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If P (2, 3) is midpoint of the line segment
find coordinates of A and B and show that OP=AP=BP
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If P 2 3 Is Midpoint Of The Line Segmentfind Coordinates Of A And B And Show That OPAPBP class=

Sagot :

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

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