Question

If $A\left( 2,1 \right),B\left( -2,3 \right)$ and $C\left( 4,5 \right)$ are the vertices of $\Delta ABC$, then find the equation of
(i) The median through \[C\]
(ii) The altitude through \[B\]
(iii) The perpendicular bisector of side \[BC\]

Answer 1

Hint: Before starting the calculation mark the given points on the coordinate plane. Go step by step and find things that you require for finding out the equation. For finding an equation you at least require two points on the line or a point on that line and the slope of that line.

Complete step by step solution:
Before starting the question we must mark the given points in the coordinate system for a better understanding.
Let’s analyse (i) from the question.
Median means a line passing through \[C\] and through the midpoint of the side \[AB\]. Name the midpoint as \[D\].
In this part, you need to first figure out the coordinates of the mid-point $\left( D \right)$ of the side \[AB\] of the triangle. Then write the equation of a line passing through point \[C\] and \[D\].
Using section formula, we know that the coordinates of a midpoint of a line segment are the mean of the abscissa (\[x\]coordinate of \[A\] and \[B\]) and ordinates(\[y\] coordinate of \[A\] and \[B\]) of its endpoint.
$\Rightarrow $ Coordinates of point \[D\] $=\left( \dfrac{{{x}_{A}}+{{x}_{B}}}{2},\dfrac{{{y}_{A}}+{{y}_{B}}}{2} \right)=\left( \dfrac{2-2}{2},\dfrac{1+3}{2} \right)=\left( 0,2 \right)$
And since we know that the equation of a line passing through two points is: $\left( \dfrac{y-{{y}_{1}}}{x-{{x}_{1}}} \right)=\left( \dfrac{{{y}_{1}}-{{y}_{2}}}{{{x}_{1}}-{{x}_{2}}} \right)$ where $\left( {{x}_{1}},{{y}_{1}} \right)$ and $\left( {{x}_{2}},{{y}_{2}} \right)$are the two given points.
$\Rightarrow $ The required equation will be: $\left( \dfrac{y-{{y}_{C}}}{x-{{x}_{C}}} \right)=\left( \dfrac{{{y}_{C}}-{{y}_{D}}}{{{x}_{C}}-{{x}_{D}}} \right)\Rightarrow \left( \dfrac{y-5}{x-4} \right)=\left( \dfrac{5-2}{4-0} \right)$
$\Rightarrow 4\left( y-5 \right)=3\left( x-4 \right)$
$\Rightarrow 4y-20-3x+12=0$
$\Rightarrow 4y-3x-8=0$
Therefore, we get the equation $4y-3x-8=0$ as our required equation through point \[C\].
Now, let’s move to part (ii), which asks us to find an equation of the altitude on the side \[AC\].
This can be done by finding the slope of side \[AC\] and then finding the equation of a line with a slope perpendicular to side \[AC\] and passing through point \[B\].
$\Rightarrow$Slope of \[AC\] $=\dfrac{{{y}_{A}}-{{y}_{C}}}{{{x}_{A}}-{{x}_{C}}}=\dfrac{1-5}{2-4}=\dfrac{-4}{-2}=2$
We also know that the product of the slope of two perpendicular lines is always $-1$ .
Therefore, the slope of altitude of \[AC\] $=\left( -1 \right)\times \dfrac{1}{2}=\dfrac{-1}{2}$
So, now we have the slope and one point passing through our required line.
According to the given information, the equation of the line will be:
$\Rightarrow Slope=\left( \dfrac{y-{{y}_{B}}}{x-{{x}_{B}}} \right)\Rightarrow \dfrac{-1}{2}=\dfrac{y-3}{x-\left( -2 \right)}\Rightarrow -1\left( x+2 \right)=2\left( y-3 \right)$
$\Rightarrow 2y-6+x+2=0$
$\Rightarrow 2y+x-4=0$
Therefore, we get the equation $2y+x-4=0$ as the required equation of altitude through \[B\].
Now, let’s move to part (iii), which asks us to find the equation of a line which is a perpendicular bisector to side \[BC\]. And it means we need to find the equation of a line that passes through the midpoint of \[BC\left( P \right)\] and have a slope perpendicular to the slope of \[BC\].
So, for that, we at least need one point and slope. The slope can be found by finding the slope of side \[BC\] first and one point on the required line can be obtained from the midpoint of side \[BC\].
From section formula, coordinates of \[P\]$=\left( \dfrac{{{x}_{B}}+{{x}_{C}}}{2},\dfrac{{{y}_{B}}+{{y}_{C}}}{2} \right)=\left( \dfrac{-2+4}{2},\dfrac{3+5}{2} \right)=\left( \dfrac{2}{2},\dfrac{8}{2} \right)=\left( 1,4 \right)$
For the slope of \[BC\] \[=\dfrac{{{y}_{B}}-{{y}_{C}}}{{{x}_{B}}-{{x}_{C}}}=\dfrac{3-5}{-2-4}=\dfrac{-2}{-6}=\dfrac{1}{3}\]
Hence, the slope of a line perpendicular to \[BC\] \[=\left( -1 \right)\times \] Reciprocal of the slope of \[BC\]
$\Rightarrow $ The slope of the required line$=-1\times \dfrac{1}{\dfrac{1}{3}}=-3$
We have the slope of the required line and a point on that line, so the equation for this line will be:
$\Rightarrow Slope=\dfrac{y-{{y}_{P}}}{x-{{x}_{P}}}\Rightarrow -3=\dfrac{y-4}{x-1}$
\[\Rightarrow -3\left( x-1 \right)=y-4\]
\[\Rightarrow y-4+3x-3=0\]
\[\Rightarrow y+3x-7=0\]

Therefore, we have the equation of the perpendicular bisector of \[BC\] as $y+3x-7=0$.

Note: The median of the triangle is nothing but a line which joins the vertex to the opposite side of that vertex at the midpoint. We know that every triangle has three medians and that three medians always met at a common single point. Where all these three pints of median met are called the centroid of the triangle. The altitude is nothing but a line which starts from the vertex and meets at the right angle on the opposite side of that vertex. As same as the median, there are three altitudes of triangle and it’s also met on the common single point. The all three altitudes meet at the point is called as ortho-centre.