Find the equation of the straight line which passes through the point \((1, -2)\) and cuts off equal intercepts from axes.
\(x + y + 1 = 0\)
Find the equation of the line passing through the point \((5, 2)\) and perpendicular to the line joining the points \((2, 3)\) and \((3, -1)\).
\(x - 4y + 3 = 0\)
Find the angle between the lines \(y = (2 - \sqrt{3})(x + 5)\) and \(y = (2 + \sqrt{3})(x - 7)\).
60° or 120°
Find the equation of the lines which pass through the point \((3, 4)\) and cut off intercepts from the coordinate axes such that their sum is 14.
\(x + y = 7\)
or
\(\dfrac{x}{6} + \dfrac{y}{8} = 1\)
Find the points on the line \(x + y = 4\) which lie at a unit distance from the line \(4x + 3y = 10\).
\((3, 1), (-7, 11)\)
Show that the tangent of an angle between the lines \(\dfrac{x}{a} + \dfrac{y}{b} = 1\) and \(\dfrac{x}{a} - \dfrac{y}{b} = 1\) is \(\dfrac{2ab}{a^2 - b^2}\).
\(\dfrac{2ab}{a^2 - b^2}\)
Find the equation of lines passing through \((1, 2)\) and making angle 30° with the y-axis.
\(y - \sqrt{3}x - 2 + \sqrt{3} = 0\)
Find the equation of the line passing through the point of intersection of \(2x + y = 5\) and \(x + 3y + 8 = 0\) and parallel to the line \(3x + 4y = 7\).
\(3x + 4y + 3 = 0\)
For what values of \(a\) and \(b\) are the intercepts cut off on the coordinate axes by the line \(ax + by + 8 = 0\) equal in length but opposite in signs to those cut off by the line \(2x - 3y + 6 = 0\)?
\(a = -\dfrac{8}{3},\ b = 4\)
If the intercept of a line between the coordinate axes is divided by the point \((-5, 4)\) in the ratio 1 : 2, then find the equation of the line.
\(8x - 5y + 60 = 0\)
Find the equation of a straight line on which the length of the perpendicular from the origin is four units and the line makes an angle of 120° with the positive direction of the x-axis.
\(\sqrt{3}x + y = 8\)
Find the equation of one of the sides of an isosceles right-angled triangle whose hypotenuse is given by \(3x + 4y = 4\) and the opposite vertex of the hypotenuse is \((2, 2)\).
\(x - 7y - 12 = 0\)
If the equation of the base of an equilateral triangle is \(x + y = 2\) and the vertex is \((2, -1)\), find the length of the side of the triangle.
The distance from the vertex \((2, -1)\) to the line \(x + y = 2\) is the altitude of the equilateral triangle.
Using the perpendicular distance formula:
\[ d = \dfrac{|2 + (-1) - 2|}{\sqrt{1^2 + 1^2}} = \dfrac{1}{\sqrt{2}} \]
In an equilateral triangle, altitude \(h = \dfrac{\sqrt{3}}{2} a\), where \(a\) is the side.
So, \( \dfrac{\sqrt{3}}{2} a = \dfrac{1}{\sqrt{2}} \Rightarrow a = \sqrt{\dfrac{2}{3}} \).
Therefore, the side length is \( \sqrt{\dfrac{2}{3}} \).
A variable line passes through a fixed point \(P\). The algebraic sum of the perpendiculars drawn from the points \((2,0)\), \((0,2)\), and \((1,1)\) on the line is zero. Find the coordinates of point \(P\).
Let the line through \(P(x_1,y_1)\) be \(y - y_1 = m(x - x_1)\).
Convert to normal form and compute perpendicular distances from \((2,0)\), \((0,2)\), and \((1,1)\).
Using the condition that the algebraic sum of perpendiculars is zero gives:
\[ (2 - x_1)m - (y_1) + (0 - x_1)m + (2 - y_1) + (1 - x_1)m + (1 - y_1) = 0 \]
Simplifying yields \(x_1 = 1, y_1 = 1\).
Thus, the fixed point is \((1,1)\).
In what direction should a line be drawn through the point \((1, 2)\) so that its point of intersection with the line \(x + y = 4\) is at a distance \(\dfrac{\sqrt{6}}{3}\) from the given point?
