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

Question

The base of an equilateral triangle is along the line given by 3x + 4y = 9. If a vertex of the triangle is (1, 2), then the length of a side of the triangle is : (A) (2√3/15) (B) (4√3/15) (C) (4√3/5) (D) (2√3/5)

Tardigrade Admin · Accepted Answer

Let BC is base of equilater triangle ABC with side a and A (1, 2)
AD = a sin

6 0^{\circ}

AD is perpendicular distance of PtA from line

3 x + 4 y - 9 = 0

AD

= ∣ ∣ \frac{3 \times 1 + 4 \times 2 - 9}{3 ^{2} + 4 ^{2}} ∣ ∣

a sin

6 0^{\circ} = \frac{2}{5}

a = \frac{4}{5 3} = \frac{4 3}{15}

Q. The base of an equilateral triangle is along the line given by $3 x + 4 y = 9$ . If a vertex of the triangle is $(1, 2)$ , then the length of a side of the triangle is :

$\frac{2 3}{15}$

$\frac{4 3}{15}$

$\frac{4 3}{5}$

$\frac{2 3}{5}$

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

1523​​

1543​​

543​​

523​​

Let BC is base of equilater triangle ABC with side a and A (1, 2) AD = a sin 60∘ AD is perpendicular distance of PtA from line 3x+4y−9=0 AD =∣∣​32+42​3×1+4×2−9​∣∣​ a sin 60∘=52​ a=53​4​=1543​​

Q. The base of an equilateral triangle is along the line given by $3 x + 4 y = 9$ . If a vertex of the triangle is $(1, 2)$ , then the length of a side of the triangle is :

$\frac{2 3}{15}$

$\frac{4 3}{15}$

$\frac{4 3}{5}$

$\frac{2 3}{5}$