The pair of straight lines $ {{x}^{2}}-3{{y}^{2}}=0 $ and the line $ x=1 $ form a triangle which is
- right angled
- isosceles
- scalene
- equilateral
The Correct Option is D
Solution and Explanation
and $ x=1 $
$ \Rightarrow $ $ (x-\sqrt{3}y)=0,\,(x+\sqrt{3}\,y)=0 $
and $ x=1 $
Here, $ AB=\sqrt{{{(0-1)}^{2}}+{{\left( 0-\frac{1}{\sqrt{3}} \right)}^{2}}} $
$ =\sqrt{1+\frac{1}{3}}=\frac{2}{\sqrt{3}} $
$ BC=\sqrt{{{(1-1)}^{2}}+{{\left( \frac{1}{\sqrt{3}}+\frac{1}{\sqrt{3}} \right)}^{2}}} $
$ =\sqrt{{{\left( \frac{2}{\sqrt{3}} \right)}^{2}}}=\frac{2}{\sqrt{3}} $
$ CA=\sqrt{{{(1-0)}^{2}}+{{\left( -\frac{1}{\sqrt{3}}-0 \right)}^{2}}} $
$ =\sqrt{1+\frac{1}{3}}=\frac{2}{\sqrt{3}} $
Hence, all the three sides of $ \Delta \,ABC $ are equal. Therefore, the triangle is equilateral.
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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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