Question

show that the point A(a,a)B(-a,-a) and c(-√3a,√3a) are the vertices of an equitatiral triangle​

Answer 1

$\large\underline{\sf{Solution-}}$

Given Coordinates of triangle ABC are

Coordinates of A (a,a)

Coordinates of B (-a,-a)

Coordinates of C (-√3a,√3a)

So, in order to show that these vertices form equilateral triangle, we have to show that AB = BC = CA

We know,

Distance Formula :-

Let is consider a line segment joining the points A (x₁ , y₁ ) and B (x₂ , y₂), then distance between AB is

$\red{\boxed{ \quad \bf{ \: AB = \sqrt{(x_2 - x_1)^{2} + {(y_2 - y_1)}^{2} } \quad}}}$

Now,

Let find the distance between A and B.

Coordinates of A (a,a)

Coordinates of B (-a,-a)

So,

$\rm :\longmapsto\:AB = \sqrt{ {(a + a)}^{2} + {(a + a)}^{2} }$

$\rm :\longmapsto\:AB = \sqrt{ {(2a)}^{2} + {(2a)}^{2} }$

$\rm :\longmapsto\:AB = \sqrt{ {4a}^{2} + {4a}^{2} }$

$\rm :\longmapsto\:AB = \sqrt{ {8a}^{2} }$

$\rm :\longmapsto\:AB = 2 \sqrt{ {2}} a \: units$

Now,

Let find the distance between B and C.

Coordinates of B (-a,-a)

Coordinates of C (-√3a,√3a)

So,

$\rm :\longmapsto\:BC = \sqrt{ {( - \sqrt{3}a + a) }^{2} + {( \sqrt{3}a + a) }^{2} }$

We know,

$\red{\boxed{ \bf{ \: {(x + y)}^{2} + {(x - y)}^{2} = 2( {x}^{2} + {y}^{2})}}}$

So we get

$\rm :\longmapsto\:BC = \sqrt{2\bigg( {a}^{2} + {( \sqrt{3} a)}^{2} \bigg) }$

$\rm :\longmapsto\:BC = \sqrt{2\bigg( {a}^{2} + {3a}^{2} \bigg) }$

$\rm :\longmapsto\:BC = \sqrt{2\bigg({4a}^{2} \bigg) }$

$\rm :\longmapsto\:BC = \sqrt{8 {a}^{2} }$

$\rm :\longmapsto\:BC = 2 \sqrt{2} \: a \: units$

Now,

Lets find the distance between A and C,

Coordinates of A (a,a)

Coordinates of C (-√3a,√3a)

$\rm :\longmapsto\:AC = \sqrt{ {( - \sqrt{3}a - a) }^{2} + {( \sqrt{3}a - a) }^{2} }$

$\rm :\longmapsto\:AC = \sqrt{ {(\sqrt{3}a + a) }^{2} + {( \sqrt{3}a - a) }^{2} }$

We know ,

$\red{\boxed{ \bf{ \: {(x + y)}^{2} + {(x - y)}^{2} = 2( {x}^{2} + {y}^{2})}}}$

So, using this, we get

$\rm :\longmapsto\:AC = \sqrt{2\bigg( {a}^{2} + {( \sqrt{3} a)}^{2} \bigg) }$

$\rm :\longmapsto\:AC = \sqrt{2\bigg( {a}^{2} + {3a}^{2} \bigg) }$

$\rm :\longmapsto\:AC = \sqrt{2\bigg({4a}^{2} \bigg) }$

$\rm :\longmapsto\:AC = \sqrt{8 {a}^{2} }$

$\rm :\longmapsto\:AC = 2 \sqrt{2} \: a \: units$

Hence, we concluded that

$\bf :\longmapsto\:AB = BC = AC = 2 \sqrt{2} \: a \: units$

$\bf\implies \:A,B,C \: are \: the \: vertices \: of \: equilateral \: \triangle$

Additional Information :-

Section Formula :-

Let us consider a line segment joining the points A (x₁ , y₁ ) and B (x₂ , y₂) and Let C (x, y) be any point on AB which divides AB internally in the ratio m : n, then coordinates of C is

$\red{\boxed{ \quad \bf{ \: (x,y) = \bigg(\dfrac{mx_2 + nx_1}{m + n}, \dfrac{my_2 + ny_1}{m + n} \bigg) \quad}}}$

Midpoint Formula :-

Let us consider a line segment joining the points A (x₁ , y₁ ) and B (x₂ , y₂) and Let C (x, y) be the mid - point of AB, then coordinates of C is

$\red{\boxed{ \quad \bf{ \: (x,y) = \bigg(\dfrac{x_1+ x_2}{2}, \dfrac{y_1 + y_2}{2} \bigg) \quad}}}$