Let the required line through \((1,2)\) make angle \(\theta\) with the positive x-axis.
Its parametric form is:
\[ (x, y) = (1, 2) + t(\cos\theta, \sin\theta). \]
Substitute into \(x + y = 4\):
\[ 1 + 2 + t(\cos\theta + \sin\theta) = 4 \Rightarrow t = \dfrac{1}{\cos\theta + \sin\theta}. \]
The distance from \((1,2)\) to the intersection point is:
\[ |t| = \dfrac{1}{\cos\theta + \sin\theta} = \dfrac{\sqrt{6}}{3}. \]
Solving this gives \(\theta = 15^\circ\) or \(75^\circ\).
A straight line moves so that the sum of the reciprocals of its intercepts on the coordinate axes is constant. Show that the line passes through a fixed point.
The intercept form of the line is:
\[ \dfrac{x}{a} + \dfrac{y}{b} = 1. \]
Given that
\[ \dfrac{1}{a} + \dfrac{1}{b} = k, \]
a constant.
Rewrite the line equation:
\[ bx + ay = ab. \]
Using \(ab(\dfrac{1}{a} + \dfrac{1}{b}) = abk\):
\[ b + a = abk. \]
Hence the equation becomes:
\[ bx + ay = a + b. \]
Dividing both sides by \(k\):
\[ b(x - k) + a(y - k) = 0. \]
This holds for every valid choice of \(a\) and \(b\), implying all such lines pass through the fixed point:
\[ (k, k). \]
Find the equation of the line which passes through the point \((-4, 3)\) and the portion of the line intercepted between the axes is divided internally in the ratio \(5:3\) by this point.
Let the line intersect the axes at \((a,0)\) and \((0,b)\).
The point \((-4,3)\) divides the intercepts in the ratio \(5:3\):
\[ \dfrac{-4}{a} = \dfrac{5}{5+3} = \dfrac{5}{8}, \quad \dfrac{3}{b} = \dfrac{3}{8}. \]
Thus, \(a = -\dfrac{32}{5}\) and \(b = 8\).
The intercept form of the line becomes:
\[ \dfrac{x}{-32/5} + \dfrac{y}{8} = 1. \]
Simplifying yields the equation:
\[ 9x - 20y + 96 = 0. \]
Find the equations of the lines through the point of intersection of the lines \(x - y + 1 = 0\) and \(2x - 3y + 5 = 0\) and whose distance from the point \((3,2)\) is \(\dfrac{7}{5}\).
Let the required line be:
\[ (x - y + 1) + \lambda (2x - 3y + 5) = 0. \]
Simplify to obtain:
\[ (1 + 2\lambda)x + (-1 - 3\lambda)y + (1 + 5\lambda) = 0. \]
Apply perpendicular distance formula to point \((3,2)\) and equate to \(\dfrac{7}{5}\).
Solving for \(\lambda\) produces two values, giving two lines:
\[ 3x - 4y + 6 = 0, \]
\[ 4x - 3y + 1 = 0. \]
If the sum of the distances of a moving point in a plane from the axes is 1, find the locus of the point.
The distance of \((x,y)\) from the x-axis is \(|y|\), and from the y-axis is \(|x|\).
Given:
\[ |x| + |y| = 1. \]
This is the equation of a square of side 1, oriented at 45° with respect to the coordinate axes.
\(P_1, P_2\) are points on either of the two lines \(y - \sqrt{3}|x| = 2\) at a distance of 5 units from their point of intersection. Find the coordinates of the foot of perpendiculars drawn from \(P_1, P_2\) on the bisector of the angle between the given lines.
The lines are:
\[ y = \sqrt{3}x + 2 \quad (x \ge 0), \]
\[ y = -\sqrt{3}x + 2 \quad (x < 0). \]
Their intersection is at \((0,2)\).
Points \(P_1, P_2\) lie 5 units away from \((0,2)\) on these lines.
The angle bisector is the y-axis. The foot of perpendicular from \((x_1,y_1)\) on the y-axis has x-coordinate 0 and y-coordinate equal to the point’s y-value.
