Question

COORDINATE GEOMETRY
CLASS 10
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•Please answer the question in the picture.
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THANKS...​ ​

Answer 1

Corrected Question: Show that the points (a, a), (-a, -a) & (-√3a, √3a) are the vertices of an equilateral triangle.

Solution:

For a triangle to be equal, the length of all it's sides must be equal.

Let (a, a), (-a, -a) & (-√3a, √3a) be named A, B & C.

So, we have to prove AB = BC = CA by using the Distance formula.

Finding AB's Distance:

A(a, a)

B(-a, -a)

$\sf \Longrightarrow AB = \sqrt{\big(x_1 - x_2 \big)^2 + \big(y_1 - y_2 \big)^2}$

$\sf \Longrightarrow AB = \sqrt{\big(a - (-a) \big)^2 + \big(a - (-a)\big)^2}$

$\sf \Longrightarrow AB = \sqrt{\big(a + a\big)^2 + \big(a + a\big)^2}$

$\sf \Longrightarrow AB = \sqrt{\big(2a\big)^2 + \big(2a\big)^2}$

$\sf \Longrightarrow AB = \sqrt{4a^2 + 4a^2}$

$\sf \Longrightarrow AB = \sqrt{8a^2}$

$\sf \Longrightarrow AB = 2\sqrt{2}a$

Finding BC's Distance:

B(-a, -a)

C(-√3a, √3a)

$\sf \Longrightarrow BC = \sqrt{\big(x_1 - x_2 \big)^2 + \big(y_1 - y_2 \big)^2}$

$\sf \Longrightarrow BC = \sqrt{\big(-a - (-\sqrt{3}a) \big)^2 + \big(-a - \sqrt{3}a \big)^2}$

$\sf \Longrightarrow BC = \sqrt{\big(-a + \sqrt{3}a\big)^2 + \big(-a - \sqrt{3}a \big)^2}$

$\sf \Longrightarrow BC = \sqrt{\big(-a\big)^2 + \big(\sqrt{3}a\big)^2 + 2\big(-a\big) \big(\sqrt{3a}\big) + \big(-a - \sqrt{3}a \big)^2}$

$\sf \Longrightarrow BC = \sqrt{a^2 + 3a^2 - 2\sqrt{3}a^2 + \big(-a - \sqrt{3}a \big)^2}$

$\sf \Longrightarrow BC = \sqrt{4a^2 - 2\sqrt{3}a^2 + \big(-a - \sqrt{3}a \big)^2}$

$\sf \Longrightarrow BC = \sqrt{4a^2 - 2\sqrt{3}a^2 + \big(-a\big)^2 + \big(\sqrt{3}a\big)^2 - 2\big(-a\big) \big(\sqrt{3}a\big)}$

$\sf \Longrightarrow BC = \sqrt{4a^2 - 2\sqrt{3}a^2 + a^2 + 3a^2 + 2\sqrt{3}a^2}$

$\sf \Longrightarrow BC = \sqrt{4a^2 + a^2 + 3a^2}$

$\sf \Longrightarrow BC = \sqrt{8a^2}$

$\sf \Longrightarrow BC = 2\sqrt{2}a$

Finding AC's Distance:

A(a, a)

C(-√3a, √3a)

$\sf \Longrightarrow AC = \sqrt{\big(x_1 - x_2 \big)^2 + \big(y_1 - y_2 \big)^2}$

$\sf \Longrightarrow AC = \sqrt{\big(a - (-\sqrt{3}a ) \big)^2 + \big(a - \sqrt{3}a\big)^2}$

$\sf \Longrightarrow AC = \sqrt{\big(a + \sqrt{3}a \big)^2 + \big(a - \sqrt{3}a\big)^2}$

$\sf \Longrightarrow AC = \sqrt{\big(a\big)^2 + \big(\sqrt{3}a\big)^2 + 2 \big(a\big)\big(\sqrt{3}a\big) + \big(a\big)^2 + \big(\sqrt{3}a\big)^2 - 2 \big(a\big)\big(\sqrt{3}a\big)}$

$\sf \Longrightarrow AC = \sqrt{a^2 + 3a^2 + 2\sqrt{3}a^2 + a^2 + 3a^2 - 2\sqrt{3}a^2}$

$\sf \Longrightarrow AC = \sqrt{a^2 + 3a^2 + a^2 + 3a^2}$

$\sf \Longrightarrow AC = \sqrt{8a^2}$

$\sf \Longrightarrow AC = 2\sqrt{2}a$

AB = BC = AB = 2√2a

∴ ABC is an equilateral triangle.

Answer 2

Step-by-step explanation:

Your question has √3 in place of those boxes.

Here, using distance formula:,

side = √{ ( a + a )² + ( a + a )²}

side = √{ (2a)² + (2a)² } = √{ 4a² + 4a² }

side = √(8a)²

side² = 8a²

If this is a equilateral triangle, area should be √(3)/4 side² = √3/4 * 8a² Area should be 2√3 a².

Area of this triangle:-

= > 1/2 | a( - a - √3a ) - a( √3a - a ) - √3a( a + a ) |

= > 1/2 | - a² - √3a² - √3a² + a² - √3a² - √3a² |

= > 1/2 | - 4√3 a² |

= > 1/2 * 4√3 a²

= > 2√3 a²

As area is as same as given above, this is an equilateral triangle.