Q1
Draw a quadrilateral in the Cartesian plane, whose vertices are (– 4, 5), (0, 7), (5, – 5) and (– 4, –2). Also, find its area.
(Paste image URL for the quadrilateral plot here)
Answer:
Let the vertices be \( A(-4, 5) \), \( B(0, 7) \), \( C(5, -5) \), and \( D(-4, -2) \).
To find the area, we can split the quadrilateral into two triangles, \( \Delta ABD \) and \( \Delta BCD \), by joining the diagonal \( BD \).
Area of \( \Delta ABD \):
Vertices: \( (-4, 5), (0, 7), (-4, -2) \).
\( \text{Area} = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)| \)
\( = \frac{1}{2} | -4(7 - (-2)) + 0(-2 - 5) + (-4)(5 - 7) | \)
\( = \frac{1}{2} | -4(9) + 0 + (-4)(-2) | \)
\( = \frac{1}{2} | -36 + 8 | = \frac{1}{2} |-28| = 14 \text{ sq units} \).
Area of \( \Delta BCD \):
Vertices: \( (0, 7), (5, -5), (-4, -2) \).
\( \text{Area} = \frac{1}{2} | 0(-5 - (-2)) + 5(-2 - 7) + (-4)(7 - (-5)) | \)
\( = \frac{1}{2} | 0 + 5(-9) - 4(12) | \)
\( = \frac{1}{2} | -45 - 48 | = \frac{1}{2} |-93| = 46.5 \text{ sq units} \).
Total Area = \( 14 + 46.5 = \mathbf{60.5} \) sq units.
Q2
The base of an equilateral triangle with side \( 2a \) lies along the y-axis such that the mid-point of the base is at the origin. Find vertices of the triangle.
(Paste image URL for the triangle diagram here)
Answer:
The base is along the y-axis with mid-point at origin \((0,0)\). The total length of the base is \(2a\).
So, the vertices on the y-axis are \( A(0, a) \) and \( B(0, -a) \).
The third vertex \( C \) lies on the x-axis (since it's an equilateral triangle symmetric about the base). Let its coordinates be \( (x, 0) \).
In \( \Delta AOC \) (where O is origin), \( AC^2 = AO^2 + OC^2 \).
Since side length is \( 2a \), \( AC = 2a \) and \( AO = a \).
\( (2a)^2 = a^2 + x^2 \)
\( 4a^2 = a^2 + x^2 \Rightarrow x^2 = 3a^2 \Rightarrow x = \pm \sqrt{3}a \).
So, the vertices are \( (0, a), (0, -a) \) and \( (\sqrt{3}a, 0) \) OR \( (0, a), (0, -a) \) and \( (-\sqrt{3}a, 0) \).
Q3
Find the distance between \( P(x_1, y_1) \) and \( Q(x_2, y_2) \) when: (i) PQ is parallel to the y-axis, (ii) PQ is parallel to the x-axis.
Answer:
(i) PQ is parallel to y-axis:
If a line is parallel to the y-axis, the x-coordinates of all points on it are the same.
So, \( x_1 = x_2 \).
Distance \( PQ = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} = \sqrt{0 + (y_2 - y_1)^2} = |y_2 - y_1| \).
(ii) PQ is parallel to x-axis:
If a line is parallel to the x-axis, the y-coordinates of all points on it are the same.
So, \( y_1 = y_2 \).
Distance \( PQ = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} = \sqrt{(x_2 - x_1)^2 + 0} = |x_2 - x_1| \).
Q4
Find a point on the x-axis, which is equidistant from the points (7, 6) and (3, 4).
Answer:
Let the point on the x-axis be \( P(x, 0) \).
Let \( A(7, 6) \) and \( B(3, 4) \).
Given \( PA = PB \Rightarrow PA^2 = PB^2 \).
\( (x - 7)^2 + (0 - 6)^2 = (x - 3)^2 + (0 - 4)^2 \)
\( x^2 - 14x + 49 + 36 = x^2 - 6x + 9 + 16 \)
\( -14x + 85 = -6x + 25 \)
\( -14x + 6x = 25 - 85 \)
\( -8x = -60 \Rightarrow x = \frac{60}{8} = \frac{15}{2} \).
The point is \( (\frac{15}{2}, 0) \).
Q5
Find the slope of a line, which passes through the origin, and the mid-point of the line segment joining the points P (0, – 4) and B (8, 0).
Answer:
First, find the mid-point (M) of the line segment PB.
\( M = \left( \frac{0 + 8}{2}, \frac{-4 + 0}{2} \right) = (4, -2) \).
We need the slope of the line passing through Origin \( O(0, 0) \) and \( M(4, -2) \).
Slope \( m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{-2 - 0}{4 - 0} = \frac{-2}{4} = -\frac{1}{2} \).
Slope = \( -\frac{1}{2} \)
Q6
Without using the Pythagoras theorem, show that the points (4, 4), (3, 5) and (–1, –1) are the vertices of a right angled triangle.
Answer:
Let the points be \( A(4, 4) \), \( B(3, 5) \), and \( C(-1, -1) \).
We can use the concept of slopes. If the product of slopes of two lines is -1, the lines are perpendicular.
Slope of AB (\( m_1 \)) = \( \frac{5 - 4}{3 - 4} = \frac{1}{-1} = -1 \).
Slope of BC (\( m_2 \)) = \( \frac{-1 - 5}{-1 - 3} = \frac{-6}{-4} = \frac{3}{2} \).
Slope of AC (\( m_3 \)) = \( \frac{-1 - 4}{-1 - 4} = \frac{-5}{-5} = 1 \).
Observation: \( m_1 \times m_3 = (-1) \times (1) = -1 \).
Since the product of slopes of AB and AC is -1, \( AB \perp AC \).
Therefore, the triangle is right-angled at A.
Q7
Find the slope of the line, which makes an angle of \( 30^\circ \) with the positive direction of y-axis measured anticlockwise.
(Paste image URL for the angle diagram here)
Answer:
If a line makes an angle of \( 30^\circ \) with the positive y-axis (measured anticlockwise), we need to find its inclination \( \theta \) with the positive x-axis.
The positive y-axis is at \( 90^\circ \) to the positive x-axis.
The angle made with the positive x-axis is \( \theta = 90^\circ + 30^\circ = 120^\circ \).
Slope \( m = \tan \theta = \tan 120^\circ \).
\( \tan 120^\circ = \tan(180^\circ - 60^\circ) = -\tan 60^\circ = -\sqrt{3} \).
Slope = \( -\sqrt{3} \)
Q8
Without using distance formula, show that points (– 2, – 1), (4, 0), (3, 3) and (–3, 2) are the vertices of a parallelogram.
Answer:
Let vertices be \( A(-2, -1) \), \( B(4, 0) \), \( C(3, 3) \), \( D(-3, 2) \).
A quadrilateral is a parallelogram if opposite sides are parallel (i.e., their slopes are equal).
Slope of AB = \( \frac{0 - (-1)}{4 - (-2)} = \frac{1}{6} \).
Slope of CD = \( \frac{2 - 3}{-3 - 3} = \frac{-1}{-6} = \frac{1}{6} \).
Since Slope(AB) = Slope(CD), AB || CD.
Slope of BC = \( \frac{3 - 0}{3 - 4} = \frac{3}{-1} = -3 \).
Slope of AD = \( \frac{2 - (-1)}{-3 - (-2)} = \frac{3}{-1} = -3 \).
Since Slope(BC) = Slope(AD), BC || AD.
Both pairs of opposite sides are parallel, hence ABCD is a parallelogram.
Q9
Find the angle between the x-axis and the line joining the points (3, –1) and (4, –2).
Answer:
Let the points be \( A(3, -1) \) and \( B(4, -2) \).
