Straight Lines Questions (433)

Given the points A(0,4) and B(0,-4), the equation of the locus of the point P such that |AP - BP| = 6 is -

If L is the line whose equation is \(ax + by = c\). Let M be the reflection of L through the y-axis, and let N be the reflection of L through the x-axis. Which of the following must be true about M and N for all choices of a, b and c?

The equation of locus of the foot of perpendicular drawn from $(5,6)$ on the family of lines $(x - 2) + \lambda(y - 3) = 0$ (where $\lambda \in \mathbb{R}$) is

A ray of light coming from the point $P(1,2)$ gets reflected from the point $Q$ on the $x$-axis and then passes through the point $R(4,3)$. If the point $S(h,k)$ is such that $PQRS$ is a parallelogram, then $hk^2$ is equal to:

Let C be the centroid of the triangle with vertices (3, -1), (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:

If the image of the point $(-4,5)$ in the line $x+2y=2$ lies on the circle $(x+4)^2+(y-3)^2=r^2$, then $r$ is equal to:

Given that lines \(x + (a-1)y = 1\) and \(2x + a^2 y = 1\) are perpendicular, then the product of slopes of lines is \(-1\). Find the distance from the origin to the point of intersection of the lines (taking the real value of \(a\)).

Equations of the sides of the triangle having $(3,-1)$ as a vertex, $x - 4y + 10 = 0$ and $6x + 10y - 59 = 0$ being the equations of an angle bisector and a median respectively drawn from different vertices. The equation of one side is:

The equation of a pair of straight lines is $ax^2 + 2hxy + by^2 = 0$. If the angle by which the axes be rotated so the term containing $xy$ in the equation may be removed is $\phi$, then:

The equation of straight lines passing through ordered pairs $(a,b)$ satisfying equation $\sec^2((a+1)b) + a^2 - 1 = 0$, and having slope $\frac{1}{2}$, is (are):

$A(1,2)$ and $B(7,10)$ are two points. If $P(x,y)$ is a point such that the angle $APB$ is $60°$ and the area of $\triangle APB$ is maximum, then which of given is (are) true?

A line which makes an acute angle $\theta$ with the positive direction of $x$-axis is drawn through the point $P(3,4)$ to meet the line $x = 6$ at $R$ and $y = 8$ at $S$, then:

The equation of the bisectors of the angles between the two intersecting lines $\frac{x-3}{\cos\theta} = \frac{y+5}{\sin\theta}$ and $\frac{x-3}{\cos\phi} = \frac{y+5}{\sin\phi}$ are $\frac{x-3}{\cos\alpha} = \frac{y+5}{\sin\alpha}$ and $\frac{x-3}{\cos\alpha} = \frac{y+5}{\sin\alpha}$, then:

Two roads are represented by the equation $y - x = 6$ and $x + y = 8$. An inspection bunglow has to be so constructed that it is at a distance of 100 from each of the roads. Possible location of the bunglow is given by:

Let $L_1: 3x + 4y = 1$ and $L_2: 5x - 12y + 2 = 0$ be two given lines. Let image of every point on $L_1$ with respect to a line $L$ lies on $L_2$ then possible equation of $L$ can be:

Let $A(1,1)$ and $B(3,3)$ be two fixed points and $P$ be a variable point such that area of $\triangle PAB$ remains constant equal to 1 for all position of $P$, then locus of $P$ is given by:

The vertices of a triangle are $(1,3), (5,0)$ and $(-1,2)$. Which of the following inequalities will be satisfied by all points lying inside the triangle?

$A(x_1, y_1), B(x_2, y_2), (y_1 < y_2)$ are two points on the line $x + y = 4$ from which perpendicular $AQ$ and $BP$ are drawn on line $4x + 3y = 10$ where $P$ and $Q$ are the feet of perpendicular such that $AQ = BP = 1$. Now considering $AB$ as diameter, a circle is drawn which meets the line $4x + 3y = 10$ at $C$ and $D$ such that $C$ is closer to $P$. Then which of the following statement(s) is correct?

If the two lines represented by $x^2\left(\tan^2 \theta + \cos^2 \theta\right) - 2y \tan \theta + y^2 \sin^2 \theta = 0$ make angles $\alpha, \beta$ with the x-axis, then:

If P((1,2)), Q((4,6)), R((5,7)) & S((a,b)) are the vertices of a parallelogram PQRS, then :

2.35

506. The equations of the sides of a triangle having \((4, -1)\) as a vertex, if the lines \(x - 1 = 0\) and \(x - y - 1 = 0\) are the equations of two internal bisectors of its angles, are:

Perpendicular distance of the line AB from the point (2, 2) is :

Two sides of a rhombus are along the lines, x - y + 1 = 0 and 7x - y - 5 = 0. If its diagonals intersect at (-1, -2), then which one of the following is a vertex of this rhombus?

In a \(\triangle\)ABC, if A (2 ,-1), B (-3, -2) and equation of angle bisector of \(\angle B\) of triangle is 3x - 2y + 5 = 0, then equation of BC is :

Perpendicular distance of the line AB from the point (2, 2) is -

The diagonals of a parallelogram PQRS are along the lines x + 3y = 4 and 6x - 2y = 7. Then PQRS must be a :

Lines must be parallel. If p(p2 + 1) = −(p2 + 1), then the value of p is:

Let \(L\) be the line passing through the point \(P(1, 2)\) such that its intercepted segment between the co-ordinate axes is bisected at \(P\). If \(L_1\) is the line perpendicular to \(L\) and passing through the point \((-2, 1)\), then the point of intersection of \(L\) and \(L_1\) is

Let \(A \equiv (1, \sqrt{3})\), \(B \equiv (0, 0)\), \(C \equiv (2, 0)\). The incentre of \(\triangle ABC\) is:

The sum of the coordinates of the point P if PDOC is a square is:

Let \(P = (-1,\, 0)\), \(Q = (0,\, 0)\) and \(R = (3,\, 3\sqrt{3})\) be three points. The equation of the bisector of the angle PQR is

For no solution, the lines represented by the system must be parallel. Given the system of equations with parameter k, the condition for no solution gives \(k^2 - 4k + 3 = 0\), so \(k = 1, 3\). Since \(k = 1\) gives coincident lines, how many values of k give no solution?