Question
The midpoint of the line segment $AB$ shown in the diagram is $(4, – 3).$ Write down the coordinates of $A$ and $B.$
✓
Answer
Let the coordinates of $A$ and Bare $(x, 0)$ and $(0, y).$
so $x=\frac{m x_2+n x_1}{m+n}, y=\frac{m y_2+n y_1}{m+n}$
$ m =2, n =3$
$x _1=4, x _2=4$
$y _1=-5, y _2=5$
$\therefore x=\frac{2 \times 4+3 \times 4}{2+3}=\frac{8+12}{5}=\frac{20}{5}=4$
$y=\frac{2 \times 5+3 \times-5}{2+3}$
$=\frac{10-15}{5}$
$=\frac{-5}{5}$
$=-1$
$\therefore$ Co-ordinates of $P$ are $(4, -1).$
Thus, the coordinates of midpoints of
$ AB =\left(\frac{x+2}{2}, \frac{y+0}{2}\right) $
$=\left(\frac{x}{2}, \frac{y}{2}\right)$
According to question, the coordinates of midpoint $= (4, -3)$
$\therefore \frac{x}{2}=4, x =8 $
$ \frac{y}{2}=-3, y =-6$
$\therefore$ The required points are $(9, 0)$ and $(0, -6).$
so $x=\frac{m x_2+n x_1}{m+n}, y=\frac{m y_2+n y_1}{m+n}$
$ m =2, n =3$
$x _1=4, x _2=4$
$y _1=-5, y _2=5$
$\therefore x=\frac{2 \times 4+3 \times 4}{2+3}=\frac{8+12}{5}=\frac{20}{5}=4$
$y=\frac{2 \times 5+3 \times-5}{2+3}$
$=\frac{10-15}{5}$
$=\frac{-5}{5}$
$=-1$
$\therefore$ Co-ordinates of $P$ are $(4, -1).$
Thus, the coordinates of midpoints of
$ AB =\left(\frac{x+2}{2}, \frac{y+0}{2}\right) $
$=\left(\frac{x}{2}, \frac{y}{2}\right)$
According to question, the coordinates of midpoint $= (4, -3)$
$\therefore \frac{x}{2}=4, x =8 $
$ \frac{y}{2}=-3, y =-6$
$\therefore$ The required points are $(9, 0)$ and $(0, -6).$
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