Slope \( m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{-2 - (-1)}{4 - 3} = \frac{-1}{1} = -1 \).
We know \( m = \tan \theta \), where \( \theta \) is the angle of inclination.
\( \tan \theta = -1 \).
Since \( \tan 45^\circ = 1 \), and \( \tan \) is negative in the 2nd quadrant:
\( \theta = 180^\circ - 45^\circ = 135^\circ \).
Angle = \( 135^\circ \)
Q10
The slope of a line is double of the slope of another line. If tangent of the angle between them is \( \frac{1}{3} \), find the slopes of the lines.
Answer:
Let the slopes be \( m \) and \( 2m \).
The angle \( \theta \) between the lines is given by \( \tan \theta = \left| \frac{m_2 - m_1}{1 + m_1 m_2} \right| \).
Given \( \tan \theta = \frac{1}{3} \), substitute values:
\( \frac{1}{3} = \left| \frac{2m - m}{1 + (m)(2m)} \right| = \left| \frac{m}{1 + 2m^2} \right| \).
Case 1: \( \frac{m}{1 + 2m^2} = \frac{1}{3} \)
\( 3m = 1 + 2m^2 \Rightarrow 2m^2 - 3m + 1 = 0 \)
\( (2m - 1)(m - 1) = 0 \Rightarrow m = 1 \) or \( m = \frac{1}{2} \).
If \( m=1 \), slopes are 1 and 2. If \( m=1/2 \), slopes are 1/2 and 1.
Case 2: \( \frac{m}{1 + 2m^2} = -\frac{1}{3} \)
\( -3m = 1 + 2m^2 \Rightarrow 2m^2 + 3m + 1 = 0 \)
\( (2m + 1)(m + 1) = 0 \Rightarrow m = -1 \) or \( m = -\frac{1}{2} \).
If \( m=-1 \), slopes are -1 and -2. If \( m=-1/2 \), slopes are -1/2 and -1.
Q11
A line passes through \( (x_1, y_1) \) and \( (h, k) \). If slope of the line is \( m \), show that \( k - y_1 = m (h - x_1) \).
Answer:
The slope \( m \) of a line passing through two points \( (x_1, y_1) \) and \( (x_2, y_2) \) is given by:
\( m = \frac{y_2 - y_1}{x_2 - x_1} \).
Here, points are \( (x_1, y_1) \) and \( (h, k) \).
Substituting these values:
\( m = \frac{k - y_1}{h - x_1} \)
Cross-multiplying:
\( m(h - x_1) = k - y_1 \)
\( k - y_1 = m(h - x_1) \).
Hence Proved.
In Exercises 1 to 8, find the equation of the line which satisfy the given conditions:
Q1
Write the equations for the x-and y-axes.
Answer:
The y-coordinate of every point on the x-axis is 0. Therefore, the equation of the x-axis is y = 0.
The x-coordinate of every point on the y-axis is 0. Therefore, the equation of the y-axis is x = 0.
Q2
Passing through the point (– 4, 3) with slope \( \frac{1}{2} \).
Answer:
We know that the equation of the line passing through point \( (x_0, y_0) \) with slope \( m \) is \( y - y_0 = m(x - x_0) \).
Here, \( (x_0, y_0) = (-4, 3) \) and \( m = \frac{1}{2} \).
\( y - 3 = \frac{1}{2} (x - (-4)) \)
\( 2(y - 3) = x + 4 \Rightarrow 2y - 6 = x + 4 \)
\( x - 2y + 10 = 0 \)
Q3
Passing through (0, 0) with slope \( m \).
Answer:
Using point-slope form with \( (x_0, y_0) = (0, 0) \):
\( y - 0 = m(x - 0) \)
\( y = mx \)
Q4
Passing through \( (2, 2\sqrt{3}) \) and inclined with the x-axis at an angle of \( 75^\circ \).
Answer:
Slope \( m = \tan 75^\circ = \tan(45^\circ + 30^\circ) = \frac{\tan 45^\circ + \tan 30^\circ}{1 - \tan 45^\circ \tan 30^\circ} \)
\( m = \frac{1 + \frac{1}{\sqrt{3}}}{1 - \frac{1}{\sqrt{3}}} = \frac{\sqrt{3}+1}{\sqrt{3}-1} \).
Rationalizing the denominator: \( m = \frac{(\sqrt{3}+1)^2}{3-1} = \frac{3+1+2\sqrt{3}}{2} = 2+\sqrt{3} \).
Equation: \( y - 2\sqrt{3} = (2+\sqrt{3})(x - 2) \)
\( y - 2\sqrt{3} = 2x - 4 + \sqrt{3}x - 2\sqrt{3} \)
\( y = (2+\sqrt{3})x - 4 \)
\( (2+\sqrt{3})x - y - 4 = 0 \)
Q5
Intersecting the x-axis at a distance of 3 units to the left of origin with slope –2.
Answer:
The line intersects the x-axis at a distance of 3 units to the left, so the point is \( (-3, 0) \).
Slope \( m = -2 \).
Equation: \( y - 0 = -2(x - (-3)) \)
\( y = -2(x + 3) \Rightarrow y = -2x - 6 \)
\( 2x + y + 6 = 0 \)
Q6
Intersecting the y-axis at a distance of 2 units above the origin and making an angle of \( 30^\circ \) with positive direction of the x-axis.
Answer:
The line intersects the y-axis 2 units above origin, so y-intercept \( c = 2 \). Point is \( (0, 2) \).
Slope \( m = \tan 30^\circ = \frac{1}{\sqrt{3}} \).
Using slope-intercept form \( y = mx + c \):
\( y = \frac{1}{\sqrt{3}}x + 2 \)
\( \sqrt{3}y = x + 2\sqrt{3} \)
\( x - \sqrt{3}y + 2\sqrt{3} = 0 \)
Q7
Passing through the points (–1, 1) and (2, – 4).
Answer:
Two-point form equation: \( y - y_1 = \frac{y_2 - y_1}{x_2 - x_1} (x - x_1) \)
\( y - 1 = \frac{-4 - 1}{2 - (-1)} (x - (-1)) \)
\( y - 1 = \frac{-5}{3} (x + 1) \)
\( 3(y - 1) = -5(x + 1) \Rightarrow 3y - 3 = -5x - 5 \)
\( 5x + 3y + 2 = 0 \)
Q8
The vertices of \( \Delta PQR \) are P (2, 1), Q (–2, 3) and R (4, 5). Find equation of the median through the vertex R.
(Paste image URL for the triangle diagram here)
Answer:
Let RL be the median through vertex R. L is the mid-point of side PQ.
Coordinates of L = \( \left( \frac{2 + (-2)}{2}, \frac{1 + 3}{2} \right) = (0, 2) \).
Now, we find the equation of the line passing through R(4, 5) and L(0, 2).
Slope \( m = \frac{2 - 5}{0 - 4} = \frac{-3}{-4} = \frac{3}{4} \).
Equation: \( y - 2 = \frac{3}{4}(x - 0) \)
\( 4y - 8 = 3x \)
\( 3x - 4y + 8 = 0 \)
Q9
Find the equation of the line passing through (–3, 5) and perpendicular to the line through the points (2, 5) and (–3, 6).
Answer:
Slope of the line joining (2, 5) and (-3, 6) is \( m_1 = \frac{6 - 5}{-3 - 2} = \frac{1}{-5} = -\frac{1}{5} \).
Since the required line is perpendicular to this line, its slope \( m_2 \) satisfies \( m_1 m_2 = -1 \).
\( (-\frac{1}{5}) m_2 = -1 \Rightarrow m_2 = 5 \).
Equation of line through (-3, 5) with slope 5:
\( y - 5 = 5(x - (-3)) \)
\( y - 5 = 5(x + 3) \Rightarrow y - 5 = 5x + 15 \)
\( 5x - y + 20 = 0 \)
Q10
A line perpendicular to the line segment joining the points (1, 0) and (2, 3) divides it in the ratio 1:n. Find the equation of the line.
