Question

An equilateral triangle is constructed between two parallel lines √3.x +y-6=0 and √3x +y+9=0
with base on one line and vertex on the other. Then the area of triangle is
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Answer 1

Let's first analyze the question.

There is no mention of dimensions of the triangle, We don't have vertices of the traingle, We don't have equations of its side.

We are given only, The parallel lines between which the triangle is constructed

To find the area of the triangle, I will find the distance between the parallel lines which will be our altitude.

Distance between two parallel lines ax + by + c =0, ax + by + k = 0 is

$\frac{c - k}{ \sqrt{ {a}^{2} + {b}^{2} } }$

So,

Given lines are √3x + y-6=0 and √3x +y+9=0

Distance between them

$= \frac{9 - 6}{ \sqrt{ ({ \sqrt{3}) }^{2} + ( {1}) ^{2} } } = \frac{3}{ \sqrt{4} } = \frac{3}{2}$

We know that,

Altitude of a triangle = h

$\frac{ \sqrt{3} }{2} \times side$

So,

Side =

$\frac{2h}{ \sqrt{3} }$

Now,

Area of triangle

$= \frac{1}{2} \times altitude \times side \\ \\ = \frac{1}{2} \times \: h \times \frac{2h}{ \sqrt{3} } \\ \\ = \frac{ {h}^{2} }{ \sqrt{3} } \\ \\ = \frac{( \frac{3}{2} )^{2} }{ \sqrt{3} } \\ \\ = \frac{9}{4 \sqrt{3} } = \frac{3 \sqrt{3} }{4}$

Therefore, Area of traingle is

$\frac{3 \sqrt{3} }{4}$