## NCERT Solutions Class 11th Maths Chapter 10: Straight Lines## Exercise 10.1
Let the given quadrilateral be ABCD with the points A (4, -5), B (0, -7), C (5, -5), and D (-4, -2). In order to find the area of the quadrilateral ABCD, we will draw the diagonal AC. Area of ABCD = Area of ABC + Area if BCD Ar (ABCD) = Ar (∆ ABC) + Ar (∆ BCD) We know that the area of a triangle with vertices (x 1/2 × |[x Therefore, Ar (∆ ABC) = 1/2 × |[-4 (7 - (-5)) + 0 + 5 (5 - 7)]| = 1/2 × |[-4(12) + 5(-2)]| = 1/2 × |[-48 - 10]| = 1/2 × |(-58)| = 1/2 × 58 = 29 square units Ar (∆ BCD) = 1/2 × |[-4 (-5 + 2) + 5 (-2 - 5) + (-4) (5 - (-5))]| = 1/2 × |[-4(-3) + 5(-7) - 4(10)]| = 1/2 × |[12 - 35 - 40]| = 1/2 × |[-63]| = 1/2 × 63 = 63/2 square units Therefore, Ar (ABCD) = 29 + 63/2 = (58 + 63)/2 = 121/2 square units. Hence, the area of given quadrilateral is 121/2 square units.
Let the given equilateral triangle with side 2a be ABC. It is given that the mid-point of the traingle's base is at the origin. Thus, we can make the following figure: Since, the origin O is the mid-point of BC, BO = CO Also, AB = BC = AC = 2a BC = 2a BO + CO = 2a BO + BO = 2a 2BO = 2a BO = a = CO The coordinates of C will be (0, a) and the coordinates of B will be (0, -a). The line which joins the vertex of an equilateral triangle to the mid-point of the opposite side is perpendicular to that side. So, A lies on the x-axis and ∆ AOC and ∆ AOB are right angled triangles. By applying Pythagoras Theorem in ∆ AOC, we get: AC (2a) 4a 3a AO = √3a units The coordinates of A will be (√3a, 0) Hence, the vertices of the triangle are (√3a, 0), (0, a), and (0, -a).
Therefore, distance between P (x = √[(x = √[0 + (y = √(y = |y
Therefore, distance between P (x = √[(x = √[(x = √(x = |x
Let the given points be A (7, 6), B (3, 4) and the point on x-axis which is equidistant from the two points be O (x, y). It is given that the point O lies on the x-axis. Therefore, its y-coordinate will be zero. Since, the point O is equidistant from A and B. Therefore, Distance between A and O = Distance between B and O √[(7 - x) Square both sides (7 - x) (7 - x) 49 + x 60 = 8x
Hence, the require point is (15/2, 0).
Let the mid-point of the line segment joining the points P (0, -4) and B (8, 0) be (x, y). x = (0 + 8)/2 = 8/2 = 4 y = (-4 + 0)/2 = -4/2 = -2 Therefore, the mid-point of the line PB is (4, -2). Slope of the line passing through (0, 0) and (4, -2) = m = (-2 - 0)/(4 - 0) = -1/2 Hence, 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) is -1/2.
Let the three vertices of the triangle be A (4, 4), B (3, 5) and C (-1, -1). Slope of the line passing through A (4, 4) and B (3, 5) = m Slope of the line passing through B (3, 5) and C (-1, -1) = m = (-6)/(-4) = 3/2 Slope of the line passing through A (4, 4) and C (-1, -1) = m Now, we can see that m Thus, the lines AB and AC are perpendicular to each other. Therefore, the triangle is right-angled at the point A (4, 4). Hence, (4, 4), (3, 5) and (-1, -1) are the vertices of a right angled triangle.
It is given that the line makes an angle of 30° with the positive direction of y-axis measured anticlockwise. Therefore, Angle made by the line with the positive direction of the x-axis measured anti-clock-wise = 90° + 30° = 120° Slope of the given line = m = tan 120° = tan (180° - 60°) = -tan 60° = -√3
Let the given points be A (x, -1), B (2, 1) and C (4, 5). It is given that the three given points are collinear. Therefore, Slope of line segment AB = m (1 + 1)/(2 - x) = (5 - 1)/(4 - 2) 2/(2 - x) = 4/2 2/(2 - x) = 2 2 = (2 - x)2 1 = 2 - x x + 1 = 2
Hence, x = 1.
