Answer:
See the below works.
Step-by-step explanation:
We can prove that OP = AP = BP by first finding the coordinates of A and B by using this formula:
[tex]\boxed{(x_{mid},y_{mid})=\left(\frac{x_1+x_2}{2} ,\frac{y_1+y_2}{2}\right)}[/tex]
Let:
- coordinate of A = [tex](x_A,0)[/tex] → A lies at x-axis
- coordinate of B = [tex](0,y_B)[/tex] → B lies at y-axis
Then:
[tex]\displaystyle (x_P,y_P)=\left(\frac{x_A+x_B}{2} ,\frac{y_A+y_B}{2}\right)[/tex]
[tex]\displaystyle (2,3)=\left(\frac{x_A+0}{2} ,\frac{0+y_B}{2}\right)[/tex]
[tex]\displaystyle (2,3)=\left(\frac{x_A}{2} ,\frac{y_B}{2}\right)[/tex]
- [tex]\displaystyle \frac{x_A}{2} =2\Longleftrightarrow x_A=4[/tex]
- [tex]\displaystyle \frac{y_B}{2} =3\Longleftrightarrow y_B=6[/tex]
Hence:
- [tex]A=(4,0)[/tex]
- [tex]B=(0,6)[/tex]
Next, to find the distance/length between point [tex](x_1,y_1)[/tex] and point [tex](x_2,y_2)[/tex], we use this formula:
[tex]\boxed{length=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}}[/tex]
The length of OP:
[tex]|OP|=\sqrt{(x_P-0)^2+(y_P-0)^2}[/tex]
[tex]=\sqrt{2^2+3^2}[/tex]
[tex]=\sqrt{13}[/tex]
The length of AP:
[tex]|AP|=\sqrt{(x_P-x_A)^2+(y_P-y_A)^2}[/tex]
[tex]=\sqrt{(2-4)^2+(3-0)^2}[/tex]
[tex]=\sqrt{13}[/tex]
The length of BP:
[tex]|BP|=\sqrt{(x_P-x_B)^2+(y_P-y_B)^2}[/tex]
[tex]=\sqrt{(2-0)^2+(3-6)^2}[/tex]
[tex]=\sqrt{13}[/tex]
Proven that OP = AP = BP