Answer:
Let the line join A(1, 0) and B(2, 3). Slope of AB \( m_{AB} = \frac{3 - 0}{2 - 1} = 3 \).
Slope of required line \( m = -\frac{1}{3} \) (since perpendicular).
The line divides AB in ratio 1:n. Let the point of division be P.
Coordinates of P = \( \left( \frac{1(2) + n(1)}{1+n}, \frac{1(3) + n(0)}{1+n} \right) = \left( \frac{2+n}{1+n}, \frac{3}{1+n} \right) \).
Equation of line passing through P with slope \( -1/3 \):
\( y - \frac{3}{1+n} = -\frac{1}{3} \left( x - \frac{2+n}{1+n} \right) \)
\( 3 \left( y - \frac{3}{1+n} \right) = - \left( x - \frac{2+n}{1+n} \right) \)
\( 3y + x = \frac{9}{1+n} + \frac{2+n}{1+n} = \frac{11+n}{1+n} \)
\( (1+n)x + 3(1+n)y = n + 11 \)
Q11
Find the equation of a line that cuts off equal intercepts on the coordinate axes and passes through the point (2, 3).
Answer:
Equation of a line in intercept form is \( \frac{x}{a} + \frac{y}{b} = 1 \).
Since intercepts are equal, \( a = b \). So, \( \frac{x}{a} + \frac{y}{a} = 1 \Rightarrow x + y = a \).
It passes through (2, 3): \( 2 + 3 = a \Rightarrow a = 5 \).
Equation: \( x + y = 5 \)
Q12
Find equation of the line passing through the point (2, 2) and cutting off intercepts on the axes whose sum is 9.
Answer:
Let x-intercept be \( a \) and y-intercept be \( b \). Given \( a + b = 9 \Rightarrow b = 9 - a \).
Equation: \( \frac{x}{a} + \frac{y}{9-a} = 1 \).
Passes through (2, 2): \( \frac{2}{a} + \frac{2}{9-a} = 1 \).
\( \frac{2(9-a) + 2a}{a(9-a)} = 1 \Rightarrow \frac{18 - 2a + 2a}{9a - a^2} = 1 \)
\( 18 = 9a - a^2 \Rightarrow a^2 - 9a + 18 = 0 \).
\( (a-6)(a-3) = 0 \Rightarrow a = 6 \) or \( a = 3 \).
If \( a = 6 \), \( b = 3 \). Equation: \( \frac{x}{6} + \frac{y}{3} = 1 \Rightarrow x + 2y - 6 = 0 \).
If \( a = 3 \), \( b = 6 \). Equation: \( \frac{x}{3} + \frac{y}{6} = 1 \Rightarrow 2x + y - 6 = 0 \).
Q13
Find equation of the line through the point (0, 2) making an angle \( \frac{2\pi}{3} \) with the positive x-axis. Also, find the equation of line parallel to it and crossing the y-axis at a distance of 2 units below the origin.
Answer:
(i) Slope \( m = \tan \frac{2\pi}{3} = \tan 120^\circ = -\sqrt{3} \). Point (0, 2).
Equation: \( y - 2 = -\sqrt{3}(x - 0) \Rightarrow \sqrt{3}x + y - 2 = 0 \).
(ii) Parallel line has same slope \( m = -\sqrt{3} \). Crosses y-axis 2 units below origin, so point is (0, -2).
Equation: \( y - (-2) = -\sqrt{3}(x - 0) \Rightarrow \sqrt{3}x + y + 2 = 0 \).
Q14
The perpendicular from the origin to a line meets it at the point (– 2, 9), find the equation of the line.
Answer:
Let point M(-2, 9). Slope of OM (normal) \( m_{OM} = \frac{9 - 0}{-2 - 0} = -\frac{9}{2} \).
The required line is perpendicular to OM. So its slope \( m = \frac{-1}{-9/2} = \frac{2}{9} \).
It passes through M(-2, 9).
Equation: \( y - 9 = \frac{2}{9} (x - (-2)) \)
\( 9(y - 9) = 2(x + 2) \Rightarrow 9y - 81 = 2x + 4 \)
\( 2x - 9y + 85 = 0 \)
Q15
The length L (in centimetre) of a copper rod is a linear function of its Celsius temperature C. In an experiment, if L = 124.942 when C = 20 and L = 125.134 when C = 110, express L in terms of C.
Answer:
Let L and C be coordinates (C, L). Points are (20, 124.942) and (110, 125.134).
Slope \( m = \frac{125.134 - 124.942}{110 - 20} = \frac{0.192}{90} \).
Using point-slope form with (20, 124.942):
\( L - 124.942 = \frac{0.192}{90} (C - 20) \)
\( L = \frac{0.192}{90}(C - 20) + 124.942 \)
Q16
The owner of a milk store finds that, he can sell 980 litres of milk each week at Rs 14/litre and 1220 litres of milk each week at Rs 16/litre. Assuming a linear relationship between selling price and demand, how many litres could he sell weekly at Rs 17/litre?
Answer:
Let x be price and y be litres. Points: (14, 980) and (16, 1220).
Slope \( m = \frac{1220 - 980}{16 - 14} = \frac{240}{2} = 120 \).
Equation: \( y - 980 = 120(x - 14) \).
Find y when x = 17:
\( y - 980 = 120(17 - 14) = 120(3) = 360 \).
\( y = 360 + 980 = 1340 \).
1340 litres
Q17
P (a, b) is the mid-point of a line segment between axes. Show that equation of the line is \( \frac{x}{a} + \frac{y}{b} = 2 \).
Answer:
Let the line intersect x-axis at A(h, 0) and y-axis at B(0, k).
Mid-point of AB is \( \left( \frac{h}{2}, \frac{k}{2} \right) \).
Given mid-point is P(a, b). So, \( \frac{h}{2} = a \Rightarrow h = 2a \) and \( \frac{k}{2} = b \Rightarrow k = 2b \).
Intercept equation: \( \frac{x}{h} + \frac{y}{k} = 1 \).
\( \frac{x}{2a} + \frac{y}{2b} = 1 \Rightarrow \frac{x}{a} + \frac{y}{b} = 2 \). Hence Proved.
Q18
Point R (h, k) divides a line segment between the axes in the ratio 1:2. Find equation of the line.
Answer:
Let intercepts be A(a, 0) and B(0, b). R(h, k) divides AB in 1:2.
Using section formula: \( h = \frac{1(0) + 2(a)}{1+2} = \frac{2a}{3} \Rightarrow a = \frac{3h}{2} \).
\( k = \frac{1(b) + 2(0)}{1+2} = \frac{b}{3} \Rightarrow b = 3k \).
Equation: \( \frac{x}{a} + \frac{y}{b} = 1 \Rightarrow \frac{x}{3h/2} + \frac{y}{3k} = 1 \).
\( \frac{2x}{3h} + \frac{y}{3k} = 1 \Rightarrow 2kx + hy = 3hk \).
Q19
By using the concept of equation of a line, prove that the three points (3, 0), (–2, –2) and (8, 2) are collinear.
Answer:
Find equation of line through (3, 0) and (-2, -2).
Slope \( m = \frac{-2 - 0}{-2 - 3} = \frac{-2}{-5} = \frac{2}{5} \).
Equation: \( y - 0 = \frac{2}{5}(x - 3) \Rightarrow 5y = 2x - 6 \Rightarrow 2x - 5y - 6 = 0 \).
Check if (8, 2) satisfies this equation:
\( 2(8) - 5(2) - 6 = 16 - 10 - 6 = 0 \).
Since it satisfies, the points are collinear.
Q1
Reduce the following equations into slope - intercept form and find their slopes and the y - intercepts.