Let the given points be A (-2, -1), B (4, 0), C (3, 3) and D (-3, 2). Slope of the line AB = m Slope of the line CD = m m Thus, AB ∥ CD. Slope of the line BC = m Slope of the line AD = m m Thus, BC ∥ AD. Now, we can see that opposite sides of the quadrilateral ABCD are parallel. Therefore, ABCD is a parallelogram. Hence, the given vertices (-2, -1), (4, 0), (3, 3) and (-3, 2) are the vertices of a parallelogram.
Slope of the line joining the points (3, -1) and (4, -2) = (-2 + 1)/(4 - 3) = (-1)/1 = -1 Let the angle of inclination of the line joining the points (3, -1) and (4, -2) be θ. So, tan θ = -1 tan θ = -tan 45° tan θ = tan (90° + 45°) Therefore, θ = 135° Hence, the angle between x-axis and the line joining the points (3, -1) and (4, -2) is 135°.
Let us consider m We know that if θ is the angle between the lines l tan θ = |(m It is given that the tangent of the angle between the two lines is 1/3. Therefore, 1/3 = |(m - 2m)/(1 + 2m 1/3 = |-m/(1 + 2m CASE I: 1/3 = -m/(1 + 2m 1 + 2m 2m 2m (m + 1) + 1(m + 1) = 0 (2m + 1) (m + 1)= 0 m = -1 or -1/2 If m = -1, then the slope of the lines are -1 and -2 If m = -1/2, then the slope of the lines are -1/2 and -1 CASE II: 1/3 = m/(1 + 2m 2m 2m 2m (m - 1) - 1(m - 1) = 0 m = 1 or 1/2 If m = 1, then the slope of the lines are 1 and 2 If m = 1/2, then the slope of the lines are 1/2 and 1 Hence, the slope of the lines are [-1 and -2] or [-1/2 and -1] or [1 and 2] or [1/2 and 1]
It is given that slope of the line is 'm'. Also, the slope of the line passing through (x So, (k - y (k - y Hence, proved.
Let us consider that the given points A (h, 0), B (a, b) and C (0, k) lie on a line. Then, the slope of AB = slope of BC (b - 0)/(a - h) = (k - b)/(0 - a) -ab = (k - b) (a - h) -ab = ka - kh - ab + bh ka + bh = kh Divide both sides by kh; we get ka/kh + bh/kh = kh/kh a/h + b/k = 1 Hence, proved.
We know that line AB passes through points A (1985, 92) and B (1995, 97). Its slope will be (97 - 92)/(1995 - 1985) = 5/10 = 1/2 Let y be the population in the year 2010. Then, according to the given graph, AB must pass through point C (2010, y) So now, slope of AB = slope of BC 1/2 = (y - 97)/(2010 - 1995) 15/2 = y - 97 y = 7.5 + 97 = 104.5 Hence, the slope of line AB is 1/2, while in the year 2010, the population will be 104.5 crores. ## EXERCISE 10.2
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 y = 0.
It is given that the line passes through the point (-4, 3) and slope is m = 1/2 We know that the point (x, y) lies on the line with slope m through the fixed point (x So, y - 3 = 1/2 (x - (-4)) y - 3 = 1/2 (x + 4) 2(y - 3) = x + 4 2y - 6 = x + 4 x + 4 - (2y - 6) = 0 x + 4 - 2y + 6 = 0 x - 2y + 10 = 0 Hence, the equation of the line is x - 2y + 10 = 0
It is given that the line passes through the point (0, 0) and slope is m. We know that the point (x, y) lies on the line with slope m through the fixed point (x So, y - 0 = m (x - 0) y = mx y - mx = 0 Hence, the equation of the line is y - mx = 0.
