Question

Q fmd value of K for which $6x^2+11xy-10y^2+x+31y+k=0$ rep. Pair of STL.

Answer 1

-15

Answer 2

The given equation is $6x^2+11xy-10y^2+x+31y+k=0$. This is a general second-degree equation of the form $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$. Comparing the given equation with the general form, we identify the coefficients: $a = 6$

$2h = 11 \implies h = \frac{11}{2}$

$b = -10$

$2g = 1 \implies g = \frac{1}{2}$

$2f = 31 \implies f = \frac{31}{2}$

$c = k$

A general second-degree equation represents a pair of straight lines if and only if the determinant of the matrix of coefficients is zero:

$\begin{vmatrix} a & h & g \\ h & b & f \\ g & f & c \end{vmatrix} = 0$

Expanding the determinant, the condition is $abc + 2fgh - af^2 - bg^2 - ch^2 = 0$.

Substitute the values of the coefficients into this condition: $a=6, b=-10, c=k, h=\frac{11}{2}, g=\frac{1}{2}, f=\frac{31}{2}$

$abc = (6)(-10)(k) = -60k$

$2fgh = 2 \left(\frac{31}{2}\right) \left(\frac{1}{2}\right) \left(\frac{11}{2}\right) = \frac{31 \times 11}{4} = \frac{341}{4}$

$af^2 = 6 \left(\frac{31}{2}\right)^2 = 6 \times \frac{961}{4} = \frac{3 \times 961}{2} = \frac{2883}{2}$

$bg^2 = (-10) \left(\frac{1}{2}\right)^2 = -10 \times \frac{1}{4} = -\frac{10}{4} = -\frac{5}{2}$

$ch^2 = k \left(\frac{11}{2}\right)^2 = k \times \frac{121}{4} = \frac{121k}{4}$

Substitute these terms into the determinant condition: $-60k + \frac{341}{4} - \frac{2883}{2} - \left(-\frac{5}{2}\right) - \frac{121k}{4} = 0$

To eliminate fractions, multiply the entire equation by 4: $4(-60k) + 4\left(\frac{341}{4}\right) - 4\left(\frac{2883}{2}\right) - 4\left(-\frac{5}{2}\right) - 4\left(\frac{121k}{4}\right) = 0$

$-240k + 341 - 2(2883) - 2(-5) - 121k = 0$

$-240k + 341 - 5766 + 10 - 121k = 0$

Group the terms with $k$ and the constant terms: $(-240k - 121k) + (341 + 10 - 5766) = 0$

$-361k + (351 - 5766) = 0$

$-361k - 5415 = 0$

$-361k = 5415$

$k = -\frac{5415}{361}$

To simplify the fraction, we can test for common factors. We know $361 = 19^2$. Divide 5415 by 19: $5415 \div 19 = 285$. Divide 285 by 19: $285 \div 19 = 15$. So, $5415 = 19 \times 285 = 19 \times 19 \times 15 = 361 \times 15$.

Therefore, $k = -\frac{361 \times 15}{361} = -15$.

The value of $k$ for which the given equation represents a pair of straight lines is $-15$.