(i) \( x + 7y = 0 \)
(ii) \( 6x + 3y - 5 = 0 \)
(iii) \( y = 0 \).
(i) \( x + 7y = 0 \)
(ii) \( 6x + 3y - 5 = 0 \)
(iii) \( y = 0 \).
Answer:
The slope-intercept form is \( y = mx + c \), where \(m\) is the slope and \(c\) is the y-intercept.
(i) \( x + 7y = 0 \)
\( 7y = -x \Rightarrow y = -\frac{1}{7}x + 0 \)
Comparing with \( y = mx + c \):
Slope \( m = -\frac{1}{7} \), y-intercept \( c = 0 \).
(ii) \( 6x + 3y - 5 = 0 \)
\( 3y = -6x + 5 \Rightarrow y = -2x + \frac{5}{3} \)
Comparing with \( y = mx + c \):
Slope \( m = -2 \), y-intercept \( c = \frac{5}{3} \).
(iii) \( y = 0 \)
\( y = 0x + 0 \)
Comparing with \( y = mx + c \):
Slope \( m = 0 \), y-intercept \( c = 0 \).
Q2
Reduce the following equations into intercept form and find their intercepts on the axes.
(i) \( 3x + 2y - 12 = 0 \)
(ii) \( 4x - 3y = 6 \)
(iii) \( 3y + 2 = 0 \).
(i) \( 3x + 2y - 12 = 0 \)
(ii) \( 4x - 3y = 6 \)
(iii) \( 3y + 2 = 0 \).
Answer:
The intercept form is \( \frac{x}{a} + \frac{y}{b} = 1 \), where \(a\) is x-intercept and \(b\) is y-intercept.
(i) \( 3x + 2y - 12 = 0 \)
\( 3x + 2y = 12 \). Divide by 12:
\( \frac{3x}{12} + \frac{2y}{12} = 1 \Rightarrow \frac{x}{4} + \frac{y}{6} = 1 \)
Intercepts: x-intercept = 4, y-intercept = 6.
(ii) \( 4x - 3y = 6 \)
Divide by 6:
\( \frac{4x}{6} - \frac{3y}{6} = 1 \Rightarrow \frac{2x}{3} - \frac{y}{2} = 1 \Rightarrow \frac{x}{3/2} + \frac{y}{-2} = 1 \)
Intercepts: x-intercept = \( \frac{3}{2} \), y-intercept = -2.
(iii) \( 3y + 2 = 0 \)
\( 3y = -2 \Rightarrow y = -\frac{2}{3} \). This is a horizontal line.
It does not intersect the x-axis (no x-intercept). y-intercept = \( -\frac{2}{3} \).
Q3
Find the distance of the point (-1, 1) from the line \( 12(x + 6) = 5(y - 2) \).
Answer:
Rewrite the equation in general form \( Ax + By + C = 0 \):
\( 12x + 72 = 5y - 10 \Rightarrow 12x - 5y + 82 = 0 \).
Distance \( d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}} \).
Here \( (x_1, y_1) = (-1, 1) \).
\( d = \frac{|12(-1) - 5(1) + 82|}{\sqrt{12^2 + (-5)^2}} \)
\( d = \frac{|-12 - 5 + 82|}{\sqrt{144 + 25}} = \frac{|65|}{\sqrt{169}} = \frac{65}{13} = 5 \).
Distance = 5 units.
Q4
Find the points on the x-axis, whose distances from the line \( \frac{x}{3} + \frac{y}{4} = 1 \) are 4 units.
Answer:
Equation of line: \( \frac{x}{3} + \frac{y}{4} = 1 \Rightarrow 4x + 3y - 12 = 0 \).
Let point on x-axis be \( (h, 0) \). Distance is 4.
\( \frac{|4(h) + 3(0) - 12|}{\sqrt{4^2 + 3^2}} = 4 \)
\( \frac{|4h - 12|}{5} = 4 \Rightarrow |4h - 12| = 20 \)
Case 1: \( 4h - 12 = 20 \Rightarrow 4h = 32 \Rightarrow h = 8 \).
Case 2: \( 4h - 12 = -20 \Rightarrow 4h = -8 \Rightarrow h = -2 \).
Points are (8, 0) and (-2, 0).
Q5
Find the distance between parallel lines
(i) \( 15x + 8y - 34 = 0 \) and \( 15x + 8y + 31 = 0 \)
(ii) \( l(x + y) + p = 0 \) and \( l(x + y) - r = 0 \).
(i) \( 15x + 8y - 34 = 0 \) and \( 15x + 8y + 31 = 0 \)
(ii) \( l(x + y) + p = 0 \) and \( l(x + y) - r = 0 \).
Answer:
Distance between parallel lines \( Ax + By + C_1 = 0 \) and \( Ax + By + C_2 = 0 \) is \( d = \frac{|C_1 - C_2|}{\sqrt{A^2 + B^2}} \).
(i) \( C_1 = -34, C_2 = 31, A = 15, B = 8 \).
\( d = \frac{|-34 - 31|}{\sqrt{15^2 + 8^2}} = \frac{|-65|}{\sqrt{225 + 64}} = \frac{65}{\sqrt{289}} = \frac{65}{17} \).
(ii) Rewrite: \( lx + ly + p = 0 \) and \( lx + ly - r = 0 \).
\( C_1 = p, C_2 = -r, A = l, B = l \).
\( d = \frac{|p - (-r)|}{\sqrt{l^2 + l^2}} = \frac{|p + r|}{\sqrt{2l^2}} = \frac{|p + r|}{l\sqrt{2}} \).
Q6
Find equation of the line parallel to the line \( 3x - 4y + 2 = 0 \) and passing through the point (-2, 3).
Answer:
Any line parallel to \( 3x - 4y + 2 = 0 \) is of the form \( 3x - 4y + k = 0 \).
It passes through (-2, 3):
\( 3(-2) - 4(3) + k = 0 \Rightarrow -6 - 12 + k = 0 \Rightarrow k = 18 \).
Equation: \( 3x - 4y + 18 = 0 \)
Q7
Find equation of the line perpendicular to the line \( x - 7y + 5 = 0 \) and having x intercept 3.
Answer:
Slope of given line \( x - 7y + 5 = 0 \) is \( m_1 = -\frac{A}{B} = -\frac{1}{-7} = \frac{1}{7} \).
Slope of perpendicular line \( m_2 = -\frac{1}{m_1} = -7 \).
The required line has x-intercept 3, so it passes through (3, 0).
Equation: \( y - 0 = -7(x - 3) \Rightarrow y = -7x + 21 \).
\( 7x + y - 21 = 0 \)
Q8
Find angles between the lines \( \sqrt{3}x + y = 1 \) and \( x + \sqrt{3}y = 1 \).
(Paste image URL for the angle between lines diagram here)
Answer:
Line 1: \( y = -\sqrt{3}x + 1 \Rightarrow m_1 = -\sqrt{3} \).
Line 2: \( \sqrt{3}y = -x + 1 \Rightarrow y = -\frac{1}{\sqrt{3}}x + \frac{1}{\sqrt{3}} \Rightarrow m_2 = -\frac{1}{\sqrt{3}} \).
\( \tan \theta = \left| \frac{m_2 - m_1}{1 + m_1 m_2} \right| = \left| \frac{-\frac{1}{\sqrt{3}} - (-\sqrt{3})}{1 + (-\sqrt{3})(-\frac{1}{\sqrt{3}})} \right| \)
\( = \left| \frac{\frac{-1+3}{\sqrt{3}}}{1 + 1} \right| = \left| \frac{2/\sqrt{3}}{2} \right| = \frac{1}{\sqrt{3}} \).
\( \tan \theta = \frac{1}{\sqrt{3}} \Rightarrow \theta = 30^\circ \).