It is given that the line passes through the point (2, 2√3) and inclined with x-axis at angle of θ = 75° Slope of the line = m = tan θ = tan 75° = tan (45° + 30°) Using the formula tan (A + B) = (tan A + tan B)/(1 - tan A tan B), we get: tan (45° + 30°) = (tan 45° + tan 30°)/(1 - tan 45° tan 30°) = (1 + 1/√3)/(1 - 1 × 1/√3) = (1 + 1/√3)/(1 - 1/√3) = (√3 + 1)/√3 × √3/(√3 - 1) = (√3 + 1)/(√3 - 1) Rationalise, tan 75° = (√3 + 1)/(√3 - 1) × (√3 + 1)/(√3 + 1) = (√3 + 1) = (3 + 2√3 + 1)/2 = (4 + 2√3)/2 = 2(2 + √3)/2 = √3 + 2 We know that the point (x, y) lies on the line with slope m through the fixed point (x So, y - 2√3 = (√3 + 2) (x - 2) y - 2√3 = x√3 + 2x - 2√3 - 4 y = x√3 + 2x - 4 y = x(√3 + 2) - 4 y - x(√3 + 2) + 4 = 0 Hence, the equation of the line is y - x(√3 + 2) + 4 = 0.
It is given that the slope of the line is m = -2 and the line intersects the x-axis at a distance of 3 units to the left of origin. We know that if a line with slope m makes x-intercept d, then the equation of the line is y = m(x - d) d = -3 y = (-2)(x - (-3)) y = (-2) (x + 3) y = -2x - 6 2x + y + 6 = 0 Hence, the equation of the line is 2x + y + 6 = 0.
It is given that the line makes θ = 30° with the positive direction of x-axis and intersects the y-axis at a distance of 2 units above the origin. Slope of the line = m = tan θ = tan 30° m = 1/√3 We know that if a line with slope m makes x-intercept c, then the equation of the line is y = m(x - c) c = 2 y = (1/√3)(x - 2) √3y = x - 2 √3y - x + 2 = 0 Hence, the equation of the line is √3y - x + 2 = 0.
It is given that the line passes through the points (-1, 1) and (2, -4) We know that the equation of a line passing through two points (x y - y y - 1 = (-4 - 1)/(2 - (-1)) × (x - (-1)) y - 1 = (-5)/(2 + 1) × (x + 1) y - 1 = (-5x - 5)/3 3y - 3 = -5x - 5 5x + 3y + 2 = 0 Hence, the equation of the line is 5x + 3y + 2 = 0.
It is given that the perpendicular distance between the line and the origin is p = 5 units and the angle made by the perpendicular with the positive x-axis is ω = 30°. We know that the equation of the line having normal distance p from the origin and angle ω, which the normal makes with the positive direction of the x-axis, is given by x cos ω + y sin ω = p x cos 30° + y sin 30° = 5 x (√3/2) + y(1/2) = 5 √3x + y = 5(2) √3x + y = 10 √3x + y - 10 = 0 Hence, the equation of the line is √3x + y - 10 = 0.
Let RL be the median of the vertex R. Since, RL is a median. Therefore, L (x Using the midpoint formula, we have: x = (2 - 2)/2 = 0 y = (1 + 3)/2 = 4/2 = 2 So, L is (0, 2). We know that the equation of a line passing through two points (x y - y y - 5 = (2 - 5)/(0 - 4) × (x - 4) y - 5 = (-3)/(-4) × (x - 4) y - 5 = 3(x - 4)/4 4(y - 5) = 3(x - 4) 4y - 20 = 3x - 12 3x - 4y + 8 = 0 Hence, the equation of median through the vertex R is 3x - 4y + 8 = 0.
