Question

Find the valuc oſk, if thc equation 2x + kxy - 6y +3x + y +1 = 0 represents a pair of straight
lines. Find the point of intersection of the lines and the angle between the straight lines for this
value of'k'.​

Answer 1

Step-by-step explanation:

Given:

$2 {x}^{2} + kxy - 6 {y}^{2} + 3x + y + 1 = 0$

To find: Value of k?

Solution:

Tip: Compare the given equation with standard equation ,find coefficient and put in condition of straight line.

Standard equation:

$\boxed{a {x}^{2} + 2hxy + b {y}^{2} + 2gx + 2fy + c = 0 }$

compare given equation and find coefficients

$a = 2 \\ h = \frac{k}{2} \\ b = - 6 \\ g = \frac{3}{2} \\ f = \frac{1}{2} \\ c = 1$

The equation represent pair of straight line,so

$abc + 2fgh - a {f}^{2} - b {g}^{2} - c {h}^{2} = 0 \\ \\ put \: all \: the \: values \: here \\ \\ (2)( - 6)(1) + 2( \frac{1}{2})( \frac{3}{2}) ( \frac{k}{2})- (2) {( \frac{1}{2} })^{2} - ( - 6){ (\frac{3}{2} )}^{2} - (1){( \frac{k}{2}) }^{2} = 0 \\ \\ - 12 + \frac{3k}{4} - \frac{1}{2} + \frac{27}{2} - \frac{ {k}^{2} }{4} = 0 \\ \\ \frac{ - 48 + 3k - 2 + 54 - {k}^{2} }{4} = 0 \\ \\ - {k}^{2} + 3k + 4 = 0 \\ \\ or \\ \\ {k}^{2} - 3k - 4 = 0$

Solve by factorisation

${k}^{2} - 3k - 4 = 0 \\ \\ {k}^{2} - 4k + k - 4 = 0 \\ \\ k(k - 4) + 1(k - 4) = 0 \\ \\ (k - 4)(k + 1) = 0 \\ \\ k - 4 = 0 \\ \\ k = 4 \\ \\ or \\ \\ k + 1 = 0 \\ \\ k = - 1$

Thus,

Values of k should be 4 and -1.

so that,given equation represent straight line.

Hope it helps you.

To learn more on brainly:

1) Values of 'k' so that equation 16^2 + 24xy + ly^2 + kx - 12y - 21 =0, represents a

pair of parallel lines ...

https://brainly.in/question/21375638

Answer 2

Answer:

Values of k should be 4 and -1.