Question

The triangle formed by ${x^2} - 9{y^2} = 0$ and $x = 4$ is
A). Isosceles
B). Equilateral
C). Right-angled
D). None of these

Answer 1

Here they have given the equation of the lines of a triangle from these we will first find the three intersection points then we will find the distance between those points to get the length of each side of a triangle. Then by comparing the length we are able to find which kind of triangle is this.
Formula Used: The distance between the points $({x_1},{y_1})$ and $({x_2},{y_2})$ is $\sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}}$

Complete step-by-step solution:
It is given that the triangle is formed by the equations ${x^2} - 9{y^2} = 0$ and $x = 4$. We know that a triangle has three sides.
Consider ${x^2} - 9{y^2} = 0$ and factorize this to get the equation of another line.
$\Rightarrow {x^2} - {\left( {3y} \right)^2} = 0 \Rightarrow \left( {x - 3y} \right)\left( {x + 3y} \right) = 0$
Thus, the line equations of the three sides of the triangle are

$$x - 3y = 0.....(1) \\\ x + 3y = 0.....(2) \\\ x = 4.............(3) \\\$$

Now we will find the vertices of the triangle by finding the intersection points of these three lines.
Solving the equations $(1)\& (2)$ we get
$(1) + (2) \Rightarrow 2x = 0 \Rightarrow x = 0$
Substituting this in the equation $(1)$ we get
$(1) \Rightarrow 0 - 3y = 0 \Rightarrow y = 0$
Thus, the lines $x - 3y = 0$ and $x + 3y = 0$ intersect at a point $(0,0)$.
Solving the equations $(2)\& (3)$ we get
$(2) - (3) \Rightarrow 3y = - 4 \Rightarrow y = \dfrac{{ - 4}}{3}$
Substituting this in the equation $(2)$ we get
$(2) \Rightarrow x + 3\left( {\dfrac{{ - 4}}{3}} \right) = 0 \Rightarrow x = 4$
Thus, the lines $x + 3y = 0$ and $x = 4$ intersect at a point $\left( {4,\dfrac{{ - 4}}{3}} \right)$.
Solving the equations $(1)\& (3)$ we get
$(1) - (3) \Rightarrow - 3y = - 4 \Rightarrow y = \dfrac{4}{3}$
Substituting this in the equation $(1)$ we get
$(1) \Rightarrow x - 3\left( {\dfrac{4}{3}} \right) = 0 \Rightarrow x = 4$
Thus, the lines $x = 4$ and $x - 3y = 0$ intersect at a point $\left( {4,\dfrac{4}{3}} \right)$.
These intersection points are the vertices of the triangle. Let $A = \left( {0,0} \right),B = \left( {4,\dfrac{{ - 4}}{3}} \right)$ and $C = \left( {4,\dfrac{4}{3}} \right)$ be the vertices of the triangle. Now we will find the length of each side of the triangle.
We know that the distance between two points formula is $\sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}}$ where $({x_1},{y_1})$ and $({x_2},{y_2})$ are the two points.
The length of the side $AB = \sqrt {{{\left( {\dfrac{{ - 4}}{3} - 0} \right)}^2} + {{\left( {4 - 0} \right)}^2}}$
$= \sqrt {\dfrac{{16}}{9} + 16}$
$= \sqrt {\dfrac{{160}}{9}}$
$AB = \dfrac{{4\sqrt {10} }}{3}$
The length of the side $BC = \sqrt {{{\left( {4 - 4} \right)}^2} + {{\left( {\dfrac{4}{3} - \left( {\dfrac{{ - 4}}{3}} \right)} \right)}^2}}$
$= \sqrt {{{\left( {4 - 4} \right)}^2} + {{\left( {\dfrac{4}{3} + \dfrac{4}{3}} \right)}^2}}$
$= \sqrt {0 + {{\left( {\dfrac{8}{3}} \right)}^2}}$
$BC = \dfrac{8}{3}$
The length of the side $CA = \sqrt {{{\left( {0 - 4} \right)}^2} + {{\left( {0 - \dfrac{4}{3}} \right)}^2}}$
$= \sqrt {{{\left( 4 \right)}^2} + {{\left( {\dfrac{{ - 4}}{3}} \right)}^2}}$
$= \sqrt {16 + \dfrac{{16}}{9}}$
$= \sqrt {\dfrac{{160}}{9}}$
$CA = \dfrac{{4\sqrt {10} }}{3}$
Thus, we have found the length of all sides of the triangle $AB = \dfrac{{4\sqrt {10} }}{3}$, $BC = \dfrac{8}{3}$ and $CA = \dfrac{{4\sqrt {10} }}{3}$. Here we can see that $AB = CA$. Thus, the triangle formed by ${x^2} - 9{y^2} = 0$ and $x = 4$ is isosceles.
Now let us see the options, option (a) isosceles is the correct option as we got the same answer from our calculation.
Option (b) equilateral is an incorrect option as we got ‘isosceles’ as the correct answer.
Option (c) right-angled is an incorrect option as we got ‘isosceles’ as the correct answer.
Option (d) None of these is an incorrect option as we got the option (a) as the correct answer.
Hence, option (a) Isosceles is the correct option.

Note: A length of a line can be determined by calculating the distance between their endpoints. An isosceles triangle is a triangle in which any two sides of that triangle will be equal, here we got the sides $AB$ and $CA$ are equal.