Question

The equations of the straight lines are \[x + y = 0\], \[5x + y = 4\] and \[x + 5y = 4\]. Then what is the type of a triangle formed by these lines?
A. an isosceles triangle
B. an equilateral triangle
C. a scalene triangle
D. a right-angled triangle

Answer 1

Hint: In the given question, three equations of the straight lines are given. By solving the equations, we will find the points of intersections of the lines. Then by using the distance formula, we will find the distance between two vertices of the formed triangle.
Formula Used:
The distance between the two points \[A\left( {{x_1},{y_1}} \right)\] and \[B\left( {{x_2},{y_2}} \right)\] is:
\[d = \sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}} \]
Complete step by step solution:
The given equations of straight lines are:
\[x + y = 0\] …….(1)
\[5x + y = 4\] …….(2)
\[x + 5y = 4\] …….(3)
Let’s solve the equations and find the points of intersections of the lines.
Rewrite equation in term \[y\].
\[y = - x\] …..(4)
Now solve the equations (4) and (2).
Substitute \[x = - y\] in equation (2).
\[5\left( { - y} \right) + y = 4\]
\[ \Rightarrow \]\[ - 4y = 4\]
\[ \Rightarrow \]\[y = - 1\]
Then, \[x = 1\].
The point of intersection of \[x + y = 0\] and \[5x + y = 4\] is \[\left( {1, - 1} \right)\].

Now solve the equations (4) and (3).
Substitute \[x = - y\] in equation (3).
\[ - y + 5y = 4\]
\[ \Rightarrow \]\[4y = 4\]
\[ \Rightarrow \]\[y = 1\]
Then, \[x = - 1\].
The point of intersection of \[x + y = 0\] and \[x + 5y = 4\] is \[\left( { - 1,1} \right)\].

Now solve the equations (2) and (3).
Substitute \[x = 4 - 5y\] in equation (2).
\[5\left( {4 - 5y} \right) + y = 4\]
\[ \Rightarrow \]\[20 - 25y + y = 4\]
\[ \Rightarrow \]\[ - 24y = - 16\]
\[ \Rightarrow \]\[y = \dfrac{2}{3}\]
Then, \[x = \dfrac{2}{3}\].
The point of intersection of \[5x + y = 4\] and \[x + 5y = 4\] is \[\left( {\dfrac{2}{3},\dfrac{2}{3}} \right)\].


Now apply distance formula to calculate the distance between the two vertices \[A\left( { - 1,1} \right)\] and \[B\left( {\dfrac{2}{3},\dfrac{2}{3}} \right)\].
\[AB = \sqrt {{{\left( {\dfrac{2}{3} + 1} \right)}^2} + {{\left( {\dfrac{2}{3} - 1} \right)}^2}} \]
\[ \Rightarrow \]\[AB = \sqrt {{{\left( {\dfrac{5}{3}} \right)}^2} + {{\left( {\dfrac{{ - 1}}{3}} \right)}^2}} \]
\[ \Rightarrow \]\[AB = \sqrt {\dfrac{{26}}{9}} \]

Now apply distance formula to calculate the distance between the two vertices \[A\left( { - 1,1} \right)\] and \[C\left( {1, - 1} \right)\].
\[AC = \sqrt {{{\left( {1 + 1} \right)}^2} + {{\left( { - 1 - 1} \right)}^2}} \]
\[ \Rightarrow \]\[AC = \sqrt {{{\left( 2 \right)}^2} + {{\left( 2 \right)}^2}} \]
\[ \Rightarrow \]\[AC = \sqrt 8 \]
\[ \Rightarrow \]\[AC = 2\sqrt 2 \]
Now apply distance formula to calculate the distance between the two vertices \[B\left( {\dfrac{2}{3},\dfrac{2}{3}} \right)\] and \[C\left( {1, - 1} \right)\].
\[BC = \sqrt {{{\left( {\dfrac{2}{3} - 1} \right)}^2} + {{\left( {\dfrac{2}{3} + 1} \right)}^2}} \]
\[ \Rightarrow \]\[BC = \sqrt {{{\left( {\dfrac{{ - 1}}{3}} \right)}^2} + {{\left( {\dfrac{5}{3}} \right)}^2}} \]
\[ \Rightarrow \]\[BC = \sqrt {\dfrac{{26}}{9}} \]
Since, the length of two sides of a triangle is equal. i.e., \[AB = BC\].
So, the triangle formed by the lines is an isosceles triangle.
Hence the correct option is A.
Note: Students are often confused with the formula \[d = \sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}} \] and \[d = \sqrt {{{\left( {{x_2} + {x_1}} \right)}^2} - {{\left( {{y_2} + {y_1}} \right)}^2}} \] . But the correct distance formula is \[d = \sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}} \].