Question

Assertion :The equation of the pair of straight lines passing through the origin and perpendicular to the straight lines $2x^2+5xy+2y^2+10x+5y=0$ is $2x^2-5xy+2y^2=0$ Reason: If $m$ be the slope of a given line, then the slope of a line ${\perp }^{er}$ to it is $-\frac{1}{m}$

• A. Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
• B. Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
• C. Assertion is correct but Reason is incorrect
• D. Both Assertion and Reason are incorrect

Answer 1

The correct option is A Both Assertion and Reason are correct and Reason is the correct explanation for Assertion

We have $2x^2+5xy+2y^2+10x+5y=0....(\ast )$ The homogenous part of $(\ast )$ is $2x^2+5xy+2y^2$ factorizing the homogeneous part, we have

$2x^2+5xy+2y^2=(2x+y)(x+2y)$ so the lines represented by $(\ast )$ be $2x+y+c_1=0$ & $x+2y+c_2=0$ Now line $\perp$ er to $2x+y+c_1=0$ is given by $x-2y+k_1=0$ & & line $\perp$ er to $x+2y+c_2=0$ is given by

$2x-y+k_2=0$ $\therefore$ Combined equation of lines perpendicular to $(\ast )$ is given by

$(x-2y+k_1)(2x-y+k_2)=0$ but lines passes through origin so $k_1=0,k_2=0$ we get $(x-2y)(2x-y)=0\Rightarrow 2x^2-5xy+2y^2=0$ $\therefore$ Assertion (A) is true Now among two perpetrdicular lines if m is the slope of one line then the slope ofa line perpendicular to it is $-\frac{1}{m}$ It is always true