A is at( − 1, 7) and B is at (4, 8) P is at (1,y) and PA =PB. Find the value of y.
Answers
Answer:
10
Step-by-step explanation:
Formula used
For two points (x_1, y_1)(x
1
,y
1
)left parenthesis, x, start subscript, 1, end subscript, comma, y, start subscript, 1, end subscript, right parenthesis and (x_2, y_2)(x
2
,y
2
)left parenthesis, x, start subscript, 2, end subscript, comma, y, start subscript, 2, end subscript, right parenthesis, we have a formula that gives the distance between them -−minus
\text{Dist} = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}Dist=
(x
1
−x
2
)
2
+(y
1
−y
2
)
2
start text, D, i, s, t, end text, equals, square root of, left parenthesis, x, start subscript, 1, end subscript, minus, x, start subscript, 2, end subscript, right parenthesis, squared, plus, left parenthesis, y, start subscript, 1, end subscript, minus, y, start subscript, 2, end subscript, right parenthesis, squared, end square root
Hint #22 / 4
Forming the equation
As PA = PBPA=PBP, A, equals, P, B, we can now equate the distances between these pairs of points.
Plugging PPP (1,y)(1,y)left parenthesis, 1, comma, y, right parenthesis, AAA (-1, 7)(−1,7)left parenthesis, minus, 1, comma, 7, right parenthesis, and BBB (4,8)(4,8)left parenthesis, 4, comma, 8, right parenthesis, we get:
\begin{aligned} PA &= PB\\\\ \sqrt{(1-(-1))^2 + (y - 7)^2} &= \sqrt{(1 - 4)^2 + (y - 8)^2} \\\\ 4 + (y-7)^2 &= 9 + (y -8)^2 \end{aligned}
PA
(1−(−1))
2
+(y−7)
2
4+(y−7)
2
=PB
=
(1−4)
2
+(y−8)
2
=9+(y−8)
2
Solving the equation
On solving, we get y = 10y=10y, equals, 10.
[Show me the steps.]
So, our point PPP is (1,10)(1,10)left parenthesis, 1, comma, 10, right parenthesis.
Hint #44 / 4
In conclusion,
y = 10y=10y, equals, 10.