Math, asked by GouthamGS3902, 9 months ago

If A(4,3) B(2,1),C(6,-1) and D(8,1) be four points, shows that AC and AD bisect eachother

Answers

Answered by Tomboyish44
7

Correction: If A(4, 3) B(2, 1), C(6, -1) and D(8, 1) be four points, shows that AC and BD bisect each other

We have four points A(4, 3), B(2, 1), C(6, -1) and D(8,1).

And, we have to prove AC & BD bisect each other, i.e, we have to prove that AO = CO & BO = OD.

For any two lines to bisect each other, they must have the same midpoint.

So we'll find the midpoints of both AC & BD, compare them and see if they're the same coordinates.

Mid-point of AC:

Let the midpoint of AC be O.

x₁ ⇒ 4

x₂ ⇒ 6

y₁ ⇒ 3

y₂ ⇒ -1

\sf \Longrightarrow O(x, y) = \Bigg\{ \dfrac{x_1 + x_2}{2} , \dfrac{y_1 + y_2}{2} \Bigg\}

\sf \Longrightarrow O(x, y) = \Bigg\{ \dfrac{4 + 6}{2} , \dfrac{3 - 1}{2} \Bigg\}

\sf \Longrightarrow O(x, y) = \Bigg\{ \dfrac{10}{2} , \dfrac{2}{2} \Bigg\}

\sf \Longrightarrow O(x, y) = \Big\{5 , 1\Big\}

Mid-point of BD:

Let the midpoint of BD be O'.

x₁ ⇒ 2

x₂ ⇒ 8

y₁ ⇒ 1

y₂ ⇒ 1

\sf \Longrightarrow O'(x, y) = \Bigg\{ \dfrac{x_1 + x_2}{2} , \dfrac{y_1 + y_2}{2} \Bigg\}

\sf \Longrightarrow O'(x, y) = \Bigg\{ \dfrac{2 + 8}{2} , \dfrac{1 + 1}{2} \Bigg\}

\sf \Longrightarrow O'(x, y) = \Bigg\{ \dfrac{10}{2} , \dfrac{2}{2} \Bigg\}

\sf \Longrightarrow O'(x, y) = \Big\{5 , 1\Big\}

∴ O(x, y) = O'(x, y)

Hence, they bisect each other.

Attachments:
Similar questions
Math, 4 months ago