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Question: Show that the equation of a perpendicular bisector of the point (t, t+1) and (3t, t+3) is \( y + tx ...

Show that the equation of a perpendicular bisector of the point (t, t+1) and (3t, t+3) is y+tx=2t2+t+2y + tx = 2{t^2} + t + 2 . If this perpendicular bisector passes through the point (5, 2), what is the unknown value of t?

Explanation

Solution

Hint : Since in the question they are talking about the perpendicular bisector of the two points, try to first find the slope of the line that will pass through the given points. From this slope, you can find the slope of the perpendicular bisector passing through the midpoint of the two given points, and you will get the equation of the perpendicular bisector.

Complete step-by-step answer :
We have been given two points as (t, t+1) and (3t, t+3). We will first try to find the midpoint of these two points.
The midpoints of this two points will be (x1+x22,y1+y22)\left( {\dfrac{{{x_1} + {x_2}}}{2},\dfrac{{{y_1} + {y_2}}}{2}} \right)
⇒(3t+t2,(t+1)+(t+3)2) ⇒(4t2,2t+42) ⇒(2t,t+2)   \Rightarrow \left( {\dfrac{{3t + t}}{2},\dfrac{{\left( {t + 1} \right) + \left( {t + 3} \right)}}{2}} \right) \\\ \Rightarrow \left( {\dfrac{{4t}}{2},\dfrac{{2t + 4}}{2}} \right) \\\ \Rightarrow \left( {2t,t + 2} \right) \;
hence the midpoint of the two points is (2t,t+2)\left( {2t,t + 2} \right) .
Also the slope of line between the two points (t, t+1) and (3t, t+3) will be
(y2−y1x2−x1)\left( {\dfrac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}} \right)
⇒(t+3)−(t+1)3t−t ⇒t+3−t−12t ⇒1t   \Rightarrow \dfrac{{\left( {t + 3} \right) - \left( {t + 1} \right)}}{{3t - t}} \\\ \Rightarrow \dfrac{{t + 3 - t - 1}}{{2t}} \\\ \Rightarrow \dfrac{1}{t} \;
The line perpendicular to the above line will have a gradient of −t- t
Hence the line is of the form y=−tx+cy = - tx + c and it passes through the midpoint (2t,t+2)\left( {2t,t + 2} \right) .
Substitute the values of midpoint in y=−tx+cy = - tx + c , we get
t+2=−t×2t+c t+2=−2t2+c 2t2+t+2=c   t + 2 = - t \times 2t + c \\\ t + 2 = - 2{t^2} + c \\\ 2{t^2} + t + 2 = c \;
Hence the equation of the line according to y=mx+cy = mx + c will be
y=−tx+2t2+t+2y = - tx + 2{t^2} + t + 2
⇒y+tx=2t2+t+2\Rightarrow y + tx = 2{t^2} + t + 2
Hence, the equation of a perpendicular bisector of the point (t, t+1) and (3t, t+3) is y+tx=2t2+t+2y + tx = 2{t^2} + t + 2
Now, if the line passes through (5, 2) then
y+tx=2t2+t+2 ⇒2+5t=2t2+t+2 ⇒2t2−4t=0 ⇒2t(t−2)=0   y + tx = 2{t^2} + t + 2 \\\ \Rightarrow 2 + 5t = 2{t^2} + t + 2 \\\ \Rightarrow 2{t^2} - 4t = 0 \\\ \Rightarrow 2t\left( {t - 2} \right) = 0 \;
⇒t=0\Rightarrow t = 0 or t=2t = 2
So, the correct answer is “ t=2t = 2 OR t=0t = 0 ”.

Note : This kind of sum carries multiple steps. The main trick would be to first link all the steps to the final output required and then apply all the formulas and results to find the lines and equations that pass through the given point. If not this you won't be able to reach the required solution.