Find a point on Y-axis which is at equal distances from points P(3, 2) and Q(-1, -5).
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Find a point on Y-axis which is at equal distances from points P(3, 2) and Q(-1, -5)
Find a point on Y-axis which is at equal distances from points P(3, 2) and Q(-1, -5).
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GIVEN :–
• Two points are P(3, 2) and Q(-1, -5) at equal distance from Y-axis.
TO FIND :–
• Point on Y-axis = ?
SOLUTION :–
• Let a point on Y-axis is A(0,y).
• We know that Distance between two points [tex] \bf (x_1, y_1) \: and \: (x_2, y_2) [/tex] is –
[tex] \\ \large\implies{ \boxed{\bf Distance = \sqrt{ {(x_2 - x_1)}^{2} + {(y_2 -y_1)}^{2} }}}\\ [/tex]
• Now Distance between A(0,y) and P(3, 2) is –
[tex] \\ \implies\bf AP = \sqrt{ {(3-0)}^{2} + {(2-y)}^{2} }\\ [/tex]
[tex] \\ \implies\bf AP = \sqrt{ {(3)}^{2} + {(2-y)}^{2} }\\ [/tex]
[tex] \\ \implies\bf AP = \sqrt{9+ 4 + {y}^{2} - 4y}\\ [/tex]
[tex] \\ \implies\bf AP = \sqrt{1 3+ {y}^{2} - 4y}\\ [/tex]
[tex] \\ \implies\bf AP = \sqrt{{y}^{2} - 4y + 13}\\ [/tex]
• And Distance between A(0,y) and Q(-1,-5) is –
[tex] \\ \implies\bf AQ= \sqrt{ {( - 1-0)}^{2} + {( - 5-y)}^{2} }\\ [/tex]
[tex] \\ \implies\bf AQ= \sqrt{ {( - 1)}^{2} + {( - 5-y)}^{2} }\\ [/tex]
[tex] \\ \implies\bf AQ= \sqrt{1+ {(5 + y)}^{2} }\\ [/tex]
[tex] \\ \implies\bf AQ= \sqrt{1+25 + {y}^{2} + 10y}\\ [/tex]
[tex] \\ \implies\bf AQ= \sqrt{{y}^{2} + 10y + 26}\\ [/tex]
• According to the question –
[tex] \\\pink\dashrightarrow\bf{\pink{\: AP = AQ}}\\ [/tex]
• So that –
[tex] \\\implies\bf \sqrt{{y}^{2} - 4y + 13} = \sqrt{{y}^{2} + 10y + 26}\\ [/tex]
• Square on both sides –
[tex] \\\implies\bf {y}^{2} - 4y + 13={y}^{2} + 10y + 26\\ [/tex]
[tex] \\\implies\bf - 4y + 13= 10y + 26\\ [/tex]
[tex] \\\implies\bf - 10y - 4y= 26 - 13\\ [/tex]
[tex] \\\implies\bf - 14y= 13\\ [/tex]
[tex] \\\implies \large{ \boxed{\bf y= - \dfrac{13}{14}}}\\ [/tex]
• Hence , The point on Y-axis is (0,-13/14).
Step-by-step explanation:
Two points are P(3, 2) and Q(-1, -5) at equal distance from Y-axis.
TO FIND :–
• Point on Y-axis = ?
SOLUTION :–
• Let a point on Y-axis is A(0,y).
• We know that Distance between two points \bf (x_1, y_1) \: and \: (x_2, y_2)(x
1
,y
1
)and(x
2
,y
2
) is –
\begin{gathered}\\ \large\implies{ \boxed{\bf Distance = \sqrt{ {(x_2 - x_1)}^{2} + {(y_2 -y_1)}^{2} }}}\\\end{gathered}
⟹
Distance=
(x
2
−x
1
)
2
+(y
2
−y
1
)
2
• Now Distance between A(0,y) and P(3, 2) is –
\begin{gathered}\\ \implies\bf AP = \sqrt{ {(3-0)}^{2} + {(2-y)}^{2} }\\\end{gathered}
⟹AP=
(3−0)
2
+(2−y)
2
\begin{gathered}\\ \implies\bf AP = \sqrt{ {(3)}^{2} + {(2-y)}^{2} }\\\end{gathered}
⟹AP=
(3)
2
+(2−y)
2
\begin{gathered}\\ \implies\bf AP = \sqrt{9+ 4 + {y}^{2} - 4y}\\\end{gathered}
⟹AP=
9+4+y
2
−4y
\begin{gathered}\\ \implies\bf AP = \sqrt{1 3+ {y}^{2} - 4y}\\\end{gathered}
⟹AP=
13+y
2
−4y
\begin{gathered}\\ \implies\bf AP = \sqrt{{y}^{2} - 4y + 13}\\\end{gathered}
⟹AP=
y
2
−4y+13
• And Distance between A(0,y) and Q(-1,-5) is –
\begin{gathered}\\ \implies\bf AQ= \sqrt{ {( - 1-0)}^{2} + {( - 5-y)}^{2} }\\\end{gathered}
⟹AQ=
(−1−0)
2
+(−5−y)
2
\begin{gathered}\\ \implies\bf AQ= \sqrt{ {( - 1)}^{2} + {( - 5-y)}^{2} }\\\end{gathered}
⟹AQ=
(−1)
2
+(−5−y)
2
\begin{gathered}\\ \implies\bf AQ= \sqrt{1+ {(5 + y)}^{2} }\\\end{gathered}
⟹AQ=
1+(5+y)
2
\begin{gathered}\\ \implies\bf AQ= \sqrt{1+25 + {y}^{2} + 10y}\\\end{gathered}
⟹AQ=
1+25+y
2
+10y
\begin{gathered}\\ \implies\bf AQ= \sqrt{{y}^{2} + 10y + 26}\\\end{gathered}
⟹AQ=
y
2
+10y+26
• According to the question –
\begin{gathered}\\\pink\dashrightarrow\bf{\pink{\: AP = AQ}}\\\end{gathered}
⇢AP=AQ
• So that –
\begin{gathered}\\\implies\bf \sqrt{{y}^{2} - 4y + 13} = \sqrt{{y}^{2} + 10y + 26}\\\end{gathered}
⟹
y
2
−4y+13
=
y
2
+10y+26
• Square on both sides –
\begin{gathered}\\\implies\bf {y}^{2} - 4y + 13={y}^{2} + 10y + 26\\\end{gathered}
⟹y
2
−4y+13=y
2
+10y+26
\begin{gathered}\\\implies\bf - 4y + 13= 10y + 26\\\end{gathered}
⟹−4y+13=10y+26
\begin{gathered}\\\implies\bf - 10y - 4y= 26 - 13\\\end{gathered}
⟹−10y−4y=26−13
\begin{gathered}\\\implies\bf - 14y= 13\\\end{gathered}
⟹−14y=13
\begin{gathered}\\\implies \large{ \boxed{\bf y= - \dfrac{13}{14}}}\\\end{gathered}
⟹
y=−
14
13
• Hence , The point on Y-axis is (0,-13/14).