Straight Lines
Straight Line
Allen Star Batch
Grade 11
Question:
The centroid of an equilateral triangle is $(0,0)$. If two vertices of the triangle lies on $x + y - 2 = 0$, then:
Area of triangle is $6\sqrt{3}$ square units
vertex not lying on the line is $(-2,-2)$
foot of the perpendicular from $(0,0)$ to the line is $(1,1)$
vertices on the given line are $\left(1 + \sqrt{3}, 1 - \sqrt{3}\right)$ and $\left(1 - \sqrt{3}, 1 + \sqrt{3}\right)$
Step-by-Step Solution
Key Concept: For an equilateral triangle, if one vertex and the opposite side are known, all dimensions can be calculated using distance formulas and trigonometry.
The distance from $(0,0)$ to line $x+y-2=0$ is $GD = \frac{|0+0-2|}{\sqrt{2}} = \sqrt{2}$. Using $\tan\alpha = -1$, we get $\cos\alpha = -\frac{1}{\sqrt{2}}$ and $\sin\alpha = -\frac{1}{\sqrt{2}}$. Point $D$ is at $(1,1)$ and point $A$ is at $(-2,-2)$, giving $AD = 3\sqrt{2}$. Since $AD = \frac{\sqrt{3}}{2}a$ for an equilateral triangle, $a = 2\sqrt{6}$.
Correct Answer: 1,2,3,4