Question

hey guys! I hope you can help me with this question.: find a point on the y axis which is equidistant from the point A(5,6) and (3,-4). Thanking in advance​

Answer 1

Given :

• Two points (5,6) and (3,-4)

To Find :

• A point on the y axis which is equidistant from the point A(5,6) and (3,-4).

To Find :

Firstly we will find The Distance of AP :

$\longmapsto\tt{{x}_{1}=5}$

$\longmapsto\tt{{y}_{1}=6}$

$\longmapsto\tt{{x}_{2}=0}$

$\longmapsto\tt{{y}_{2}=y}$

Using Formula :

$\longmapsto\tt\boxed{Distance\:Formula=\sqrt{({{x}_{2}-{x}_{1)}}^{2}+({{y}_{2}-{y}_{1)}}^{2}}}$

Putting Values :

$\longmapsto\tt{\sqrt{({0-5)}^{2}+({y-6)}^{2}}}$

$\longmapsto\tt\bf{\sqrt{25+{y}^{2}+36-12y}}$

Similarly ,

For PB :

$\longmapsto\tt{{x}_{1}=0}$

$\longmapsto\tt{{y}_{1}=y}$

$\longmapsto\tt{{x}_{2}=3}$

$\longmapsto\tt{{y}_{2}=-4}$

Using Formula :

$\longmapsto\tt\boxed{Distance\:Formula=\sqrt{({{x}_{2}-{x}_{1)}}^{2}+({{y}_{2}-{y}_{1)}}^{2}}}$

Putting Values :

$\longmapsto\tt{\sqrt{({3-0)}^{2}+({-4-y)}^{2}}}$

$\longmapsto\tt\bf{\sqrt{9+16+{y}^{2}-8y}}$

Now ,

$\longmapsto\tt{AP=PB}$

$\longmapsto\tt{\sqrt{25+{y}^{2}+36-12y}=\sqrt{9+16+{y}^{2}-8y}}$

$\longmapsto\tt{25+{{\cancel{y}^{2}}}+36-12y-9-16-{{\cancel{y}^{2}}}+8y=0}$

$\longmapsto\tt{25-9-16+36-12y+8y=0}$

$\longmapsto\tt{16-16+36-4y=0}$

$\longmapsto\tt{36=4y}$

$\longmapsto\tt{\cancel\dfrac{36}{4}=y}$

$\longmapsto\tt\bf{9=y}$

So ,The Points are (0,9) .

Answer 2

Answer:

Question :-

• find a point on the y axis which is equidistant from the point A(5,6) and (3,-4).

Given :-

• Point are (5,6) and (3,-4).

Find Out :-

• Find a point on the y axis which is equidistant from the point.

Solution :-

Let the point on y-axis be (0,y).

As we know that :

$\red{ \boxed{\sf{Distance\: Formula =\: \sqrt{\bigg(x_{2}-x_{1}\bigg)^{2}+\bigg(y_{2}-y_{1}\bigg)^{2}} }}}$

☣ Distance of AP :

We have :

• x₁ = 5
• x₂ = 0
• y₁ = 6
• y₂ = y

By using formula we get,

➙ $\sf \sqrt{(0 - 5)^2 + (y - 6)^2}$

➙ $\sf \sqrt{(25 + y^2 + 36 - 12y)}$

☣ Distance of PB :-

We have :

• x₁ = 0
• x₂ = 3
• y₁ = y
• y₂ = - 4

➙ $\sf \sqrt{(3 - 0)^2 + (- 4 - y)^2}$

➙ $\sf \sqrt{(9 + 16 + y^2 - 8y)}$

According to the question,

➙ $\sf \sqrt{(25 + y^2 + 36 - 12y)} =\: \sqrt{(9 + 16 + y^2 - 8y)}$

➙ 25 + y² + 36 - 12y - 9 - 16 - y² + 8y = 0

➙ 36 = 4y

➙ y = $\sf \dfrac{36}{4}$

➙ ${\small{\bold{\purple{\underline{y = 9}}}}}$

Henceforth, the required points are (0,9).

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$\qquad\qquad\underline{\textsf{\textbf{ \color{magenta}{Brainly\: Extra\: Shots :-} }}}$

Distance Formula :-

• The Distance Formula is a useful tool in finding the distance between two points which can be arbitrarily represented as points (x₁,y₁) and (x₂,y₂).
• The formula for distance formula is √(x₂ −x₁ )² + (y₂ − y₁)².

Section Formula :-

• When the point P(x, y) divides the line segment into two halves, we may say that P(x, y) is the midpoint of the line segment. By the use of section and midpoint formula: m : n = 1 : 1.
• The formula for section formula is { (m₁x₂ + m₂x₁)/(m₁ + m₂ ) , (m₁y₂ + m₂y₁)/(m₁ + m₂ ) } .