) - (-11)(-7)} = \frac{1}{(6)(20) - (-11)(-11)}$ $\frac{x}{-143 + 140} = \frac{-y}{78 - 77} = \frac{1}{120 - 121}$ $\frac{x}{-3} = \frac{-y}{1} = \frac{1}{-1}$ $x = 3$ and $y = 1$. Thus,the point of intersection is $(3, 1)$.

The point of intersection of the lines represented by the equation $3x^2 - 11xy + 10y^2 - 7x + 13y + 4 = 0$ is

Similar Questions

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

The straight line passing through $(-1, 1)$ and parallel to the common line of the pairs of lines given by $6x^2 - xy - 12y^2 = 0$ and $15x^2 + 14xy - 8y^2 = 0$ is:

The area of the triangle formed by the line $x + y = 3$ and the angle bisectors of the pair of lines $x^2 - y^2 + 2y = 1$ is:

The line $x - 2y = 0$ will be a bisector of the angle between the lines represented by the equation ${x^2} - 2hxy - 2{y^2} = 0$,if $h = $

The combined equation of lines parallel to the coordinate axes and passing through the point of intersection of lines represented by $x^2 - 6xy + 5y^2 + 10x - 14y + 9 = 0$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module