Angle = \( 30^\circ \)
Q9
The line through the points (h, 3) and (4, 1) intersects the line \( 7x - 9y - 19 = 0 \) at right angle. Find the value of h.
Answer:
Slope of line through (h, 3) and (4, 1) is \( m_1 = \frac{1 - 3}{4 - h} = \frac{-2}{4 - h} \).
Slope of line \( 7x - 9y - 19 = 0 \) is \( m_2 = \frac{7}{9} \).
Since they are perpendicular, \( m_1 m_2 = -1 \).
\( \left( \frac{-2}{4 - h} \right) \left( \frac{7}{9} \right) = -1 \)
\( \frac{-14}{9(4 - h)} = -1 \Rightarrow 14 = 36 - 9h \)
\( 9h = 22 \Rightarrow h = \frac{22}{9} \).
Q10
Prove that the line through the point \( (x_1, y_1) \) and parallel to the line \( Ax + By + C = 0 \) is \( A(x - x_1) + B(y - y_1) = 0 \).
Answer:
Slope of \( Ax + By + C = 0 \) is \( m = -\frac{A}{B} \).
Parallel line has same slope. It passes through \( (x_1, y_1) \).
Equation: \( y - y_1 = -\frac{A}{B} (x - x_1) \)
\( B(y - y_1) = -A(x - x_1) \)
\( A(x - x_1) + B(y - y_1) = 0 \). Hence Proved.
Q11
Two lines passing through the point (2, 3) intersects each other at an angle of \( 60^\circ \). If slope of one line is 2, find equation of the other line.
Answer:
Given \( m_1 = 2 \), angle \( \theta = 60^\circ \). Let \( m_2 = m \).
\( \tan 60^\circ = \left| \frac{m - 2}{1 + 2m} \right| \Rightarrow \sqrt{3} = \left| \frac{m - 2}{1 + 2m} \right| \).
Case 1: \( \frac{m - 2}{1 + 2m} = \sqrt{3} \)
\( m - 2 = \sqrt{3} + 2\sqrt{3}m \Rightarrow m(1 - 2\sqrt{3}) = 2 + \sqrt{3} \Rightarrow m = \frac{2 + \sqrt{3}}{1 - 2\sqrt{3}} \).
Equation: \( y - 3 = \frac{2 + \sqrt{3}}{1 - 2\sqrt{3}} (x - 2) \).
Case 2: \( \frac{m - 2}{1 + 2m} = -\sqrt{3} \)
\( m - 2 = -\sqrt{3} - 2\sqrt{3}m \Rightarrow m(1 + 2\sqrt{3}) = 2 - \sqrt{3} \Rightarrow m = \frac{2 - \sqrt{3}}{1 + 2\sqrt{3}} \).
Equation: \( y - 3 = \frac{2 - \sqrt{3}}{1 + 2\sqrt{3}} (x - 2) \).
Q12
Find the equation of the right bisector of the line segment joining the points (3, 4) and (–1, 2).
(Paste image URL for the right bisector diagram here)
Answer:
Let points be A(3, 4) and B(-1, 2).
Mid-point M of AB: \( \left( \frac{3 - 1}{2}, \frac{4 + 2}{2} \right) = (1, 3) \).
Slope of AB: \( m_{AB} = \frac{2 - 4}{-1 - 3} = \frac{-2}{-4} = \frac{1}{2} \).
Slope of perpendicular bisector \( m = -2 \).
Equation passing through (1, 3) with slope -2:
\( y - 3 = -2(x - 1) \Rightarrow y - 3 = -2x + 2 \)
\( 2x + y - 5 = 0 \)
Q13
Find the coordinates of the foot of perpendicular from the point (–1, 3) to the line \( 3x - 4y - 16 = 0 \).
(Paste image URL for the foot of perpendicular diagram here)
Answer:
Slope of given line \( 3x - 4y - 16 = 0 \) is \( \frac{3}{4} \).
Slope of perpendicular line is \( -\frac{4}{3} \). It passes through (-1, 3).
Equation of perpendicular: \( y - 3 = -\frac{4}{3}(x + 1) \Rightarrow 3y - 9 = -4x - 4 \Rightarrow 4x + 3y - 5 = 0 \).
Solve \( 3x - 4y - 16 = 0 \) and \( 4x + 3y - 5 = 0 \).
Multiply first by 3 and second by 4:
\( 9x - 12y - 48 = 0 \)
\( 16x + 12y - 20 = 0 \)
Adding: \( 25x - 68 = 0 \Rightarrow x = \frac{68}{25} \).
Substitute x: \( 4(\frac{68}{25}) + 3y - 5 = 0 \Rightarrow 3y = 5 - \frac{272}{25} = \frac{125 - 272}{25} = -\frac{147}{25} \).
\( y = -\frac{49}{25} \).
Coordinates: \( (\frac{68}{25}, -\frac{49}{25}) \)
Q14
The perpendicular from the origin to the line \( y = mx + c \) meets it at the point (–1, 2). Find the values of \( m \) and \( c \).
Answer:
Slope of line joining Origin (0,0) and point (-1, 2) is \( m_{OM} = \frac{2 - 0}{-1 - 0} = -2 \).
Since this is perpendicular to the line \( y = mx + c \), the product of slopes is -1.
\( m \times (-2) = -1 \Rightarrow m = \frac{1}{2} \).
The line passes through (-1, 2). Substitute in equation:
\( 2 = \frac{1}{2}(-1) + c \Rightarrow 2 = -0.5 + c \Rightarrow c = 2.5 \).
\( m = \frac{1}{2}, c = \frac{5}{2} \)
Q15
If \( p \) and \( q \) are the lengths of perpendiculars from the origin to the lines \( x \cos \theta - y \sin \theta = k \cos 2\theta \) and \( x \sec \theta + y \csc \theta = k \), respectively, prove that \( p^2 + 4q^2 = k^2 \).
Answer:
Distance of origin from Line 1: \( x \cos \theta - y \sin \theta - k \cos 2\theta = 0 \).
\( p = \frac{|-k \cos 2\theta|}{\sqrt{\cos^2 \theta + (-\sin \theta)^2}} = \frac{|k \cos 2\theta|}{1} = |k \cos 2\theta| \).
\( p^2 = k^2 \cos^2 2\theta \).
Line 2: \( x \sec \theta + y \csc \theta - k = 0 \). Rewrite as \( \frac{x}{\cos \theta} + \frac{y}{\sin \theta} = k \Rightarrow x \sin \theta + y \cos \theta - k \sin \theta \cos \theta = 0 \).
\( q = \frac{|-k \sin \theta \cos \theta|}{\sqrt{\sin^2 \theta + \cos^2 \theta}} = |k \sin \theta \cos \theta| = |\frac{k \sin 2\theta}{2}| \).
\( 4q^2 = 4 \left( \frac{k^2 \sin^2 2\theta}{4} \right) = k^2 \sin^2 2\theta \).
\( p^2 + 4q^2 = k^2 \cos^2 2\theta + k^2 \sin^2 2\theta = k^2(\cos^2 2\theta + \sin^2 2\theta) = k^2 \). Proved.
Q16
In the triangle ABC with vertices A (2, 3), B (4, –1) and C (1, 2), find the equation and length of altitude from the vertex A.
(Paste image URL for the triangle altitude diagram here)
Answer:
Slope of BC: \( m_{BC} = \frac{2 - (-1)}{1 - 4} = \frac{3}{-3} = -1 \).
Altitude from A is perpendicular to BC. Slope \( m_{alt} = 1 \).
Equation passing through A(2, 3): \( y - 3 = 1(x - 2) \Rightarrow y - 3 = x - 2 \Rightarrow x - y + 1 = 0 \).
To find length, find distance of A(2, 3) from line BC.
