The lines p(p^2 + 1)x - y + q = 0 and (p^2 + | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) If two lines are perpendicular to a common line,they must be parallel to each other.The slope of the first line $m_1$ is given by $m_1 = p(p^2 + 1

Vedclass · Accepted Answer

For exactly one value of $p$

Vedclass · Answer

(D) If two lines are perpendicular to a common line,they must be parallel to each other.The slope of the first line $m_1$ is given by $m_1 = p(p^2 + 1)$.The slope of the second line $m_2$ is given by $m_2 = -\frac{(p^2 + 1)^2}{p^2 + 1} = -(p^2 + 1)$.For the lines to be parallel,$m_1 = m_2$.$p(p^2 + 1) = -(p^2 + 1)$.Since $p^2 + 1 
eq 0$ for any real $p$,we can divide both sides by $(p^2 + 1)$:$p = -1$.Thus,the lines are perpendicular to a common line for exactly one value of $p$.

The lines $p(p^2 + 1)x - y + q = 0$ and $(p^2 + 1)^2x + (p^2 + 1)y + 2q = 0$ are perpendicular to a common line for which value of $p$?

Similar Questions

The lines $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$ are perpendicular to each other,if

Consider the two curves $y = 2x$ and $x^2 - xy + 2y^2 = 28$. The absolute value of the tangent of the angle between the two curves at the points where they meet is:

If straight lines $\alpha^2x + \alpha y = 9$ and $3x + 2y = 5$ are perpendicular,then the value of $\alpha$ is

$A$ straight line $L$ through the point $(3,-2)$ is inclined at an angle of $60^{\circ}$ to the line $\sqrt{3} x+y=1$. If $L$ also intersects the $X$-axis,then the equation of $L$ is

The obtuse angle between the lines $y = -2$ and $y = x + 2$ is .....$^o$

Vedclass Test Series

Exam Paper Generator

Online Exam Module