The y-coordinates are:
\[ y = 2 \pm 5\cos 30^\circ = 2 \pm \dfrac{5\sqrt{3}}{2}. \]
Thus the feet of perpendiculars are:
\[ \left(0, 2 + \dfrac{5\sqrt{3}}{2}\right), \, \left(0, 2 - \dfrac{5\sqrt{3}}{2}\right). \]
If \(p\) is the length of perpendicular from the origin on the line \(\dfrac{x}{a} + \dfrac{y}{b} = 1\) and \(a^2, p^2, b^2\) are in A.P., show that \(a^4 + b^4 = 0\).
The perpendicular distance from origin to the line \(\dfrac{x}{a} + \dfrac{y}{b} = 1\) is:
\[ p = \dfrac{ab}{\sqrt{a^2 + b^2}}. \]
Given \(a^2, p^2, b^2\) in A.P., so:
\[ 2p^2 = a^2 + b^2. \]
Substitute the expression for \(p^2\):
\[ 2 \left( \dfrac{a^2 b^2}{a^2 + b^2} \right) = a^2 + b^2. \]
Cross-multiplying gives:
\[ 2a^2 b^2 = (a^2 + b^2)^2. \]
Expanding the right-hand side:
\[ 2a^2 b^2 = a^4 + 2a^2 b^2 + b^4. \]
Simplifying:
\[ a^4 + b^4 = 0. \]
A line cutting off intercept \(-3\) from the y-axis and the tangent of angle to the x-axis is \(\tfrac{3}{?}\), its equation is
5y - 3x + 15 = 0
3y - 5x + 15 = 0
5y - 3x - 15 = 0
None of these
Slope of a line which cuts off intercepts of equal lengths on the axes is
-1
0
2
\(\sqrt{3}\)
The equation of the straight line passing through the point \((3,2)\) and perpendicular to the line \(y = x\) is
x - y = 5
x + y = 5
x + y = 1
x - y = 1
The equation of the line passing through the point \((1,2)\) and perpendicular to the line \(x + y + 1 = 0\) is
y - x + 1 = 0
y - x - 1 = 0
y - x + 2 = 0
y - x - 2 = 0
The tangent of angle between the lines whose intercepts on the axes are \(a, -b\) and \(b, -a\), respectively, is
\(\dfrac{a^2 - b^2}{ab}\)
\(\dfrac{b^2 - a^2}{2}\)
\(\dfrac{b^2 - a^2}{2ab}\)
None of these
If the line \(\dfrac{x}{a} + \dfrac{y}{b} = 1\) passes through the points \((2,-3)\) and \((4,-5)\), then \((a,b)\) is
(1, 1)
(-1, 1)
(1, -1)
(-1, -1)
The distance of the point of intersection of the lines \(2x - 3y + 5 = 0\) and \(3x + 4y = 0\) from the line \(5x - 2y = 0\) is
\(\dfrac{130}{17\sqrt{29}}\)
\(\dfrac{13}{7\sqrt{29}}\)
\(\dfrac{130}{7}\)
None of these
The equations of the lines which pass through the point \((3,-2)\) and are inclined at \(60^\circ\) to the line \(\sqrt{3}\,x + y = 1\) is
\(y + 2 = 0\) and \(\sqrt{3}x - y - 2 - 3\sqrt{3} = 0\)
\(x - 2 = 0\) and \(\sqrt{3}x - y + 2 + 3\sqrt{3} = 0\)
\(\sqrt{3}x - y - 2 - 3\sqrt{3} = 0\)
None of these
The equations of the lines passing through the point \((1,0)\) and at a distance \(\dfrac{\sqrt{3}}{2}\) from the origin are
\(\sqrt{3}x + y - \sqrt{3} = 0\), \(\sqrt{3}x - y - \sqrt{3} = 0\)
\(\sqrt{3}x + y + \sqrt{3} = 0\), \(\sqrt{3}x - y + \sqrt{3} = 0\)
x + \sqrt{3}y - \sqrt{3} = 0, \; x - \sqrt{3}y - \sqrt{3} = 0
None of these
The distance between the lines \(y = mx + c_1\) and \(y = mx + c_2\) is
\(\dfrac{c_1 - c_2}{\sqrt{m^2 + 1}}\)
\(\dfrac{|c_1 - c_2|}{\sqrt{1 + m^2}}\)
\(\dfrac{c_2 - c_1}{\sqrt{1 + m^2}}\)
0
The coordinates of the foot of perpendiculars from the point \((2,3)\) on the line \(y = 3x + 4\) is given by
\(\left(\dfrac{37}{10}, -\dfrac{1}{10}\right)\)
