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# If (2, 1), (-1, -2), (3, 3) are midpoints of sides BC, CA, AB of triangle ABC, then the equation of AB isA). $x-y= \dfrac{1}{2}$B). $x+y=1$C). $x-y=0$D). $x=y$

Last updated date: 23rd Jul 2024
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Hint: In this question, we’ll first use the midpoint formula and solve the equations to get the vertex of the triangle. Just to understand better we’ll also draw the diagram. Then we'll get the equation of the line by using the formula of line equation passing through 2 points.

Complete step by step Solution:

As we can in the above figure, we have assumed the vertex of the triangle as (a, b), (p, q), and (h, k). Now we’ll use the midpoint formula in order to get these vertices. i.e.
x-coordinate of midpoint = $\dfrac{{{\text{x - coordinate of first point + x - coordinate of second point}}}}{2}$
Similarly, y-coordinate of midpoint = $\dfrac{{{\text{y - coordinate of first point + y - coordinate of second point}}}}{2}$
Also, we need to get the equation of line BC. That means we need to get (p, q) and (h, k). Hence, we’ll start with the mid-point of line BC only.
$2 = \dfrac{{p + h}}{2}$and $1 = \dfrac{{q + k}}{2}$
On solving above equations, we get,
$p + h = 4 - - - - (1) \\ q + k = 2 - - - - (2)$
Now we’ll do same for midpoint of AC
$3 = \dfrac{{a + h}}{2}$ and $3 = \dfrac{{b + k}}{2}$
On Solving we get,
$a + h = 6 - - - - - (3) \\ b + k = 6 - - - - - (4)$
And for the midpoint of AB.
$- 1 = \dfrac{{a + p}}{2}$ and $- 2 = \dfrac{{b + q}}{2}$
On Solving,
$a + p = - 2 - - - - - (5) \\ b + q = - 4 - - - - - (6)$
Now, we’ll solve these equations for the unknowns a, b, h, k, p, q.
Observe from equation (1), (3), and (5). If we put the value of h and p from equation (3) and (5) in equation (1) then we’ll get the value of a.
From (3), $h = 6 - a$ and from (5) $p = - 2 - a$. On putting these values in equation (1) we get,
$p + h = 4 \\ \Rightarrow - 2 - a + 6 - a = 4 \\ \Rightarrow - 2a + 4 = 4 \\ \Rightarrow - 2a = 0 \\ \Rightarrow a = 0$
Since, we have got the value of a, on putting it back to equation (3) and (5) we’ll get p, h.
From (3), $0 + h = 6 \Rightarrow h = 6$
From (5), $0 + p = - 2 \Rightarrow p = - 2$
Overserve that we have got x- coordinates of all three points.
Now for y-coordinate, we’ll be dealing with b, q, k. It means we’ll solve equations (2), (4), and (6).
Notice that we can put the value of k and q from equation (4) and (6) respectively, in equation (2)
From (4), $k = 6 - b$ and from (6) $q = - 4 - b$. On putting these values in equation (2) we get,
$q + k = 2 \\ \Rightarrow - 4 - b + 6 - b = 2 \\ \Rightarrow - 2b + 2 = 2 \\ \Rightarrow - 2b = 2 - 2 \\ \Rightarrow - 2b = 0 \\ \Rightarrow b = 0$
Now, we’ll put this value back to equation (4) and (6) we’ll get q, k as
From (4) $0 + k = 6 \Rightarrow k = 6$
From (6) $0 + q = - 4 \Rightarrow q = - 4$
Now, we have got the y-coordinates of all three points.
Hence, vertices of the triangle are (a, b) as (0, 0), (h, k) as (6, 6) and (p, q) as (-2, -4).
Now we’ll use the formula of a line passing through points (p, q) and (h, k). that is
$y - q = \left( {\dfrac{{k - q}}{{h - p}}} \right)\left( {x - p} \right)$
Using this formula and putting the values we get,
$y + 4 = \left( {\dfrac{{6 + 4}}{{6 + 2}}} \right)\left( {x + 2} \right) \\ \Rightarrow y + 4 = \left( {\dfrac{{10}}{8}} \right)\left( {x + 2} \right) \\ \Rightarrow y + 4 = \left( {\dfrac{5}{4}} \right)\left( {x + 2} \right) \\ \Rightarrow 4y + 16 = 5x + 10 \\ \Rightarrow 4y - 5x + 6 = 0$
Hence, $4y - 5x + 6 = 0$ is the required equation of the line BC.

Note: This question can also be solved by an alternative method. What we can do it we’ll find the equation of the line passing through (-1, -2) and (3, 3). Since these lines are midpoint and we know that, a line joining the midpoint of two lines of a triangle will be parallel to the third one. So, this line will be parallel to the required line BC. Also, parallel lines have the same slope so we can get the slope of the required line BC. We know the line BC passing through (2, 1) hence we can use the formula to get the equation of line which is $y - {y_1} = m\left( {x - {x_1}} \right)$ where m is the slope of line BC.