The base of an equilateral triangle isalong the line given by 3x + 4y

1.	The base of an equilateral triangle isalong the line given by 3x + 4y =9. If avertex of the triangle is (1, 2), then thelength of a side of the triangle is :
Answer» Answer: The length of each side of the TRIANGLE is 4√3 / 15. Step-by-step explanation: Given data: Let the base BC of an equilateral ∆ ABC be along the line given by 3X + 4y = 9 Coordinates of Vertex A of the triangle is (1,2) To find: length of a side of the triangle. Considering the length of each side of equilateral triangle be “a”. Let us draw a PERPENDICULAR from A to BC at D such that BD = DC = a/2 Therefore, by Pythagoras theorem, Length of AD = √[AB² - BD²] = √[a²-(a/2)² = [√3] a / 2 …. (i) Also, The SHORTEST distance, AD of A(1,2) from 3x+4y=9 is given by AD = \| [ax1+by1+c] / [√[a²-b²] \| Or, AD = \| [(31)+(42)-9] / [√[3²+4²] \| Or, AD = \| 2/5 \| ….. (II) Thus, equating (i) & (ii), we get [√3] a / 2 = 2/5 Or, a = 4/(5√3) = [4/(5√3)] * [√3/√3] = 4√3 / 15

The base of an equilateral triangle isalong the line given by 3x + 4y =9. If avertex of the triangle is (1, 2), then thelength of a side of the triangle is :

Answer»

Answer:

The length of each side of the TRIANGLE is 4√3 / 15.

Step-by-step explanation:

Given data:

Let the base BC of an equilateral ∆ ABC be along the line given by 3X + 4y = 9

Coordinates of Vertex A of the triangle is (1,2)

To find: length of a side of the triangle.

Considering the length of each side of equilateral triangle be “a”.

Let us draw a PERPENDICULAR from A to BC at D such that BD = DC = a/2

Therefore, by Pythagoras theorem,

Length of AD = √[AB² - BD²] = √[a²-(a/2)² = [√3] a / 2 …. (i)

Also,

The SHORTEST distance, AD of A(1,2) from 3x+4y=9 is given by

AD = | [ax1+by1+c] / [√[a²-b²] |

Or, AD = | [(3*1)+(4*2)-9] / [√[3²+4²] |

Or, AD = | 2/5 | ….. (II)

Thus, equating (i) & (ii), we get

[√3] a / 2 = 2/5

Or, a = 4/(5*√3) = [4/(5*√3)] * [√3/√3] = 4√3 / 15