Question:
The distance between the lines represented by
\[
4x^2+20xy+25y^2+2x+5y-12=0
\]
is equal to
The distance between the lines represented by
\[
4x^2+20xy+25y^2+2x+5y-12=0
\]
is equal to
Show Hint
When the quadratic part is a perfect square, convert the equation into a quadratic in a linear expression. Then factorize to get two parallel lines.
Updated On: Jun 26, 2026
- \(\dfrac{7}{\sqrt{29}}\)
- \(0\)
- \(\dfrac{7}{29}\)
- \(\dfrac{49}{29}\)
Show Solution
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The Correct Option is A
Solution and Explanation
Step 1: Observe the quadratic part.
The given equation is \[ 4x^2+20xy+25y^2+2x+5y-12=0 \] The quadratic part is \[ 4x^2+20xy+25y^2 \] This can be written as \[ (2x+5y)^2 \] because \[ (2x+5y)^2=4x^2+20xy+25y^2 \]
Step 2: Rewrite the full equation.
So the given equation becomes \[ (2x+5y)^2+(2x+5y)-12=0 \]
Step 3: Substitute a new variable.
Let \[ u=2x+5y \] Then the equation becomes \[ u^2+u-12=0 \]
Step 4: Factorize the quadratic equation.
\[ u^2+u-12=0 \] \[ (u+4)(u-3)=0 \] Hence, \[ u=-4 \] or \[ u=3 \]
Step 5: Convert back to equations of lines.
Since \[ u=2x+5y, \] the two lines are \[ 2x+5y=-4 \] and \[ 2x+5y=3 \] That is, \[ 2x+5y+4=0 \] and \[ 2x+5y-3=0 \]
Step 6: Find the distance between the parallel lines.
The distance between two parallel lines \[ ax+by+c_1=0 \] and \[ ax+by+c_2=0 \] is \[ \frac{|c_1-c_2|}{\sqrt{a^2+b^2}} \] Here, \[ a=2,\quad b=5,\quad c_1=4,\quad c_2=-3 \] Therefore, \[ \text{Distance} = \frac{|4-(-3)|}{\sqrt{2^2+5^2}} \] \[ = \frac{7}{\sqrt{4+25}} \] \[ = \frac{7}{\sqrt{29}} \]
Step 7: Final conclusion.
Hence, the distance between the lines is \[ \boxed{\frac{7}{\sqrt{29}}} \]
The given equation is \[ 4x^2+20xy+25y^2+2x+5y-12=0 \] The quadratic part is \[ 4x^2+20xy+25y^2 \] This can be written as \[ (2x+5y)^2 \] because \[ (2x+5y)^2=4x^2+20xy+25y^2 \]
Step 2: Rewrite the full equation.
So the given equation becomes \[ (2x+5y)^2+(2x+5y)-12=0 \]
Step 3: Substitute a new variable.
Let \[ u=2x+5y \] Then the equation becomes \[ u^2+u-12=0 \]
Step 4: Factorize the quadratic equation.
\[ u^2+u-12=0 \] \[ (u+4)(u-3)=0 \] Hence, \[ u=-4 \] or \[ u=3 \]
Step 5: Convert back to equations of lines.
Since \[ u=2x+5y, \] the two lines are \[ 2x+5y=-4 \] and \[ 2x+5y=3 \] That is, \[ 2x+5y+4=0 \] and \[ 2x+5y-3=0 \]
Step 6: Find the distance between the parallel lines.
The distance between two parallel lines \[ ax+by+c_1=0 \] and \[ ax+by+c_2=0 \] is \[ \frac{|c_1-c_2|}{\sqrt{a^2+b^2}} \] Here, \[ a=2,\quad b=5,\quad c_1=4,\quad c_2=-3 \] Therefore, \[ \text{Distance} = \frac{|4-(-3)|}{\sqrt{2^2+5^2}} \] \[ = \frac{7}{\sqrt{4+25}} \] \[ = \frac{7}{\sqrt{29}} \]
Step 7: Final conclusion.
Hence, the distance between the lines is \[ \boxed{\frac{7}{\sqrt{29}}} \]
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