NIMCET 2007 — Mathematics PYQ

Formula and Concept

The standard equation for a pair of straight lines passing through the origin is given by:

$$a{x}^2+2hxy+b{y}^2=0$$

The length of the product of perpendiculars drawn from any point $(x_1,y_1)$ to this pair of straight lines is given by the standard formula:

$$P=\frac{|a{x}_1^2+2h{x}_1{y}_1+b{y}_1^2|}{\sqrt{(a-b{)}^2+(2h{)}^2}}$$

Step 1: Identify the Coefficients

Compare the given pair of straight lines equation $2x^2+6xy+y^2=0$ with the standard form:

• $a=2$

• $2h=6$

• $b=1$

The given point is $(x_1,y_1)=(2,-1)$.

Step 2: Evaluate the Numerator

Substitute the point $(2,-1)$ directly into the expression $a{x}_1^2+2h{x}_1{y}_1+b{y}_1^2$:

$$\text{Numerator}=|2(2{)}^2+6(2)(-1)+1(-1{)}^2|$$

$$\text{Numerator}=|2(4)-12+1(1)|$$

$$\text{Numerator}=|8-12+1|$$

$$\text{Numerator}=|-3|=3$$

Step 3: Evaluate the Denominator

Substitute the values of $a$, $b$, and $2h$ into the denominator expression $\sqrt{(a-b{)}^2+(2h{)}^2}$:

$$\text{Denominator}=\sqrt{(2-1{)}^2+(6{)}^2}$$

$$\text{Denominator}=\sqrt{(1{)}^2+36}$$

$$\text{Denominator}=\sqrt{1+36}=\sqrt{37}$$

Step 4: Calculate the Product of Perpendiculars ($P$)

Combine the evaluated numerator and denominator:

$$P=\frac{3}{\sqrt{37}}$$

Conclusion

The product of the perpendiculars from the point $(2,-1)$ to the given pair of straight lines is $\frac{3}{\sqrt{37}}$.

Correct Option: (d) $\frac{3}{\sqrt{37}}$