It is given that the line passes through the point (-3, 5) and the perpendicular to this line passes through the points (2, 5) and (-3, 6). Therefore, slope of the perpendicular is m = (6 - 5)/(-3 - 2) = 1/(-5) = -1/5 We know that two non-vertical lines are perpendicular to each other only if their slopes are negative reciprocals of each other. Therefore, the slope of the line is M = -m = -(1/m) = -(-5/1) = 5 We know that the point (x, y) lies on the line with slope m through the fixed point (x Then, y - 5 = 5(x - (-3)) y - 5 = 5x + 15 5x + 15 - y + 5 = 0 5x - y + 20 = 0 Hence, the equation of the line is 5x - y + 20 = 0
We know that the coordinates of a point dividing the line segment joining the points (x x-coordinate = (mx = (1(2) + n(1)/(1 + n) = (2 + n)/(1 + n) and y-coordinate = (my = (1(3) + n(0))/(1 + n) = 3/(1 + n) We know that slope of a line is m = (y = (3 - 0)/(2 - 1) = 3/1 = 3 We know that two non-vertical lines are perpendicular to each other only if their slopes are negative reciprocals of each other. M = -m = -(1/m) = -(1/3) = -1/3 We know that the point (x, y) lies on the line with slope m through the fixed point (x y - 3(1 + n) = -1/3 × (x - (2 + n)/(1 + n)) 3((1 + n) y - 3) = (-(1 + n) x + 2 + n) 3(1 + n) y - 9 = -(1 + n) x + 2 + n (1 + n) x + 3(1 + n) y - n - 9 - 2 = 0 (1 + n) x + 3(1 + n) y - n - 11 = 0 Hence, the equation of the line is (1 + n) x + 3(1 + n) y - n - 11 = 0
We know that equation of the line intercepts a and b on the x-and the y-axis, respectively, which is x/a + y/b = 1 It is given that the line cuts off equal intercepts on the coordinate axes. Therefore, a = b x/a + y/a = 1 1/a × (x + y) = 1 x + y = a Put x = 2 and y = 3 as the given points is (2, 3). 2 + 3 = a a = 5 So, the equation will be x + y - 5 = 0
We know that equation of the line-making intercepts a and b on the x-and the y-axis, respectively, is x/a + y/b = 1 It is given that the sum of intercepts is 9. Therefore, a + b = 9 a = 9 - b So, x/a + y/b = 1 x/(9 - b) + y/b = 1 [bx + y(9 - b)]/b(9 - b) = 1 bx + 9y - by = 9b - b b Put x = 2 and y = 2 as the line passes through (2, 2). b b b b b(b - 3) - 6(b - 3) = 0 (b - 3) (b - 6) = 0 (b - 3) = 0 ⇒ OR (b - 6) = 0 ⇒ When b = 3, a = 9 - b = 9 - 3
Then, the equation will be x/3 + y/6 = 1 (2x + y)/6 = 1 2x + y = 6
When b = 6, a = 9 - b = 9 - 6
The, the equation will be x/6 + y/3 = 1 (x + 2y)/6 = 1 x + 2y = 6
Hence, the equation of the line is 2x + y - 6 = 0 or x + 2y - 6 = 0.
It is given that the line passes through the point (0, 2) and makes an angle θ = 2π/3 with the positive x-axis. Slope of the line = m = tan θ = tan 2π/3 m = -√3 We know that the point (x, y) lies on the line with slope m through the fixed point (x y - 2 = -√3(x - 0) y - 2 = -√3x √3x + y - 2 = 0 The line which is parallel to the line in above equation crosses the y-axis at a distance of 2 units below the origin. Therefore, It passes through the point (0, -2) and its slope is m = -√3. Thus, its equation will be y - (-2) = -√3(x - 0) y + 2 = -√3x √3x + y + 2 = 0 Hence, the equation of the line is √3x + y - 2 = 0, and that of the line parallel to it is √3x + y + 2 = 0.
The perpendicular to the required line passes through the points (0, 0) and (-2, 9). Slope of this perpendicular m m We know that two non-vertical lines are perpendicular to each other only if their slopes are negative reciprocals of each other. m m We know that the point (x, y) lies on the line with slope m through the fixed point (x y - 9 = 2/9 × (x - (-2)) 9(y - 9) = 2(x + 2) 9y - 91 = 2x + 4 2x + 4 - 9y + 81 = 0 2x - 9y + 85 = 0 Hence, the equation of the line is 2x - 9y + 85 = 0.