\(\left(-\dfrac{1}{10}, \dfrac{37}{10}\right)\)
\(\left(\dfrac{10}{37}, -10\right)\)
\(\dfrac{2}{3}, -\dfrac{1}{3}\)
If the coordinates of the middle point of the portion of a line intercepted between the coordinate axes is \((3,2)\), then the equation of the line will be
2x + 3y = 12
3x + 2y = 12
4x - 3y = 6
5x - 2y = 10
Equation of the line passing through \((1,2)\) and parallel to the line \(y = 3x - 1\) is
y + 2 = x + 1
y + 2 = 3(x + 1)
y - 2 = 3(x - 1)
y - 2 = x - 1
Equations of diagonals of the square formed by the lines \(x = 0\), \(y = 0\), \(x = 1\) and \(y = 1\) are
y = x, \; y + x = 1
y = x, \; x + y = 2
2y = x, \; y + x = \dfrac{1}{3}
y = 2x, \; y + 2x = 1
For specifying a straight line, how many geometrical parameters should be known?
1
2
4
3
The point \((4,1)\) undergoes the following two successive transformations:
(i) Reflection about the line \(y = x\)
(ii) Translation through a distance 2 units along the positive x-axis
Then the final coordinates of the point are
(4, 3)
(3, 4)
(1, 4)
\(\left(\tfrac{7}{2},\tfrac{7}{2}\right)\)
A point equidistant from the lines \(4x + 3y + 10 = 0\), \(5x - 12y + 26 = 0\) and \(7x + 24y - 50 = 0\) is
(1, -1)
(1, 1)
(0, 0)
(0, 1)
A line passes through \((2,2)\) and is perpendicular to the line \(3x + y = 3\). Its y-intercept is
\(\dfrac{1}{3}\)
\(\dfrac{2}{3}\)
1
\(\dfrac{4}{3}\)
The ratio in which the line \(3x + 4y + 2 = 0\) divides the distance between the lines \(3x + 4y + 5 = 0\) and \(3x + 4y - 5 = 0\) is
1 : 2
3 : 7
2 : 3
2 : 5
One vertex of the equilateral triangle with centroid at the origin and one side as \(x + y - 2 = 0\) is
(-1, -1)
(2, 2)
(-2, -2)
(2, -2)
If \(a, b, c\) are in A.P., then the straight lines \(ax + by + c = 0\) will always pass through ____.
(1, -2)
The line which cuts off equal intercept from the axes and passes through the point \((1, -2)\) is ____.
x + y + 1 = 0
Equations of the lines through the point \((3, 2)\) and making an angle of \(45^\circ\) with the line \(x - 2y = 3\) are ____.
3x - y - 7 = 0, x + 3y - 9 = 0
The points \((3, 4)\) and \((2, -6)\) are situated on the ____ of the line \(3x - 4y - 8 = 0\).
opposite sides
A point moves so that square of its distance from the point \((3, -2)\) is numerically equal to its distance from the line \(5x - 12y = 3\). The equation of its locus is ____.
13(x^2 + y^2) - 83x + 64y + 182 = 0
Locus of the mid-points of the portion of the line \(x \, \sin\theta + y \, \cos\theta = p\) intercepted between the axes is ____.
4x^2 y^2 = p^2 (x^2 + y^2)
If the vertices of a triangle have integral coordinates, then the triangle can not be equilateral.
True
The points A \((-2,1)\), B \((0,5)\), C \((-1,2)\) are collinear.
False
Equation of the line passing through the point \((a\cos^3\theta, a\sin^3\theta)\) and perpendicular to the line \(x \sec\theta + y \csc\theta = a\) is \(x \cos\theta - y \sin\theta = a \sin 2\theta\).
False
The straight line \(5x + 4y = 0\) passes through the point of intersection of the straight lines \(x + 2y - 10 = 0\) and \(2x + y + 5 = 0\).
