Question

The vertices of the base of an isosceles triangle lie on a parabola \[{{y}^{2}}=4x\] and the base is a part of the line \[y=2x-4\]. If the third vertex of the triangle lies on the x-axis, its coordinates are
(a) \[\left( \dfrac{5}{2},\,0 \right)\]
(b) \[\left( \dfrac{7}{2},\,0 \right)\]
(c) \[\left( \dfrac{9}{2},\,0 \right)\]
(d) \[\left( \dfrac{11}{2},\,0 \right)\]

Answer 1

Hint:We will substitute \[y=2x-4\] in \[{{y}^{2}}=4x\] because the vertices of the base of an isosceles triangle lie on both the line and the parabola. Then to find the third vertex we will use the isosceles triangle definition that the third vertex is equidistant from the other two vertices.

Complete step-by-step answer:

It is mentioned in the question that the two vertices of the base of an isosceles triangle lie on a parabola \[{{y}^{2}}=4x.......(1)\] and also on the line \[y=2x-4.........(2)\]. So substituting y from equation (2) in equation (1) we get,
 \[\,\Rightarrow {{(2x-4)}^{2}}=4x........(3)\]
Squaring the left hand side of the equation (3) and rearranging equation (3) we get,
 \[\,\Rightarrow 4{{x}^{2}}-16x+16=4x........(4)\]
Cancelling similar terms and simplifying equation (4) we get,
 \[\begin{align}
  & \,\Rightarrow 4{{x}^{2}}-20x+16=0 \\
 & \,\Rightarrow {{x}^{2}}-5x+4=0.........(5) \\
\end{align}\]
Now factorizing equation (5) and solving for x we get,
 \[\begin{align}
  & \,\Rightarrow {{x}^{2}}-4x-x+4=0 \\
 & \,\Rightarrow x(x-4)-1(x-4)=0 \\
 & \,\Rightarrow (x-4)\,(x-1)=0 \\
 & \,\Rightarrow x=1,\,4 \\
\end{align}\]
Now substituting \[x=1\] in equation (2) we get \[y=2\times 1-4=-2\].
And substituting \[x=4\] in equation (2) we get \[y=2\times 4-4=4\].
So the corresponding coordinates of the points of the base of the isosceles triangle lying on the parabola are (1, -2) and (4, 4).
Now let the third vertex on the x-axis be (h, 0). Also in the question it is mentioned that the triangle is isosceles which means the third vertex (h, 0) is equidistant from (1, -2) and (4, 4).
So now applying the distance formula between (h, 0) and (1, -2), also between (h, 0) and (4, 4) and equating the both distances we get,
 \[\Rightarrow \sqrt{{{(h-1)}^{2}}+{{(0+2)}^{2}}}=\sqrt{{{(h-4)}^{2}}+{{(0-4)}^{2}}}........(6)\]
Squaring both sides of equation (6) and rearranging the terms we get,
\[\begin{align}
  & \,\Rightarrow {{(h-1)}^{2}}+{{(0+2)}^{2}}={{(h-4)}^{2}}+{{(0-4)}^{2}} \\
 & \,\Rightarrow {{h}^{2}}+1-2h+4={{h}^{2}}+16-8h+16..........(7) \\
\end{align}\]
Cancelling similar terms in equation (7) and bringing like terms on either side we get,
 \[\begin{align}
  & \,\Rightarrow 8h-2h=16+16-4-1 \\
 & \,\Rightarrow 6h=27 \\
 & \,\Rightarrow h=\dfrac{27}{6}=\dfrac{9}{2} \\
\end{align}\]
The coordinates of the third vertex is \[\left( \dfrac{9}{2},\,0 \right)\]. So the correct answer is option (c).

Note: We should remember the basic definition of an isosceles triangle and also the distance formula. Distance between two points \[({{x}_{1}},\,{{x}_{2}})\] and \[({{y}_{1}},\,{{y}_{2}})\] is \[d=\sqrt{{{({{x}_{1}}-{{x}_{2}})}^{2}}+{{({{y}_{1}}-{{y}_{2}})}^{2}}}\]. We in a hurry can commit a mistake in solving equation (5) and equation (6) so we need to be careful while doing these steps. Also we are substituting x values to get corresponding y values because then only we get coordinates of the two vertices.