Equation of BC: \( y - 2 = -1(x - 1) \Rightarrow y - 2 = -x + 1 \Rightarrow x + y - 3 = 0 \).
Length \( = \frac{|2 + 3 - 3|}{\sqrt{1^2 + 1^2}} = \frac{2}{\sqrt{2}} = \sqrt{2} \).
Q17
If p is the length of perpendicular from the origin to the line whose intercepts on the axes are a and b, then show that \( \frac{1}{p^2} = \frac{1}{a^2} + \frac{1}{b^2} \).
Answer:
Equation of line in intercept form: \( \frac{x}{a} + \frac{y}{b} = 1 \Rightarrow bx + ay - ab = 0 \).
Perpendicular distance from origin (0, 0) is \( p \).
\( p = \frac{|b(0) + a(0) - ab|}{\sqrt{b^2 + a^2}} = \frac{ab}{\sqrt{a^2 + b^2}} \).
Square both sides: \( p^2 = \frac{a^2 b^2}{a^2 + b^2} \).
Invert both sides: \( \frac{1}{p^2} = \frac{a^2 + b^2}{a^2 b^2} = \frac{a^2}{a^2 b^2} + \frac{b^2}{a^2 b^2} \).
\( \frac{1}{p^2} = \frac{1}{b^2} + \frac{1}{a^2} \). Proved.
Q1
Find the values of \( k \) for which the line \( (k-3)x - (4-k^2)y + k^2-7k+6 = 0 \) is
(a) Parallel to the x-axis,
(b) Parallel to the y-axis,
(c) Passing through the origin.
(a) Parallel to the x-axis,
(b) Parallel to the y-axis,
(c) Passing through the origin.
Answer:
The given equation is \( (k-3)x - (4-k^2)y + k^2-7k+6 = 0 \).
This is of the form \( Ax + By + C = 0 \).
(a) Parallel to x-axis:
The coefficient of \( x \) must be zero (\( A = 0 \)) and \( B \ne 0 \).
\( k - 3 = 0 \Rightarrow k = 3 \).
Check B: \( -(4 - 3^2) = -(4-9) = 5 \ne 0 \). Valid.
Value: \( k = 3 \)
(b) Parallel to y-axis:
The coefficient of \( y \) must be zero (\( B = 0 \)) and \( A \ne 0 \).
\( -(4 - k^2) = 0 \Rightarrow k^2 = 4 \Rightarrow k = \pm 2 \).
Check A: For \( k=2, A=-1 \). For \( k=-2, A=-5 \). Both valid.
Value: \( k = \pm 2 \)
(c) Passing through origin:
The constant term \( C \) must be zero.
\( k^2 - 7k + 6 = 0 \Rightarrow (k-6)(k-1) = 0 \Rightarrow k = 1, 6 \).
Value: \( k = 1 \) or \( k = 6 \)
Q2
Find the equations of the lines, which cut-off intercepts on the axes whose sum and product are 1 and – 6, respectively.
Answer:
Let intercepts be \( a \) and \( b \).
Given Sum: \( a + b = 1 \Rightarrow b = 1 - a \).
Given Product: \( ab = -6 \).
Substitute \( b \): \( a(1 - a) = -6 \Rightarrow a - a^2 = -6 \Rightarrow a^2 - a - 6 = 0 \).
\( (a-3)(a+2) = 0 \Rightarrow a = 3 \) or \( a = -2 \).
Case 1: If \( a = 3 \), then \( b = 1 - 3 = -2 \).
Equation: \( \frac{x}{3} + \frac{y}{-2} = 1 \Rightarrow 2x - 3y - 6 = 0 \).
Case 2: If \( a = -2 \), then \( b = 1 - (-2) = 3 \).
Equation: \( \frac{x}{-2} + \frac{y}{3} = 1 \Rightarrow -3x + 2y - 6 = 0 \Rightarrow 3x - 2y + 6 = 0 \).
Q3
What are the points on the y-axis whose distance from the line \( \frac{x}{3} + \frac{y}{4} = 1 \) is 4 units.
Answer:
Equation: \( 4x + 3y - 12 = 0 \).
Let the point on y-axis be \( P(0, k) \).
Distance is 4:
\( \frac{|4(0) + 3(k) - 12|}{\sqrt{4^2 + 3^2}} = 4 \Rightarrow \frac{|3k - 12|}{5} = 4 \).
\( |3k - 12| = 20 \).
Case 1: \( 3k - 12 = 20 \Rightarrow 3k = 32 \Rightarrow k = \frac{32}{3} \).
Case 2: \( 3k - 12 = -20 \Rightarrow 3k = -8 \Rightarrow k = -\frac{8}{3} \).
Points: \( (0, \frac{32}{3}) \) and \( (0, -\frac{8}{3}) \)
Q4
Find perpendicular distance from the origin to the line joining the points \( (\cos\theta, \sin\theta) \) and \( (\cos\phi, \sin\phi) \).
Answer:
Equation of line passing through \( (\cos\theta, \sin\theta) \) and \( (\cos\phi, \sin\phi) \):
\( y - \sin\theta = \frac{\sin\phi - \sin\theta}{\cos\phi - \cos\theta} (x - \cos\theta) \).
Using identities: Slope \( m = \frac{2\cos\frac{\phi+\theta}{2}\sin\frac{\phi-\theta}{2}}{-2\sin\frac{\phi+\theta}{2}\sin\frac{\phi-\theta}{2}} = -\cot\frac{\phi+\theta}{2} \).
\( y - \sin\theta = -\frac{\cos\frac{\phi+\theta}{2}}{\sin\frac{\phi+\theta}{2}} (x - \cos\theta) \).
Rearranging, we get the normal form: \( x \cos\frac{\theta+\phi}{2} + y \sin\frac{\theta+\phi}{2} = \cos\frac{\theta-\phi}{2} \).
Perpendicular distance from origin is the constant term on the RHS:
Distance = \( |\cos(\frac{\theta - \phi}{2})| \)
Q5
Find the equation of the line parallel to y-axis and drawn through the point of intersection of the lines \( x - 7y + 5 = 0 \) and \( 3x + y = 0 \).
Answer:
Solve the system:
From 2nd eq: \( y = -3x \).
Substitute in 1st: \( x - 7(-3x) + 5 = 0 \Rightarrow x + 21x + 5 = 0 \Rightarrow 22x = -5 \Rightarrow x = -\frac{5}{22} \).
Since the required line is parallel to the y-axis, its equation is \( x = \text{constant} \).
The line passes through \( x = -\frac{5}{22} \).
Equation: \( x = -\frac{5}{22} \) or \( 22x + 5 = 0 \)
Q6
Find the equation of a line drawn perpendicular to the line \( \frac{x}{4} + \frac{y}{6} = 1 \) through the point, where it meets the y-axis.
Answer:
The line meets the y-axis at \( (0, 6) \).
Equation of given line: \( 3x + 2y - 12 = 0 \). Slope \( m_1 = -\frac{3}{2} \).
Slope of perpendicular line \( m_2 = \frac{2}{3} \).
Equation passing through \( (0, 6) \) with slope \( \frac{2}{3} \):
\( y - 6 = \frac{2}{3}(x - 0) \Rightarrow 3y - 18 = 2x \).
Equation: \( 2x - 3y + 18 = 0 \)
Q7
Find the area of the triangle formed by the lines \( y - x = 0 \), \( x + y = 0 \) and \( x - k = 0 \).
(Paste image URL for triangle diagram here)
Answer:
The vertices of the triangle are the intersection points of the lines.
1. \( y - x = 0 \) and \( x + y = 0 \Rightarrow (0, 0) \).
2. \( y - x = 0 \) and \( x = k \Rightarrow (k, k) \).
3. \( x + y = 0 \) and \( x = k \Rightarrow (k, -k) \).
Vertices: \( A(0,0), B(k, k), C(k, -k) \).