Let us assume L along x-axis and C along y-axis. It is given that L = 124.942 when C = 20 and L = 125.134 when C = 110 so we have two points (124.942, 20) and (125.134, 110) in XY-plane. We know that the equation of the line passing through the points (x y - y C - 20 = (110 - 20)/(125.134 - 124.942) × (L - 124.942) C - 20 = 90/(0.192) × (L - 124.942) 0.192(C - 20) = 90 (L - 124.942) L - 124.942 = 0.192(C - 20)/90 L = 0.192(C - 20)/90 + 124.942 Hence, the required relation between L and C is L = 0.192(C - 20)/90 + 124.942.
Let us assume the selling price for each litre along x-axis and the demand along y-axis. It is given that if selling price per litre = Rs 14 then demand = 980 litres and when selling price per litre = Rs 16 then demand = 1220 litres so we have two points (14, 980) and (16, 1220) in XY-plane. We know that the equation of the line passing through the points (x y - y y - 980 = (1220 - 980)/(16 - 14) × (x - 14) y - 980 = 240/2 × (x - 14) y - 980 = 120(x - 14) y = 120(x - 14) + 980 Put selling price per litre = x = 17 y = 120(17 - 14) + 980 y = 120(3) + 980 y = 360 + 980 = 1340 Hence, the owner can sell 1340 litres of milk each week at Rs 17/litre.
Let the required line segment be AB. A is (0, y) and B is (x, 0) as the line segment lies between the axes. Mid-point of line segment AB is given by ((x + 0)/2, (0 + y)/2). It is given that P (a, b) is the mid-point of the segment AB. Therefore, a = x/2 x = 2a AND b = y/2 y = 2b So, A is (0, 2b) and B is (2a, 0) We know that the equation of the line passing through the points (x y - y y - 2b = (0 - 2b)/(2a - 0) × (x - 0) y - 2b = (-2b)x/2a y - 2b = -bx/a 2b - y = bx/a a(2b - y) = bx 2ab - ay = bx bx + ay = 2ab Divide both side by ab bx/ab + ay/ab = 2ab/ab
Hence, proved.
Let the required line segment be 1. A is (0, y) and B is (x, 0) as the line segment lies between the axes. We know that the coordinates of a point dividing the line segment join the points (x It is given that the point R (h, k) divides the segment AB in ratio 1 : 2. Therefore, h = (1(0) + 2(x))/(1 + 2) h = 2x/3 x = 3h/2 AND k = (1(y) + 2(0))/(1 + 2) k = y/3 y = 3k
y - y y - 3k = (0 - 3k)/(3h/2 - 0) × (x - 0) y - 3k = (-3k)x/(3h/2) (3h/2) × (y - 3k) = -3kx 3hy/2 - 9hk/2 = -3kx 3hy - 9hk = -6kx 6kx + 3hy = 9hk Divide both sides by 9hk 6kx/9hk + 3hk/9hk = 9hk/9hk 2x/3h + 1/3 = 1 Hence, the equation of the line segment is given by 2x/3h + 1/3 = 1.
We need to prove that the given three points (3, 0), (-2, -2) and (8, 2) are collinear. The three points will be collinear if the line passing through the points (3, 0) and (-2, -2) also passes through (8, 2). We know that the equation of a line passing through the points (x y - y y - 0 = (-2 - 0)/(-2 - 3) × (x - 3) y = (-2)/(-5) × (x - 3) 5y = 2(x - 3) 5y = 2x - 6 2x - 5y - 6 = 0 Check LHS with x = 8 and y = 2 2(8) - 5(2) - 6 = 16 - 10 - 6 = 0 = RHS Thus, the line 2x - 5y - 6 = 0 passing through (3, 0) and (-2, -2) also passes through the point (8, 2). Hence, proved that the points (3, 0), (-2, -2) and (8, 2) are collinear. ## Exercise 10.3
The slope-intercept form is represented in the form y = mx + c, where m is the slope and c is the y-intercept. So, the above equation can be expressed as y = -1/7x + 0 Hence, the above equation is of the form y = mx + c, where m = -1/7 and c = 0.
The slope-intercept form is represented in the form y = mx + c, where m is the slope and c is the y-intercept. So, the above equation can be expressed as 3y = -6x + 5 y = -6/3x + 5/3 y = -2x + 5/3 Hence, the above equation is of the form y = mx + c, where m = -2 and c = 5/3.
