Straight lines practice · Easy

If two vertices of a triangle are $(5,-1)$ and $(-2,3)$, and if its orthocentre lies at the origin, then the coordinates of the third vertex, are

The points $(p+1,1),(2 p+1,3)$ and $(2 p+2,2 p)$ are collinear, if

Let $C$ be the centroid of the triangle with vertices (3,-1) and (1,3) and (2,4). Let $P$ be the point of intersection of the lines $x + 3y - 1 = 0$ and $3x - y + 1 = 0.$ Then the line passing through the points $C$ and $P$ also passes through the point

The distance between $4 x+3 y=11$ and $8 x+6 y=15$, is

The angle between the lines represented by the equation $a x^2+x y+b y^2=0$ will be $45^{\circ}$, if

If the straight line through the point $P(3,4)$ makes an angle $\frac{\pi}{6}$ with the $x$-axis and meets the line $12 x+5 y+10=0$ at $Q$, then the length $P Q$ is $\frac{132}{A \sqrt{3}+B}$. Find $A+B$.

If the point $(p, 5)$ lies on the line parallel to $y$-axis and passing through the intersection of the lines $2\left(a^2+1\right) x+b y+4\left(a^3+a\right)=0$ and $\left(a^2+1\right) x-6 b y+2\left(a^3+a\right)=0$, then $p$ is equal to

If the angle between the lines represented by the equation $y^2+k x y-x^2 \tan ^2 A=0$ be $2 A$, then $k=$

$P\left( {3,\;1} \right),Q\left( {6,5} \right)$ and $R\left( {x,y} \right)$ are three points such that the angle PRQ is a right angle and the area of $\Delta \;RQP = 7,$ then the number of such points R is

The equation of the straight line equally inclined to the coordinate axes and equidistant from the point $(1,-2)$ and $(3,4)$, is

A line with positive rational slope, which passes through the point $(6,0)$ and at a distance of 5 units from the point $(1,3)$ is equal to $\frac{p}{q}$ (where $p, q$ are co-prime), then find the value of $(p+q)$.

A straight line through P(1,2) is such that its intercept between the axes is bisected at P. Its equation is

$A \equiv(-4,0), B \equiv(4,0), M$ and $N$ are variable points on $Y$-axis such that $M$ lies below $N$ and $M N=4$. If the line joining $A M$ and $B N$ intersect at $P$, then locus of $P$ is

The straight line through the point of intersection of $a x+b y+c=0$ and $a^{\prime} x+b^{\prime} y+c^{\prime}=0$ are parallel to the $y$-axis has the equation

Point on line $x+y=4$ which is at a unit distance from the line $4 x+3 y-11=0$ is

The coordinates of the foot of the perpendicular from the point (2,4) on the line $x + y = 1$ are

Let $A$ be the fixed point $(0,4)$ and $B(2 t, 0)$ be a moving point. Let $M$ be the mid-point of $A B$ and let the perpendicular bisector of $A B$ meet the $y$-axis at $R$. The locus of the mid-point $P$ of $M R$, is

The image of the origin with reference to the line $4x + 3y - 25 = 0$ is

The equation of the line bisecting perpendicularly the segment joining the points (-4,6) and (8,8) is

The distance between the lines $4\;x + 3\;y = 11$ and $8\;x + 6\;y = 15,$ is

The intercept made by a line on $Y$-axis is double to the intercept made by it on $X$-axis, if it passes through $(1,2)$, then its equation is

The lines $x+y=|a|$ and $a x-y=1$ intersect each other in the first quadrant. Then the set of all possible values of a is the interval:

A line forms a triangle of area $54 \sqrt{3}$ sq-units with the coordinate axes. If the perpendicular drawn from the origin to the line makes an angle of $60^{\circ}$ with the $x$-axis, then the equation of the line is

If $p\,\mathrm{and}\,q$ are the lengths of the perpendiculars from the origin on the lines, $x\,\cos ec\,\alpha \, - \,y\,\sec \,\alpha \, = \,k\,\cot \,2\alpha \,\mathrm{and}\,x\,\sin \,\alpha \, + \,y\,\cos \,\alpha \, = \,k\,\sin \,2\alpha ,$ respectively then ${k^2}$ is equal to

The equation of straight line equally inclined to the axes and equidistance from the points $\left( {1,\; - 2} \right)$ and (3,4) is $ax + by + c = 0$, where

If the lines $3x + 4y + 1 = 0,\;5x + \lambda y + 3 = 0$ and $2x + y - 1 = 0$ are concurrent, then $\lambda$ is equal to

The equation of the bisectors of the angle between lines represented by equation $4 x^2-16 x y-7 y^2=0$ is

Let $k$ be an integer such that the triangle with vertices $(k,-3 k),(5, k)$ and $(-k, 2)$ has area 28 sq. units. Then the orthocentre of this triangle is at the point:

Two sides of an isosceles triangle are given by the equation $7x - y + 3 = 0$ and $x + y - 3 = 0.$ If its third side passes through the point (1,-10), then its equations are

The vertices of a family of triangles have integer coordinates. If two of the vertices of all the triangles are (0, 0) and (6, 8), then the least value of areas of the triangles is