The value of $k$ for which the equation
$ {{x}^{2}}-4xy-{{y}^{2}}+6x+2y+k=0 $ represents a pair of straight lines is
- $ k=4 $
- $ k=-1 $
- $ k=\frac{-4}{5} $
- $ k=\frac{-22}{5} $
The Correct Option is C
Solution and Explanation
represent the point of straight line, if $ \Delta =0, $
where $ \Delta =abc+2fgh-a{{f}^{2}}-b{{g}^{2}}-c{{h}^{2}}=0 $ ..(ii)
Comparing E (i) with the following equation
$ a{{x}^{2}}+2hxy+b{{y}^{2}}+2gx+2fy+c=0 $
We get, $ a=1,\,h=-2,\,b=-1,\,\,g=3,\,f=1,\,c=k $
From E (ii), $ (1)\,(-1)\,(k)+2(1)\,(3)\,(-2)-(1)\,{{(1)}^{2}}-(-1) $
$ {{(3)}^{2}}-(k)\,{{(-2)}^{2}}=0 $
$ \Rightarrow $ $ -k-12-1+9-4k=0 $
$ \Rightarrow $ $ -5k=4\,\,\Rightarrow \,k=-4/5 $
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Concepts Used:
Straight lines
A straight line is a line having the shortest distance between two points.
A straight line can be represented as an equation in various forms, as show in the image below:
The following are the many forms of the equation of the line that are presented in straight line-
1. Slope – Point Form
Assume P0(x0, y0) is a fixed point on a non-vertical line L with m as its slope. If P (x, y) is an arbitrary point on L, then the point (x, y) lies on the line with slope m through the fixed point (x0, y0) if and only if its coordinates fulfil the equation below.
y – y0 = m (x – x0)
2. Two – Point Form
Let's look at the line. L crosses between two places. P1(x1, y1) and P2(x2, y2) are general points on L, while P (x, y) is a general point on L. As a result, the three points P1, P2, and P are collinear, and it becomes
The slope of P2P = The slope of P1P2 , i.e.
\(\frac{y-y_1}{x-x_1} = \frac{y_2-y_1}{x_2-x_1}\)
Hence, the equation becomes:
y - y1 =\( \frac{y_2-y_1}{x_2-x_1} (x-x1)\)
3. Slope-Intercept Form
Assume that a line L with slope m intersects the y-axis at a distance c from the origin, and that the distance c is referred to as the line L's y-intercept. As a result, the coordinates of the spot on the y-axis where the line intersects are (0, c). As a result, the slope of the line L is m, and it passes through a fixed point (0, c). The equation of the line L thus obtained from the slope – point form is given by
y – c =m( x - 0 )
As a result, the point (x, y) on the line with slope m and y-intercept c lies on the line, if and only if
y = m x +c
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