The slope-intercept form is given by y = mx + c, where m is the slope and c is the y-intercept. y = 0 × x + 0 Hence, the above equation is of the form y = mx + c, where m = 0 and c = 0.
The equation of the line in intercept form is given by x/a + y/b = 1, where a and b are intercepted on the x-axis and the y-axis, respectively. So, 3x + 2y = 12 Divide both sides by 12 3x/12 + 2y/12 = 12/12 x/4 + y/6 = 1 Therefore, above equation is of the form x/a + y/b = 1, where a = 4, b = 6 Hence, the intercept on the x-axis is 4 and the intercept on the y-axis is 6.
The equation of the line in intercept form is given by x/a + y/b = 1, where a and b are intercepted on the x-axis and the y-axis, respectively. So, 4x - 3y = 6 Divide both sides by 6 4x/6 - 3y/6 = 6/6 2x/3 - y/2 = 1 x/(3/2) + y/(-2) = 1 Therefore, above equation is of the form x/a + y/b = 1, where a = 3/2, b = -2 Hence, the intercept on the x-axis is 3/2 and the intercept on the y-axis is -2.
The equation of the line in intercept form is given by x/a + y/b = 1, where a and b are intercepted on the x-axis and the y-axis, respectively. So, 3y = -2 Divide both sides by -2 3y/(-2) = -2/-2 3y/(-2) = 1 y/(-2/3) = 1 Therefore, above equation is of the form x/a + y/b = 1, where a = 0, b = -2/3 Hence, the intercept on the x-axis is 0 and the intercept on the y-axis is -2/3.
The equation of the line in normal form is given by x cos θ + y sin θ = p where θ is the angle between the perpendicular and the positive x-axis and p is the perpendicular distance from the origin. So, x - √3y + 8 = 0 x - √3y = -8 Divide both the sides by √(1 x/2 - √3y/2 = -8/2 (-1/2)x + (√3/2)y = 4 Now, x cos θ = x(-1/2) cos θ = -cos 60° cos θ = cos (180° - 60°) cos θ = cos 120° AND y sin θ = y (√3/2) sin θ = sin 60° sin θ = sin (180° - 60°) sin θ = 120° The above equation is of the form x cos θ + y sin θ = p, where θ = 120° and p = 4. Hence, perpendicular distance of the line from origin = 4 and the angle between the perpendicular and positive x-axis = 120°
The equation of the line in normal form is given by x cos θ + y sin θ = p where θ is the angle between the perpendicular and the positive x-axis and p is the perpendicular distance from the origin. So, 0.x + y = 2 Divide both the sides by √(0 0(x) + 1(y) = 2 Now, x cos θ = x(0) cos θ = cos 90° AND y sin θ = y(1) sin θ = sin 90° The above equation is of the form x cos θ + y sin θ = p, where θ = 90° and p = 2. Hence, perpendicular distance of the line from origin = 2 and the angle between the perpendicular and positive x-axis = 90°
The equation of the line in normal form is given by x cos θ + y sin θ = p where θ is the angle between the perpendicular and the positive x-axis and p is the perpendicular distance from the origin. So, x - y = 4 Divide both the sides by √(1 x/√2 - y/√2 = 4/√2 (1/√2)x + (-1/√2)y = 2√2 Now, x cos θ = x(1/√2) cos θ = cos 45° AND y sin θ = y(-1/√2) sin θ = -sin 45° sin θ = sin (360° - 45°) sin θ = sin 315° cos 45° = cos (360° + 45°) So, cos θ = cos 315° The above equation is of the form x cos θ + y sin θ = p, where θ = 315° and p = 2√2. Hence, perpendicular distance of the line from origin = 2√2 and the angle between the perpendicular and positive x-axis = 315°
12(x + 6) = 5(y - 2) 12x + 72 = 5y - 10 12x - 5y + 82 = 0 On comparing the above equation with general equation of a line Ax + By + C = 0, we get A = 12, B = -5 and C = 82. Perpendicular distance (d) of the line from point (-1, 1) will be d = |Ax d = |12 × (-1) + (-5) × 1 + 82|/√(12 = |-12 - 5 + 82|/√(144 + 25) = |65|/√169 = 5 units Hence, the distance is 5 units.
