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Line parallel and perpendicular to a given line is considered one the most difficult concept.
37 Questions around this concept.
The locus of the mid-points of the perpendiculars drawn from points on the line,x=2y to the line x=y is :
If equation represent the same lines then the values of
and
are:
Find the equation of line parallel to $2 x-y+1=0$ and pass through $
(-7,2)$.
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$
\text { Find the equation of line perpendicular to } x-2y+3=0 \text { and passing through }(1,0) \text {. }
$
The vertices of a triangle are A(-1,3), B(-2,2) and C(3,-1). A new triangle is formed by shifting the sides of the triangle by one unit inwards. Then the equation of the side of the new triangle nearest to origin is :
Find the value of d such that line $(x+1) /-5=(y+1) / 2=(z+2) / 1$ lies on the plane $x+y+z=d$
A perpendicular line will be known if we are given
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Line parallel and perpendicular to a given line
1. The equation of the line parallel to $a x+b y+c=0$ is given as $a x+b y+\lambda=0$, where $\lambda$ is some constant.
Equation of the given line is $a x+b y+c=0$
Its slope is $(-a / b)$
So, any equation of line parallel to $a x+b y+c=0$ is
$
\begin{aligned}
& y=\left(-\frac{a}{b}\right) x+c_1 \\
& a x+b y-b c_1=0 \\
& a x+b y+\lambda=0
\end{aligned}
$
2. The equation of the line perpendicular to $a x+b y+c=0$ is given as $b x-a y+\lambda=0$, where $\lambda$ is some constant.
Equation of the given line is $a x+b y+c=0$
Its slope is $(-a / b)$
Slope of perpendicular line will be (b/a)
So, any equation of line perpendicular to $a x+b y+c=0$ is
$
\begin{aligned}
& y=\left(\frac{b}{a}\right) x+c_1 \\
& b x-b y-a c_1=0 \\
& b x-a y+\lambda=0
\end{aligned}
$
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