Question

The line with equation x+3y=6 cuts the x axis at A the y axis at B .Find the equation of (I) median from origin to AB. (II) median from A to OB. (III) median from B to OA

Answer 1

Let's find the coordinates of A and B.

The interception of line x+3y=6 with x axis (y=0): x+3*0=6, so x=6 and A(6, 0).

The interception of line x+3y=6 with y axis (x=0): 0+3*y=6, so y=2 and B(0, 2).

I. The midpoint of AB is $C=(\frac{6+0}{2}, \frac{0+2}{2})=(3, 1).$

So we have to find the equation of the median OC through the origin O(0,0) and the point C(3, 1). It has a form y=mx and we have $1=m\cdot 3, m=\frac{1}{3}.$

The equation of median OC is $y=\frac{1}{3}x.$

II. The midpoint of OB is $E=(\frac{0+0}{2}, \frac{0+2}{2})=(0, 1).$

So we have to find the equation of the median AE through two points A and E. It has a form y=mx+b.

The slope $m=\frac{y_2-y_1}{x_2-x_1}=\frac{1-0}{0-6}=-\frac{1}{6}.$

The y intercept b=1.

The equation of median AE is $y=-\frac{1}{6}x+1$ .

III. The midpoint of OA is $D=(\frac{0+6}{2}, \frac{0+0}{2})=(3, 0).$

So we have to find the equation of the median BD through two points B and D. It has a form y=mx+b.

The slope $m=\frac{y_2-y_1}{x_2-x_1}=\frac{0-2}{3-0}=-\frac{2}{3}.$

The y intercept b=2.

The equation of median BD is $y=-\frac{2}{3}x+2.$

Answer. (I) $y=\frac{1}{3}x,$ (II) $y=-\frac{1}{6}x+1,$ (III) $y=-\frac{2}{3}x+2.$