x/3 + y/4 = 1 (4x + 3y)/12 = 1 4x + 3y = 12 4x + 3y - 12 = 0 On comparing the above equation with general equation of a line Ax + By + C = 0, we get A = 4, B = 3 and C = -12. Let the point on x-axis whose distance from the line is 4 units be (a, 0).
d = |(C d = |(-34 - 31)/√(15 = |-65/√(225 + 64)| = |-65/√289| = 65/17
l(x + y) + p = 0 AND l(x + y) - r = 0 lx + ly + p = 0 AND lx + ly - r = 0 The distance (d) between parallel lines Ax + By + C d = |(C d = |(p - (-r))/√(l = |(p + r)/√(2l = |p + r|/l√2
3x - 4y + 2 = 0 4y = 3x + 2 y = (3x + 2)/4 y = 3x/4 + 1/2 The above equation is of the form y = mx + c, where m is the slope of the line. Therefore, slope of the given line is 3/4. Parallel lines have same slope, so the slope of the parallel line = 3/4 The equation of line having slope m and passing through (x y - y Equation of the line having slope 3/4 and passing through (-2, 3) is y - 3 = 3/4 × (x - (-2)) 4(y - 3) = 3(x + 2) 4y - 12 = 3x + 6 3x - 4y = -18 Hence, the equation is 3x - 4y = -18.
x - 7y + 5 = 0 7y = x + 5 y = (x + 5)/7 y = x/7 + 5/7 The above equation is of the form y = mx + c, where m is the slope of the line. Therefore, slope of the given line is 1/7. Slope of the line perpendicular to the above line = -1/m = -(7/1) = -7 The equation of line having slope m and x-intercept d is given by y = m(x - d) Equation of the line having slope -7 and x-intercept 3 y = (-7)(x - 3) y = -7x + 21 7x + y = 21 Hence, the equation is 7x + y = 21.
√3x + y = 1 and x + √3y = 1 y = 1 - √3x and y = (1 - x)/√3 y = -√3x + 1 and y = -x/√3 + 1/√3 The above equations are of the form y = mx + c, where m is the slope of the line. Slope of the line √3x + y = 1 is m Let the angle between the two lines be θ tan θ = |(m tan θ = |(-√3 - (-1/√3))/(1 + (-√3)(-1/√3))| tan θ = |(-√3 + 1/√3)/(1 + 1)| tan θ = |((-3 + 1)/√3)/2| tan θ = |-2/2√3| tan θ = |-1/√3| tan θ = 1/√3 tan θ = tan 30° θ = 30° Hence, the angle between the given lines is 30°.
Let the slope of the line passing through (h, 3) and (4, 1) be m m Let the slope of line 7x - 9y - 19 = 0 be m 7x - 9y - 19 = 0 9y = 7x - 19 y = (7x - 19)/9 y = 7x/9 - 19/9 The above equation is of the form y = mx + c, where m is the slope of the line. So, m The given lines are perpendicular. Thus, m -2/(4 - h) × 7/9 = -1 -14 = -9(4 - h) 14 = 36 - 9h 9h = 22
Let the slope of line Ax + By + C = 0 be m Ax + By + C = 0 By = C - Ax y = (C - Ax)/B y = -A/Bx - C/B The above equation is of the form y = mx + c, where m is the slope of the line. So, m = -A/B Equation of the line passing through point (x y - y y - y B (y - y A(x - x So, the line through point (x Hence, proved.
Given: m Let the slope of the first line be m And let the slope of the other line be m The angle between the two lines is 60°. So, tan θ = |(m tan 60° = |(2 - m √3 = ± (2 - m When √3 = (2 - m √3 = (2 - m √3 (1 + 2m √3 + 2√3m 2√3m m
When √3 = -(2 - m √3 = -(2 - m √3 (1 + 2m √3 + 2√3m m m m
CASE I: m The equation of the line passing through point (2, 3) and slope m y - 3 = (2 - √3)/(2√3 + 1) × (x - 2) (2√3 + 1)(y - 3) = (2 - √3) × (x - 2) (2√3 + 1)y - 3(2√3 + 1) = x(2 - √3) - 2(2 - √3) (2√3 + 1)y + x(√3 - 2) = -4 + 2√3 + 6√3 - 3
Therefore, equation of the other line is (2√3 + 1)y + x(√3 - 2) = 8√3 - 1. CASE II: m y - 3 = -(2 + √3)/(2√3 - 1) × (x - 2) (2√3 - 1)(y - 3) = (2 + √3) × (2 - x) (2√3 - 1)y - 3(2√3 - 1) = 2(2 + √3) - (2 + √3)x (2√3 - 1)y - (2 + √3)x = 4 + 2√3 + 6√3 - 3
Therefore, equation of the other line is (2√3 - 1)y - (2 + √3)x = 8√3 + 1.
