The ends of the base of an isosceles triangle are at (2a, 0) and (0, a). The equation of one side is $x=2a$. The equation of the other side is
(a) $x+2y-a=0$
(b) $x+2y=2a$
(c) $3x+4y-4a=0$
(d) $3x-4y+4a=0$
VerifiedHint: Assume the two given points of the isosceles triangle as $P(2a,0)$ and $Q(0,a)$. Write the general equation of the line in the slope-intercept form which is $y=mx+c$, where $m$ is the slope and $c$ is the intercept on y-axis. Now, find the slope of the line PQ using the formula: $m=\dfrac{\Delta y}{\Delta x}$. Next, find the value of $c$ by substituting the point in the equation of line PQ. Hence, find the equation of line by substituting the value of $m$ and $c$ in the equation. Assume the third point of the triangle as $R(2a,y)$. Use distance formula: $d=\sqrt{{{({{x}_{1}}-{{x}_{2}})}^{2}}++{{({{y}_{1}}-{{y}_{2}})}^{2}}}$ to calculate the value of $y$. Hence, apply the same process to determine the equation of line QR.
Complete Step-by-Step solution:
Let us assume that the given points are $P(2a,0)$ and $Q(0,a)$. Now, we have to find the equation of the line PQ which is the base of the isosceles triangle. We know that the equation of a line is $y=mx+c$, where $m$ is the slope and $c$ is the intercept on the y-axis.
Now, using the formula, $m=\dfrac{{{y}_{1}}-{{y}_{2}}}{{{x}_{1}}-{{x}_{2}}}$, we get,
$m=\dfrac{0-a}{2a-0}=\dfrac{-a}{2a}=\dfrac{-1}{2}$
Therefore, the equation of the line becomes, $y=\dfrac{-x}{2}+c$. Since, this line passes through $Q(0,a)$, therefore, substituting the coordinates in the equation of the line we get,
$\begin{align}
& a=\dfrac{-0}{2}+c \\
& \Rightarrow c=a \\
\end{align}$
Therefore, substituting the value of $c$ in the equation of the line, we get,
$\begin{align}
& y=\dfrac{-x}{2}+a \\
& 2y=-x+2a \\
& \Rightarrow x+2y=2a......................(i) \\
\end{align}$
Now, assume the third point of the isosceles triangle as $R(2a,y)$ because it must lie on the line $x=2a$ to form an isosceles triangle. Since, PQR is an isosceles triangle and PQ is its base, therefore the other two sides must be equal.
$\therefore PR=QR$
Applying distance formula: $d=\sqrt{{{({{x}_{1}}-{{x}_{2}})}^{2}}++{{({{y}_{1}}-{{y}_{2}})}^{2}}}$, we get,
$\begin{align}
& \sqrt{{{(2a-2a)}^{2}}+{{\left( y-0 \right)}^{2}}}=\sqrt{{{(2a-0)}^{2}}+{{(y-a)}^{2}}} \\
& \sqrt{{{y}^{2}}}=\sqrt{4{{a}^{2}}+{{(y-a)}^{2}}} \\
\end{align}$
On squaring both sides we get,
$\begin{align}
& {{y}^{2}}=4{{a}^{2}}+{{(y-a)}^{2}} \\
& {{y}^{2}}=4{{a}^{2}}+{{y}^{2}}+{{a}^{2}}-2ay \\
& 0=5{{a}^{2}}-2ay \\
& y=\dfrac{5a}{2} \\
\end{align}$
Therefore, the point is $R\left( 2a,\dfrac{5a}{2} \right)$. Now, let the equation of line QR is, $y=mx+c$. Therefore,
$m=\dfrac{\dfrac{5a}{2}-a}{2a-0}=\dfrac{3a}{4a}=\dfrac{3}{4}$
Therefore, the equation of the line becomes, $y=\dfrac{3}{4}x+c$. Since, this line passes through (0, a), therefore, substituting the value of coordinates in the equation of the line, we get,
$\begin{align}
& a=\dfrac{3}{4}\times 0+c \\
& \Rightarrow c=a \\
\end{align}$
Therefore, substituting the value of $c$ in the equation of the line, we get,
$\begin{align}
& y=\dfrac{3}{4}x+a \\
& 4y=3x+4a \\
& \Rightarrow 3x-4y+4a=0..........................(ii) \\
\end{align}$
From equations (i) and (ii) we get, options (b) and (d) are the correct answers.
Note: We have to find the equations of both the sides of the triangle because the equation of only one side is given. At a point we have assumed that the third point must lie in the 1st quadrant. The only confusion you will have is, why would the point not lie in the 4th quadrant? It can lie in the fourth quadrant if $a$ is a negative value.
