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If the family of lines $tx+3y-6=0.$ where $t$ is variable intersect the lines $x-2y+3=0$ and $x-y+1=0$ at point $A$ and $B.$ Then locus of mid point of $AB$ is

what i try

Intersection of line $tx+3y-6=0$ and $x-2y+3=0$ is

$\displaystyle A\bigg(\frac{3}{3+2t},\frac{6+3t}{3+2t}\bigg)$

and Intersection of $tx+3y-6=0$ and $x-y+1=0$ is

$\displaystyle B\bigg(\frac{3}{3+t},\frac{6+t}{3+t}\bigg)$

Let locus of mid point of $AB$ is at $M(h,k)$

Then

$\displaystyle h=\frac{1}{2}\bigg(\frac{3}{3+2t}+\frac{3}{3+t}\bigg)$ and $\displaystyle k = \frac{1}{2}\bigg(\frac{6+3t}{3+2t}+\frac{6+t}{3+t}\bigg)$

How do i eliminate $t$ in an easy way

or please help me is any other easier ways to solve it thanks

jacky
  • 5,194
  • One minor problem: after eliminating $t$ you will end up with an equation that overgenerates solutions. That is, there are points that satisfy this equation that are not the midpoint of $A$ and $B$ for any value of $t$. – amd Aug 21 '19 at 00:57

2 Answers2

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Hint:

$$2k=\dfrac{6+3t}{3+2t}+\dfrac{6+t}{3+t}$$

$$2k-\dfrac32-1=\dfrac{6+3t}{3+2t}-\dfrac32+\dfrac{6+t}{3+t}-1=\dfrac{3}{2(3+2t)}+\dfrac3{3+t}$$

$$2h=\dfrac3{3+t}+\dfrac3{3+2t}$$

Solve the simultaneous equations for $\dfrac1{3+2t},\dfrac1{3+t}$

Finally use $$2(3+t)-(3+2t)=3$$

Optionally we can simplify the result

lab bhattacharjee
  • 274,582
0

Let $M(x,y).$

Thus, $$y-x=\frac{1}{2}\left(\frac{3t+3}{2t+3}+1\right)$$ or $$t=\frac{6x-6y+6}{4y-4x-5}.$$ Can you end it now?

Michael Rozenberg
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