Question #122064
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
1
Expert's answer
2020-06-16T17:42:03-0400

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=(6+02,0+22)=(3,1).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=m3,m=13.1=m\cdot 3, m=\frac{1}{3}.

The equation of median OC is y=13x.y=\frac{1}{3}x.

II. The midpoint of OB is E=(0+02,0+22)=(0,1).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=y2y1x2x1=1006=16.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=16x+1y=-\frac{1}{6}x+1 .

III. The midpoint of OA is D=(0+62,0+02)=(3,0).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=y2y1x2x1=0230=23.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=23x+2.y=-\frac{2}{3}x+2.

Answer. (I) y=13x,y=\frac{1}{3}x, (II) y=16x+1,y=-\frac{1}{6}x+1, (III) y=23x+2.y=-\frac{2}{3}x+2.


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