Courses
Courses for Kids
Free study material
Offline Centres
More
Store Icon
Store

Find the point on the y-axis which is equidistant from the points (5, -2) and (-3,2).

seo-qna
Last updated date: 17th Jul 2024
Total views: 423.3k
Views today: 12.23k
Answer
VerifiedVerified
423.3k+ views
Hint: In order to solve this problem, we need to find the distance from the first point to a certain point on the y axis and from the second point to the same point. And as the distance between them is the same we can equate them by using the distance formula. The distance formula says that $\text{Distance}=\sqrt{{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}+{{\left( {{y}_{2}}-{{y}_{1}} \right)}^{2}}}$ where ${{x}_{2}},{{x}_{1}}$ are the x coordinates and ${{y}_{1}},{{y}_{2}}$ are the y coordinates.

Complete step-by-step solution:
We need to point on the y-axis which is equidistant to point A and B
Let the point A be A (5, -2).
Let point B be B (-3,2).
Let the point on the y-axis be P.
We know the points lie on the y-axis.
All the points that lie on the y-axis have the x coordinate equal to zero.
So, the point P becomes P (0, y).
We must know the distance formula to find the distance between PA and PB.
The distance formula is given by $\text{Distance}=\sqrt{{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}+{{\left( {{y}_{2}}-{{y}_{1}} \right)}^{2}}}$ .
Let’s start by finding the distance between points P and A.
Let point P (0, y) be $P\left( {{x}_{1}},{{y}_{1}} \right)$ .
And point A (5, -2) be $A\left( {{x}_{2}},{{y}_{2}} \right)$ .
Substituting the values, we get,
$PA=\sqrt{{{\left( -2-y \right)}^{2}}+{{\left( 5-0 \right)}^{2}}}$
Solving this equation, we get,
$PA=\sqrt{{{\left( -2-y \right)}^{2}}+25}............(i)$
Let’s find the distance between points P and B.
Let point P (0, y) be $P\left( {{x}_{1}},{{y}_{1}} \right)$ .
And point B (-3, 2) be $B\left( {{x}_{2}},{{y}_{2}} \right)$ .
Substituting the values, we get,
$PB=\sqrt{{{\left( 2-y \right)}^{2}}+{{\left( -3-0 \right)}^{2}}}$
Solving this equation, we get,
$PB=\sqrt{{{\left( 2-y \right)}^{2}}+9}............(ii)$
We know that point P is at the same distance from point A and B, hence PA = PB.
Equating (i) and (ii), we get,
\[\sqrt{{{\left( -2-y \right)}^{2}}+25}=\sqrt{{{\left( 2-y \right)}^{2}}+9}\]
Squaring on both sides we get,
\[{{\left( -2-y \right)}^{2}}+25={{\left( 2-y \right)}^{2}}+9\]
Using the identity that ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$ , we get,
\[{{\left( -2 \right)}^{2}}+{{\left( -y \right)}^{2}}+\left( 2\times -y\times -2 \right)+25={{2}^{2}}+{{\left( -y \right)}^{2}}+\left( 2\times -y\times 2 \right)+9\]
Solving for y we get,
\[ {{\left( -2 \right)}^{2}}+{{\left( -y \right)}^{2}}+\left( 2\times -y\times -2 \right)+25={{2}^{2}}+{{\left( -y \right)}^{2}}+\left( 2\times -y\times 2 \right)+9 \\
\Rightarrow 4+{{y}^{2}}+4y+25=4+{{y}^{2}}-4y+9 \\
 \Rightarrow 8y=9-25 \\
\Rightarrow 8y=-16 \\
\Rightarrow y=\dfrac{-16}{8}=-2 \]
Hence, the point P is P (0, -2).

Note: All the points on the y-axis have their x coordinate zero. This is because as the x-axis passes through the origin it interests the x-axis at x coordinate = 0. Hence, its all the points have x coordinates zero. In the distance formula, we can choose any point first because the difference between them will always be squared and the square of any number is always positive.