True
The vertex of an equilateral triangle is \((2,3)\) and the equation of the opposite side is \(x + y = 2\). Then the other two sides are \(y - 3 = (2 \pm \sqrt{3})(x - 2)\).
True
The equation of the line joining the point \((3,5)\) to the point of intersection of the lines \(4x + y - 1 = 0\) and \(7x - 3y - 35 = 0\) is equidistant from the points \((0,0)\) and \((8,34)\).
True
The line \(\dfrac{x}{a} + \dfrac{y}{b} = 1\) moves in such a way that \(\dfrac{1}{a^2} + \dfrac{1}{b^2} = \dfrac{1}{c^2}\), where \(c\) is a constant. The locus of the foot of the perpendicular from the origin on the given line is \(x^2 + y^2 = c^2\).
True
The lines \(ax + 2y + 1 = 0\), \(bx + 3y + 1 = 0\) and \(cx + 4y + 1 = 0\) are concurrent if \(a, b, c\) are in G.P.
False
Line joining the points \((3,-4)\) and \((-2,6)\) is perpendicular to the line joining the points \((-3,6)\) and \((9,-18)\).
False
Match the items in Column A with Column B using the table below.
| Column A | Column B |
|---|---|
The coordinates of the points P and Q on the line \(x + 5y = 13\) which are at a distance of 2 units from the line \(12x - 5y + 26 = 0\) | (3, 1), (-7, 11) |
The coordinates of the point on the line \(x + y = 4\), which are at a unit distance from the line \(4x + 3y - 10 = 0\) | \(\left(\dfrac{1}{3}, \dfrac{11}{3}\right), \left(\dfrac{4}{3}, \dfrac{7}{3}\right)\) |
The coordinates of the point on the line joining A(-2,5) and B(3,1) such that AP = PQ = QB | \(\left(1, \dfrac{12}{5}\right), \left(-3, \dfrac{16}{5}\right)\) |
| Column A | Matched Item from Column B |
|---|---|
The coordinates of the points P and Q on the line \(x + 5y = 13\) which are at a distance of 2 units from the line \(12x - 5y + 26 = 0\) | \(\left(1, \dfrac{12}{5}\right), \left(-3, \dfrac{16}{5}\right)\) |
The coordinates of the point on the line \(x + y = 4\), which are at a unit distance from the line \(4x + 3y - 10 = 0\) | (3, 1), (-7, 11) |
The coordinates of the point on the line joining A(-2,5) and B(3,1) such that AP = PQ = QB | \(\left(\dfrac{1}{3}, \dfrac{11}{3}\right), \left(\dfrac{4}{3}, \dfrac{7}{3}\right)\) |
Match the items in Column A with Column B using the table below.
| Column A | Column B |
|---|---|
Parallel to y-axis | \(\lambda = -\dfrac{3}{4}\) |
Perpendicular to \(7x + y - 4 = 0\) | \(\lambda = \dfrac{1}{3}\) |
Passes through (1,2) | \(\lambda = \dfrac{17}{41}\) |
Parallel to x-axis | \(\lambda = 3\) |
| Column A | Matched Item from Column B |
|---|---|
Parallel to y-axis | \(\lambda = 3\) |
Perpendicular to \(7x + y - 4 = 0\) | \(\lambda = -\dfrac{3}{4}\) |
Passes through (1,2) | \(\lambda = \dfrac{1}{3}\) |
Parallel to x-axis | \(\lambda = \dfrac{17}{41}\) |
Match the items in Column A with Column B using the table below.
| Column A | Column B |
|---|---|
Through the point (2,1) | 2x - y = 4 |
Perpendicular to the line \(x + 2y + 1 = 0\) | x + y - 5 = 0 |
Parallel to the line \(3x - 4y + 5 = 0\) | x - y - 1 = 0 |
Equally inclined to the axes | 3x - 4y - 1 = 0 |
| Column A | Matched Item from Column B |
|---|---|
Through the point (2,1) | x - y - 1 = 0 |
Perpendicular to the line \(x + 2y + 1 = 0\) | 2x - y = 4 |
Parallel to the line \(3x - 4y + 5 = 0\) | 3x - 4y - 1 = 0 |
Equally inclined to the axes | x + y - 5 = 0 |