Straight Lines ...

Slope of a line is not defined if the line is

The distance of the point (α,β) from X axis is

Projection (the foot of perpendicular) from ( x , y ) on the x – axis is

Slope of any line parallel to X axis is

The straight lines x + y = 0 , 3x - y – 4 = 0 , x + 3y – 4 = 0 form a triangle which is

The angle between line:  $$r[2 \cos \theta  + 5\sin\theta ] = 3$$ and $$r[2 \sin \theta  - 5\cos\theta ]+ 4 = 0$$ is:

The lines $$( {a}+ {b}) {x}+( {a}- {b}) {y}-2 {a} {b}= 0,\ ( {a}- {b}) {x}+( {a}+ {b}) {y}-2 {a} {b}=0$$ and $$ {x}+ {y}=0$$ forms an isosceles triangle whose vertical angle is

The angle between the lines $$x\cos \alpha+y\sin\alpha={p}_{1}$$ and $$x\cos \beta +y\sin \beta=p_{2},$$ where $$\alpha>\beta$$ is

If the slope of the line passing through the points $$(2, \sin\theta)$$ and $$(1,\cos\theta)$$ is $$0,$$ then the general solution of $$\theta$$, is

$$A(1,3)$$ and $$B(7,5)$$ are two opposite vertices of a square. The equation of a side through $$A$$ is

If the angle between the lines $$ {k} {x}- {y}+6=0,\ 3 {x}-5 {y}+7=0$$ is $$\displaystyle \frac{\pi}{4},$$ then one of the value of $$ {k}=$$ 

The angle between the lines $$kx+y+9=0$$, $$y-3x =4$$ is $$45^{o},$$ then the value of $$k$$ is :

The angle between the lines $$r \cos(\theta -\alpha )= p, r\sin(\theta -\alpha )= q $$ is

The fourth vertex $$D$$ of a parallelogram $$ABCD$$ whose three vertices are $$A (2, 3), B (6, 7)$$ and $$C (8, 3)$$ is

Find the area of triangle having vertices are $$A (3, 1), B (12, 2)$$ and $$C (0, 2)$$.