Question

The equation \[{\left( {x - 5} \right)^2} + \left( {x - 5} \right)\left( {y - 6} \right) - 2{\left( {y - 6} \right)^2} = 0\], represents
A. A circle
B. Two straight lines passing through the origin
C. Two straight lines passing through the point \[\left( {5,6} \right)\]
D. None of these

Answer 1

Hint: In this question, we have to figure out what the given equation means. For this, we need to simplify the given equation and after simplification, we will check the conditions for circle and straight line and for which this equation is satisfying .

Formula used: We will use the following algebraic formula for solving this example.
 \[{\left( {a - b} \right)^2} = {a^2} - 2ab + {b^2}\]

Complete step-by-step answer:
We know that \[{\left( {x - 5} \right)^2} + \left( {x - 5} \right)\left( {y - 6} \right) - 2{\left( {y - 6} \right)^2} = 0\]
Let us simplify the above equation.
Here, we will split the term \[\left( {x - 5} \right)\left( {y - 6} \right)\] as \[2\left( {x - 5} \right)\left( {y - 6} \right) - \left( {x - 5} \right)\left( {y - 6} \right)\]
Thus, we get
 \[{\left( {x - 5} \right)^2} + 2\left( {x - 5} \right)\left( {y - 6} \right) - \left( {x - 5} \right)\left( {y - 6} \right) - 2{\left( {y - 6} \right)^2} = 0\]
By taking \[\left( {x - 5} \right)\] common from the first two terms and \[\left( {y - 6} \right)\] from the last two terms, we get
\[
   \Rightarrow \left( {x - 5} \right)\left( {\left( {x - 5} \right) + 2\left( {y - 6} \right)} \right) - \left( {y - 6} \right)\left( {\left( {x - 5} \right) + 2\left( {y - 6} \right)} \right) = 0 \\
   \Rightarrow \left( {x - 5 - \left( {y - 6} \right)} \right)\left( {\left( {x - 5} \right) + 2\left( {y - 6} \right)} \right) = 0 \\
   \Rightarrow \left( {x - 5 - y + 6} \right)\left( {x - 5 + 2y - 12} \right) = 0 \\
   \Rightarrow \left( {x - y + 1} \right)\left( {x + 2y - 17} \right) = 0 \\
 \]
That means, \[\left( {x - y + 1} \right) = 0\] or \[\left( {x + 2y - 17} \right) = 0\]
These indicate equations of two straight lines as the standard equation of the straight line is \[y = mx + c\] where, \[m\] and \[c\] are constants.
Now, we will choose the correct option from the two remaining options.
Put \[\left( {a,b} \right) \equiv \left( {0,0} \right)\] in the given equation.
So, we get
 \[
  {\left( {0 - 5} \right)^2} + \left( {0 - 5} \right)\left( {0 - 6} \right) - 2{\left( {0 - 6} \right)^2} = 0 \\
   \Rightarrow 25 + 30 - 72 = 0 \\
   \Rightarrow - 17 \ne 0 \\
 \]
That means it does not satisfy the given equation for the point \[\left( {0,0} \right)\].
Now, we will check for the point \[\left( {a,b} \right) \equiv \left( {5,6} \right)\]
Put \[\left( {a,b} \right) \equiv \left( {5,6} \right)\]in the given equation.
So, we get
 \[
  {\left( {5 - 5} \right)^2} + \left( {5 - 5} \right)\left( {6 - 6} \right) - 2{\left( {6 - 6} \right)^2} = 0 \\
   \Rightarrow 0 = 0 \\
 \]
That means the given equation satisfies the point \[\left( {5,6} \right)\].
Hence, the equation \[{\left( {x - 5} \right)^2} + \left( {x - 5} \right)\left( {y - 6} \right) - 2{\left( {y - 6} \right)^2} = 0\] represents two straight lines passing through the point \[\left( {5,6} \right)\]

Therefore, the correct option is (C).

Additional Information: A straight line equation can be expressed in a variety of ways, including point-slope form, slope-intercept form, general form, standard form, and so on. A straight line is a two-dimensional geometric element that goes indefinitely on both ends.

Note:
Many students make mistakes in the simplification part. Specifically in simplifying the given equation which is the multiplication of two brackets. They may confuse with signs so that they will not be able to determine whether the given equation is of the circle or straight lines.