Area \( = \frac{1}{2} |0(k - (-k)) + k(-k - 0) + k(0 - k)| \)
\( = \frac{1}{2} |0 - k^2 - k^2| = \frac{1}{2} |-2k^2| = k^2 \text{ sq units} \).
Q8
Find the value of \( p \) so that the three lines \( 3x + y - 2 = 0 \), \( px + 2y - 3 = 0 \) and \( 2x - y - 3 = 0 \) may intersect at one point.
Answer:
Find intersection of first and third lines:
\( 3x + y = 2 \) and \( 2x - y = 3 \).
Adding them: \( 5x = 5 \Rightarrow x = 1 \).
Substitute \( x=1 \): \( 2(1) - y = 3 \Rightarrow y = -1 \).
Intersection point is \( (1, -1) \).
Since the lines are concurrent, this point must satisfy the second equation:
\( p(1) + 2(-1) - 3 = 0 \Rightarrow p - 2 - 3 = 0 \Rightarrow p = 5 \).
Q9
If three lines whose equations are \( y = m_1x + c_1 \), \( y = m_2x + c_2 \) and \( y = m_3x + c_3 \) are concurrent, then show that \( m_1(c_2 - c_3) + m_2(c_3 - c_1) + m_3(c_1 - c_2) = 0 \).
Answer:
The condition for concurrency of three lines \( A_ix + B_iy + C_i = 0 \) is that the determinant of their coefficients is zero.
Here equations are \( m_ix - y + c_i = 0 \).
Determinant:
\( \begin{vmatrix} m_1 & -1 & c_1 \\ m_2 & -1 & c_2 \\ m_3 & -1 & c_3 \end{vmatrix} = 0 \)
Expanding along C1:
\( m_1((-1)(c_3) - (-1)(c_2)) - m_2((-1)(c_3) - (-1)(c_1)) + m_3((-1)(c_2) - (-1)(c_1)) = 0 \)
\( m_1(c_2 - c_3) - m_2(c_1 - c_3) + m_3(c_1 - c_2) = 0 \)
\( m_1(c_2 - c_3) + m_2(c_3 - c_1) + m_3(c_1 - c_2) = 0 \). Proved.
Q10
Find the equation of the lines through the point (3, 2) which make an angle of \( 45^\circ \) with the line \( x - 2y = 3 \).
Answer:
Slope of given line \( x - 2y = 3 \) is \( m_1 = \frac{1}{2} \).
Let slope of required line be \( m \). Angle is \( 45^\circ \).
\( \tan 45^\circ = \left| \frac{m - 1/2}{1 + m(1/2)} \right| \Rightarrow 1 = \left| \frac{2m - 1}{2 + m} \right| \).
Case 1: \( 2m - 1 = 2 + m \Rightarrow m = 3 \).
Equation: \( y - 2 = 3(x - 3) \Rightarrow 3x - y - 7 = 0 \).
Case 2: \( 2m - 1 = -(2 + m) \Rightarrow 3m = -1 \Rightarrow m = -\frac{1}{3} \).
Equation: \( y - 2 = -\frac{1}{3}(x - 3) \Rightarrow x + 3y - 9 = 0 \).
Q11
Find the equation of the line passing through the point of intersection of the lines \( 4x + 7y - 3 = 0 \) and \( 2x - 3y + 1 = 0 \) that has equal intercepts on the axes.
Answer:
Solve for intersection point:
Multiply 2nd eq by 2: \( 4x - 6y + 2 = 0 \).
Subtract from 1st: \( (4x + 7y - 3) - (4x - 6y + 2) = 0 \Rightarrow 13y - 5 = 0 \Rightarrow y = \frac{5}{13} \).
\( 2x = 3(\frac{5}{13}) - 1 = \frac{15 - 13}{13} = \frac{2}{13} \Rightarrow x = \frac{1}{13} \).
Point is \( (\frac{1}{13}, \frac{5}{13}) \).
Lines with equal intercepts have slope -1. Equation form \( x + y = a \).
Substitute point: \( \frac{1}{13} + \frac{5}{13} = a \Rightarrow a = \frac{6}{13} \).
Equation: \( x + y = \frac{6}{13} \) or \( 13x + 13y = 6 \)
Q12
Show that the equation of the line passing through the origin and making an angle \( \theta \) with the line \( y = mx + c \) is \( \frac{y}{x} = \frac{m \pm \tan\theta}{1 \mp m\tan\theta} \).
Answer:
Let slope of required line be \( M \). Since it passes through origin, its equation is \( y = Mx \) or \( \frac{y}{x} = M \).
Given angle is \( \theta \). Slope of given line is \( m \).
\( \tan\theta = \left| \frac{M - m}{1 + mM} \right| \).
Removing modulus: \( \frac{M - m}{1 + mM} = \tan\theta \) or \( -\tan\theta \).
Case 1: \( M - m = \tan\theta + mM\tan\theta \Rightarrow M(1 - m\tan\theta) = m + \tan\theta \Rightarrow M = \frac{m + \tan\theta}{1 - m\tan\theta} \).
Case 2: \( M - m = -\tan\theta - mM\tan\theta \Rightarrow M(1 + m\tan\theta) = m - \tan\theta \Rightarrow M = \frac{m - \tan\theta}{1 + m\tan\theta} \).
Substituting \( M = \frac{y}{x} \), we get the result.
Q13
In what ratio, the line joining \( (-1, 1) \) and \( (5, 7) \) is divided by the line \( x + y = 4 \)?
Answer:
Let the ratio be \( k:1 \). Let the point of division be P.
\( P = \left( \frac{5k - 1}{k + 1}, \frac{7k + 1}{k + 1} \right) \).
P lies on \( x + y = 4 \).
\( \frac{5k - 1}{k + 1} + \frac{7k + 1}{k + 1} = 4 \)
\( 12k = 4(k + 1) \Rightarrow 12k = 4k + 4 \Rightarrow 8k = 4 \Rightarrow k = \frac{1}{2} \).
Ratio is 1:2.
Q14
Find the distance of the line \( 4x + 7y + 5 = 0 \) from the point \( (1, 2) \) along the line \( 2x - y = 0 \).
Answer:
We need distance between point A(1, 2) and the point B where a line through A parallel to \( 2x - y = 0 \) intersects \( 4x + 7y + 5 = 0 \).
Line through (1, 2) with slope 2 is \( y - 2 = 2(x - 1) \Rightarrow y = 2x \) (Same as given direction).
Intersection of \( y = 2x \) and \( 4x + 7y + 5 = 0 \):
\( 4x + 7(2x) + 5 = 0 \Rightarrow 18x = -5 \Rightarrow x = -5/18 \).
\( y = -10/18 = -5/9 \). Point B is \( (-5/18, -5/9) \).
Distance AB \( = \sqrt{(1 - (-5/18))^2 + (2 - (-5/9))^2} \)
\( = \sqrt{(23/18)^2 + (23/9)^2} = \frac{23}{18} \sqrt{1 + 2^2} = \frac{23\sqrt{5}}{18} \).
Q15
Find the direction in which a straight line must be drawn through the point \( (-1, 2) \) so that its point of intersection with the line \( x + y = 4 \) may be at a distance of 3 units from this point.
Answer:
Let slope be \( m = \tan\theta \). Parametric point at distance \( r=3 \) from \( (-1, 2) \) is \( (-1 + 3\cos\theta, 2 + 3\sin\theta) \).
This point lies on \( x + y = 4 \).
\( (-1 + 3\cos\theta) + (2 + 3\sin\theta) = 4 \)
\( 1 + 3(\cos\theta + \sin\theta) = 4 \Rightarrow \cos\theta + \sin\theta = 1 \).
This implies either \( \cos\theta = 1, \sin\theta = 0 \) OR \( \cos\theta = 0, \sin\theta = 1 \).
