Question:

The pairs of straight lines $x^{2}-3xy+2y^{2}=0 $ and $x^{2}-3xy+2y^{2}+x-2=0$ form a

Updated On: Jun 17, 2022
  • square but not rhombus
  • rhombus
  • parallelogram
  • rectangle but not a square
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The Correct Option is C

Solution and Explanation

Given pair of lines are
$x^{2}-3xy+2y^{2}=0$
and $x^{2}-3xy+2y^{2}+x-2=0$
$\therefore\left(x-2y\right)\left(x-y\right)=0$
and $\left(x-2y+2\right)\left(x-y-1\right)=0$
$\Rightarrow x-2y=0, x-y=0$ and
$x-2y+2=0, x-y-1=0$
The lines $x-2y=0, x-2y+2=0$ and
$x-y=0, x-y-1=0$ are parallel.
Also, angle between $x - 2y = 0$ and
$x - y = 0$ is not 90$^{\circ}$.
$\therefore$ It is a parallelogram.
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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