Answer:
3.x = -3 or x =5.
4.y = -4 or y=12.
Step-by-step 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:
[tex]d =\sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}[/tex]
where:
d is the distance between the two points
[tex](x1, y1) [/tex]are the coordinates of the first point
[tex](x_2, y_2[/tex]) are the coordinates of the second point
In this case, we have:
d = 5
[tex](x_1, y_1) = (1, 2)[/tex]
[tex](x_2, y_2) = (x, 5)[/tex]
`Substituting these values into the distance formula, we get:
[tex]5 = \sqrt{(x - 1)^2 + (5 - 2)^2}[/tex]
squaring both side
[tex]25 = (x - 1)^2 + 3^2[/tex]
[tex]25 = x^2 - 2x + 1 + 9[/tex]
[tex]25 = x^2 - 2x + 10[/tex]
[tex]x^2-2x+10-25=0[/tex]
[tex]x^2-2x - 15=0[/tex]
Factoring the quadratic, we get:
[tex]x^2-(5-3)x-15=0[/tex]
[tex]x^2-5x+3x-15=0[/tex]
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
[tex](x_1, y_1) = (-1, 4)[/tex]
[tex](x_2, y_2) = (5, y)[/tex]
Substituting these values into the distance formula, we get:
[tex]10 =\sqrt{(5 - (-1))^2 + (y - 4)^2}[/tex]
[tex]10 = \sqrt{6^2 + (y - 4)^2}[/tex]
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.