So \( \theta = 0^\circ \) (Parallel to x-axis) or \( \theta = 90^\circ \) (Parallel to y-axis).
Q16
The hypotenuse of a right angled triangle has its ends at the points (1, 3) and (– 4, 1). Find the equation of the legs (perpendicular sides) of the triangle which are parallel to the axes.
Answer:
Let vertices be A(1, 3) and B(-4, 1). The third vertex C must share coordinates with A and B since legs are parallel to axes.
Possible coordinates for C are (1, 1) or (-4, 3).
Case 1: C is (1, 1).
Leg through A(1, 3) and C(1, 1) is \( x = 1 \).
Leg through B(-4, 1) and C(1, 1) is \( y = 1 \).
Case 2: C is (-4, 3).
Leg through A(1, 3) and C(-4, 3) is \( y = 3 \).
Leg through B(-4, 1) and C(-4, 3) is \( x = -4 \).
Q17
Find the image of the point (3, 8) with respect to the line \( x + 3y = 7 \) assuming the line to be a plane mirror.
Answer:
Let image be \( (h, k) \). The line is the perpendicular bisector of the segment joining point and image.
Formula: \( \frac{h - x_1}{A} = \frac{k - y_1}{B} = -2 \frac{Ax_1 + By_1 + C}{A^2 + B^2} \).
\( \frac{h - 3}{1} = \frac{k - 8}{3} = -2 \frac{1(3) + 3(8) - 7}{1^2 + 3^2} \)
\( = -2 \frac{20}{10} = -4 \).
\( h - 3 = -4 \Rightarrow h = -1 \).
\( k - 8 = 3(-4) = -12 \Rightarrow k = -4 \).
Image: (-1, -4)
Q18
If the lines \( y = 3x + 1 \) and \( 2y = x + 3 \) are equally inclined to the line \( y = mx + 4 \), find the value of \( m \).
Answer:
Slopes: \( m_1 = 3 \), \( m_2 = 1/2 \). Let slope of third line be \( m \).
\( \left| \frac{m - 3}{1 + 3m} \right| = \left| \frac{m - 1/2}{1 + m/2} \right| \Rightarrow \left| \frac{m - 3}{1 + 3m} \right| = \left| \frac{2m - 1}{2 + m} \right| \).
Solving \( \frac{m - 3}{1 + 3m} = \frac{2m - 1}{2 + m} \) gives \( m^2 + 1 = 0 \) (No real solution).
Solving \( \frac{m - 3}{1 + 3m} = -\frac{2m - 1}{2 + m} \):
\( (m-3)(m+2) = -(2m-1)(3m+1) \)
\( m^2 - m - 6 = -(6m^2 - m - 1) \)
\( 7m^2 - 2m - 7 = 0 \Rightarrow m = \frac{2 \pm \sqrt{4 + 196}}{14} = \frac{2 \pm 10\sqrt{2}}{14} = \frac{1 \pm 5\sqrt{2}}{7} \).
Q19
If sum of the perpendicular distances of a variable point P (x, y) from the lines \( x + y - 5 = 0 \) and \( 3x - 2y + 7 = 0 \) is always 10. Show that P must move on a line.
Answer:
\( \frac{|x + y - 5|}{\sqrt{2}} + \frac{|3x - 2y + 7|}{\sqrt{13}} = 10 \).
Within any specific region bounded by these lines, the signs of the expressions inside the modulus are constant.
For example, if both are positive: \( \frac{x+y-5}{\sqrt{2}} + \frac{3x-2y+7}{\sqrt{13}} = 10 \).
This simplifies to a linear equation of the form \( Ax + By + C = 0 \), which represents a straight line.
Q20
Find equation of the line which is equidistant from parallel lines \( 9x + 6y - 7 = 0 \) and \( 3x + 2y + 6 = 0 \).
Answer:
Rewrite the second equation to match coefficients: \( 3(3x + 2y + 6) = 9x + 6y + 18 = 0 \).
Lines are \( 9x + 6y - 7 = 0 \) and \( 9x + 6y + 18 = 0 \).
The equidistant parallel line lies exactly in the middle. Its constant term is the average.
\( C = \frac{-7 + 18}{2} = \frac{11}{2} \).
Equation: \( 9x + 6y + \frac{11}{2} = 0 \Rightarrow 18x + 12y + 11 = 0 \).
Q21
A ray of light passing through the point (1, 2) reflects on the x-axis at point A and the reflected ray passes through the point (5, 3). Find the coordinates of A.
Answer:
Let A be \( (x, 0) \). The reflection of point P(1, 2) in the x-axis is P'(1, -2).
Since the light reflects, the points P', A, and Q(5, 3) must be collinear.
Slope P'A = Slope AQ.
\( \frac{0 - (-2)}{x - 1} = \frac{3 - 0}{5 - x} \Rightarrow \frac{2}{x-1} = \frac{3}{5-x} \).
\( 10 - 2x = 3x - 3 \Rightarrow 5x = 13 \Rightarrow x = \frac{13}{5} \).
Coordinates of A: \( (\frac{13}{5}, 0) \)
Q22
Prove that the product of the lengths of the perpendiculars drawn from the points \( (\sqrt{a^2-b^2}, 0) \) and \( (-\sqrt{a^2-b^2}, 0) \) to the line \( \frac{x}{a}\cos\theta + \frac{y}{b}\sin\theta = 1 \) is \( b^2 \).
Answer:
Line: \( bx\cos\theta + ay\sin\theta - ab = 0 \).
Points are foci \( (\pm ae, 0) \) where \( b^2 = a^2(1-e^2) \Rightarrow ae = \sqrt{a^2-b^2} \).
\( p_1 = \frac{|b(ae)\cos\theta - ab|}{\sqrt{b^2\cos^2\theta + a^2\sin^2\theta}} \), \( p_2 = \frac{|b(-ae)\cos\theta - ab|}{\sqrt{b^2\cos^2\theta + a^2\sin^2\theta}} \).
\( p_1 p_2 = \frac{b^2 |a^2e^2\cos^2\theta - a^2|}{b^2\cos^2\theta + a^2\sin^2\theta} = \frac{a^2b^2 (1 - e^2\cos^2\theta)}{a^2\sin^2\theta + b^2\cos^2\theta} \).
Substitute \( b^2 = a^2 - a^2e^2 \) into denominator and simplify to get \( b^2 \). Proved.
Q23
A person standing at the junction (crossing) of two straight paths represented by the equations \( 2x - 3y + 4 = 0 \) and \( 3x + 4y - 5 = 0 \) wants to reach the path whose equation is \( 6x - 7y + 8 = 0 \) in the least time. Find equation of the path that he should follow.
Answer:
Intersection of paths is the starting point. Solve \( 2x - 3y = -4 \) and \( 3x + 4y = 5 \).
Multiply 1st by 4, 2nd by 3: \( 8x - 12y = -16 \), \( 9x + 12y = 15 \). Add: \( 17x = -1 \Rightarrow x = -1/17 \).
Substitute x: \( 3(-1/17) + 4y = 5 \Rightarrow 4y = 5 + 3/17 = 88/17 \Rightarrow y = 22/17 \).
Point P \( (-1/17, 22/17) \).
Least time path is perpendicular to destination line \( 6x - 7y + 8 = 0 \).
Slope of destination line is \( 6/7 \). Perpendicular slope \( m = -7/6 \).
Equation: \( y - \frac{22}{17} = -\frac{7}{6}(x + \frac{1}{17}) \).
\( 17y - 22 = -\frac{7 \times 17}{6}(x + \frac{1}{17}) \Rightarrow 6(17y - 22) = -7(17x + 1) \).
\( 102y - 132 = -119x - 7 \Rightarrow 119x + 102y - 125 = 0 \).