Question

$\Delta ABC$ is an isosceles triangle, equation of side AB= x-y=4 and coordinate of B (4, 0); C (6, 4)
Find the equation of side AC. (Altitude = median in isosceles triangle)

Answer 1

Hint: At first, use the property that its altitude is its median. Then, find the equation of altitude using formula $y-{{y}_{1}}=m\left( x-{{x}_{1}} \right)$ where m is slope and $\left( {{x}_{1}},{{y}_{1}} \right)$ is mid-point of base. Then, find coordinates of A using fact that AB and altitude intersect at A. Thus, from point A, find the equation of a line.

Complete step by step answer:
In the question, we are given an isosceles triangle ABC. We are also given an equation of AB which is $x-y=4$ and also coordinates of B and C are given as (4, 0) and (6, 4) respectively.
So at first, let's draw a figure,

In an isosceles triangle, there is a property that, the altitude from A to BC will be referred to as a median of triangle ABC through A hence, the figure can be re-drawn as,

Let the D be the point on BC such that AD is median and altitude making D as a mid-point of BC.
As we know, the coordinates of B and C is (4, 0) and (6, 4) so we can say D is $\left( \dfrac{4+6}{2},\dfrac{0+4}{2} \right)\Rightarrow \left( \dfrac{10}{2},\dfrac{4}{2} \right)\Rightarrow \left( 5,2 \right)$ by using formula of finding mid-point $\left( \dfrac{{{x}_{1}}+{{x}_{2}}}{2},\dfrac{{{y}_{1}}+{{y}_{2}}}{2} \right)$ if points are $\left( {{x}_{1}},{{y}_{1}} \right)\text{ and }\left( {{x}_{2}},{{y}_{2}} \right)$
Now, we will find slope of BC by using formula $m=\left( \dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}} \right)$ if points are $\left( {{x}_{1}},{{y}_{1}} \right)\text{ and }\left( {{x}_{2}},{{y}_{2}} \right)$
So if points are B (4, 0) and C (6, 4) then its slope is $\dfrac{4-0}{6-4}\Rightarrow \dfrac{4}{2}\Rightarrow 2$
As we know that, the product of the slope of the perpendicular line is -1. So, we can say that, if the slope of BC is 2 then the slope of AD is -1 as AD is perpendicular to BC.
Hence, we can find equation of AD by using formula $y-{{y}_{1}}=m\left( x-{{x}_{1}} \right)$ if $\left( {{x}_{1}},{{y}_{1}} \right)$ is point and m is slope.
Thus, we can say,
\[y-2=\dfrac{-1}{2}\left( x-5 \right)\]
On cross multiplication we get:
\[2y-4=-x+5\]
Which can be written as:
\[\begin{align}
  & x+2y-4-5=0 \\
 & \Rightarrow x+2y-9=0 \\
\end{align}\]
Hence, we know equation of AD is $x+2y=9$ and also we were given the equation of AB as $x-y=4$ so, there intersection is A. Hence, solving them we can find point A.
So, the equations are:
\[\begin{align}
  & x+2y=9\cdots \cdots \cdots \cdots \cdots \left( i \right) \\
 & x-y=4\cdots \cdots \cdots \cdots \cdots \left( ii \right) \\
\end{align}\]
Subtracting (ii) from (i) we get:
\[\begin{align}
  & 2y-\left( -y \right)=9-4 \\
 & \Rightarrow 3y=5 \\
 & \Rightarrow y=\dfrac{5}{3} \\
\end{align}\]
Now substituting y as $\dfrac{5}{3}$ in (i) we get:
\[\begin{align}
  & x+2\times \dfrac{5}{3}=9 \\
 & \Rightarrow x+\dfrac{10}{3}=9 \\
 & \Rightarrow x=9-\dfrac{10}{3}=\dfrac{17}{3} \\
\end{align}\]
Hence, the point A has coordinates $\left( \dfrac{17}{3},\dfrac{5}{3} \right)$
Now, we know points A and C has co-ordinates $\left( \dfrac{17}{3},\dfrac{5}{3} \right)$ and (6,4).
So, we will find slope of AC using formula $m=\left( \dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}} \right)$ if coordinate are $\left( {{x}_{1}},{{y}_{1}} \right)\text{ and }\left( {{x}_{2}},{{y}_{2}} \right)$ so we get:
\[m=\dfrac{4-\dfrac{5}{3}}{6-\dfrac{17}{3}}=\dfrac{\dfrac{7}{3}}{\dfrac{1}{3}}=7\]
Hence, slope is 7.
Thus, we will find equation using formula $y-{{y}_{1}}=m\left( x-{{x}_{1}} \right)$ so we get:
\[\begin{align}
  & y-4=7\left( x-6 \right) \\
 & \Rightarrow y-4=7x-42 \\
 & \Rightarrow 7x-42-\left( y-4 \right)=0 \\
 & \Rightarrow 7x-y-42+4=0 \\
 & \Rightarrow 7x-y-38=0 \\
\end{align}\]
Hence equation of line AC is $7x-y-38=0$

Note:
Students while solving for values of x and y from the equations of AD and AB to find co-ordinates of A should be careful about calculation because from that point A and C we will find their slope and thus find the equation of AC otherwise the solution might get wrong.