Question

Pls solve tyyyyyyyyyy

Answer 1

`\sf 2) \quad The\;coordinates\;of\;point\;B\;are\;\left(\:\boxed{2√(3)}\:, \boxed{\vphantom{√(3)}0}\:\right).`

`\sf 3) \quad The\;distance\;between\;points\;A\;and\;B\;is\; \boxed{2√(3)}\:.`

Explanation:

Question 2

From observation of the given graph, point B is the intersection of the x-axis and
`y = -√(3)x + 6`.

Any point that intersects the x-axis has a y-value of zero.

Therefore, to find the x-value of B, we can substitute y = 0 into the equation and solve for x:

`\begin{aligned} -√(3)x+6&=0\\\\ -√(3)x&=-6\\\\√(3)x&=6\\\\x&=(6)/(√(3))\\\\x&=(6)/(√(3))\cdot (√(3))/(√(3))\\\\x&=(6√(3))/(3)\\\\x&=2√(3)\end{aligned}`

Therefore, the coordinates of point B are (2√3, 0).

`\hrulefill`

Question 3

To find the distance between points A and B, we first need to determine the coordinates of point A.

Point A is the point of intersection of the two graphed lines.

To find the x-value of point A, substitute one equation into the other and solve for x:

`\begin{aligned}√(3)x&=-√(3)x+6\\\\√(3)x+√(3)x&=-√(3)x+6+√(3)x\\\\2√(3)x&=6\\\\x&=(6)/(2√(3))\\\\x&=(3)/(√(3))\\\\x&=(3)/(√(3))\cdot (√(3))/(√(3))\\\\x&=(3√(3))/(3)\\\\x&=√(3)\end{aligned}`

Substitute the found value of x into one of the equations to find the y-value of point A:

`\begin{aligned}y&=√(3) \cdot √(3)\\\\y&=3\end{aligned}`

Therefore, the coordinates of point A are (√3, 3).

To find the distance between points A and B, substitute the points into the distance formula:

`\begin{aligned}d&=√((x_B-x_A)^2+(y_B-y_A)^2)\\&=\sqrt{(2√(3)-√(3))^2+(0-3)^2}\\&=\sqrt{(√(3))^2+(-3)^2}\\&=√(3+9)\\&=√(12)\\&=√(2^2 \cdot 3)\\&=√(2^2) √(3)\\&=2√(3)\end{aligned}`

Therefore, the distance between points A and B is 2√3.