Question

NO LINKS!! URGENT HELP PLEASE!!

3. Use the distance formula to find the value of x if the distance between (1, 2) and (x, 5) is 5units.

4. Use the distance (-1, 4) and (5, y) is 10 units.

Answer 1

3) x = -3, x = 5

4) y = -4, y = 12

Explanation:

To calculate the distance between two points we can use the distance formula.

`\boxed{\begin{minipage}{7.4 cm}\underline{Distance Formula}\\\\$d=√((x_2-x_1)^2+(y_2-y_1)^2)$\\\\\\where:\\ \phantom{ww}$\bullet$ $d$ is the distance between two points. \\\phantom{ww}$\bullet$ $(x_1,y_1)$ and $(x_2,y_2)$ are the two points.\\\end{minipage}}`

Question 3

To find the values of x, given the distance between points (1, 2) and (x, 5) is 5 units, substitute the coordinates of the points into the distance formula and set d = 5.

`\begin{aligned}√((x-1)^2+(5-2)^2)&=5\\√((x-1)^2+(3)^2)&=5\\√((x-1)^2+9)&=5\\\left(√((x-1)^2+9)\right)^2&=5^2\\(x-1)^2+9&=25\\x^2-2x+1+9&=25\\x^2-2x-15&=0\\x^2-5x+3x-15&=0\\x(x-5)+3(x-5)&=0\\(x+3)(x-5)&=0\\\\(x+3)&=0 \implies x=-3\\(x-5)&=0 \implies x=5\end{aligned}`

Therefore, the values of x are:

• x = -3 or x = 5

`\hrulefill`

Question 4

To find the values of y, given the distance between points (-1, 4) and (5, y) is 10 units, substitute the coordinates of the points into the distance formula and set d = 10.

`\begin{aligned}√((5-(-1))^2+(y-4)^2)&=10\\√((6)^2+(y-4)^2)&=10\\√(36+(y-4)^2)&=10\\\left(√(36+(y-4)^2)\right)^2&=10^2\\36+(y-4)^2&=100\\36+y^2-8y+16&=100\\y^2-8y-48&=0\\y^2-12y+4y-48&=0\\y(y-12)+4(y-12)&=0\\(y+4)(y-12)&=0\\\\(y+4)&=0 \implies y=-4\\(y-12)&=0 \implies y=12\end{aligned}`

Therefore, the values of y are:

• y = -4, y = 12

Answer 2

3.x = -3 or x =5.

4.y = -4 or y=12.

Explanation:

3. Use the distance formula to find the value of x if the distance between (1, 2) and (x, 5) is 5units.

The distance formula is:

`d =√((x_1 - x_2)^2 + (y_1 - y_2)^2)`

where:

d is the distance between the two points

`(x1, y1)`are the coordinates of the first point

`(x_2, y_2`) are the coordinates of the second point

In this case, we have:

d = 5

`(x_1, y_1) = (1, 2)`

`(x_2, y_2) = (x, 5)`

`Substituting these values into the distance formula, we get:

`5 = √((x - 1)^2 + (5 - 2)^2)`

squaring both side

`25 = (x - 1)^2 + 3^2`

`25 = x^2 - 2x + 1 + 9`

`25 = x^2 - 2x + 10`

`x^2-2x+10-25=0`

`x^2-2x - 15=0`

Factoring the quadratic, we get:

`x^2-(5-3)x-15=0`

`x^2-5x+3x-15=0`

x(x-5)+3(x-5)=0

(x-5)(x+3)=0

either x=5

or x=-3

Therefore, x = -3 or x =5.

4. Use the distance (-1, 4) and (5, y) is 10 units.

The distance formula is the same as in the previous problem. In this case, we have:

d = 10

`(x_1, y_1) = (-1, 4)`

`(x_2, y_2) = (5, y)`

Substituting these values into the distance formula, we get:

`10 =√((5 - (-1))^2 + (y - 4)^2)`

`10 = √(6^2 + (y - 4)^2)`

squaring both sides

100 = 36 + (y - 4)²

100-36=(y-4)²

64 = (y - 4)²

±8 = y - 4

taking positive

8=y-4

y=8+4

y=12

again

taking negative

-8=y-4

y = -8+4

y = -4

Therefore, y = -4 or y=12.