The right bisector of a line segment bisects the line segment at 90°. End-points of the line segment AB are given as A (3, 4) and B (-1, 2). Let the mid-point of AB be (x, y). x = (3 - 1)/2 = 2/2 = 1 y = (4 + 2)/2 = 6/2 = 3 (x, y) is (1, 3). Let the slope of line AB be m m = -2/(-4) = 1/2 And let the slope of the line perpendicular to AB be m m = -1/(1/2) = -2 The equation of the line passing through (1, 3) and having a slope of -2 is (y - 3) = -2 (x - 1) y - 3 = -2x + 2 2x + y = 5 Hence, the required equation of the line is 2x + y = 5.
Let us consider the coordinates of the foot of the perpendicular from (-1, 3) to the line 3x - 4y - 16 = 0 be (a, b) Let the slope of the line joining (-1, 3) and (a, b) be m m Let the slope of the line 3x - 4y - 16 = 0 be m 4y = 3x - 16 y = (3x - 16)/4 y = 3x/4 - 4 The above equation is of the form y = mx + c, where m is the slope of the line. So, m The given two lines are perpendicular. Therefore, m (b - 3)/(a + 1) × 3/4 = -1 3(b - 3) = -4(a + 1) 3b - 9 = -4a - 4 4a + 3b = 5 4a = 5 - 3b
We know that (a, b) also lies on the line 3x - 4y = 16. So, 3a - 4b = 16 3 × (5 - 3b)/4 - 4b = 16 (15 - 9b)/4 = 16 + 4b 15 - 9b = 4(16 + 4b) 15 - 9b = 64 + 16b -49 = 25b
Then, a = (5 - 3(-49)/25)/4 = (5 + 147/25)/4 = (125 + 147)/4(25) = 272/4(25)
Hence, the coordinates of the foot of the perpendicular are (68/25, -49/25).
The perpendicular from the origin meets the given line at (-1, 2). The equation of the line is y = mx + c The line joining the points (0, 0) and (-1, 2) is perpendicular to the given line. So, the slope of the line joining (0, 0) and (-1, 2) = 2/(-1) = -2 The slope of the given line is m. m × (-2) = -1 m = 1/2 Since point (-1, 2) lies on the given line, y = mx + c 2 = 1/2 × (-1) + c c = 2 + 1/2 = 5/2 Hence, the values of m and c are 1/2 and 5/2, respectively.
x cos θ - y sin θ = k cos 2θ x sec θ + y cosec θ = k Perpendicular distance (d) of a line Ax + By + C = 0 from a point (x d = |Ax On comparing x cos θ - y sin θ - k cos 2θ = 0 to Ax + By + C = 0, we get A = cos θ, B = -sin θ, C = -k cos 2θ It is given that p is the length of perpendicular from (0, 0) to line x cos θ - y sin θ - k cos 2θ = 0 p = |cos θ (0) + (-sin θ)(0) + (-k cos 2θ)|/√(cos θ) = |-k cos 2θ|/√(cos = k cos 2θ Square both sides
On comparing x sec θ + y cosec θ - k = 0 to Ax + By + C = 0, we get A = sec θ, B = cosec θ, C = -k It is given that q is the length of perpendicular from (0, 0) to line x sec θ + y cosec θ - k = 0 q = |sec θ (0) + (cosec θ)(0) + (-k)|/√(sec θ) = |-k|/√(sec = |-k|/√(1/cos = k/√((sin = k cos θ sin θ Multiply both sides by 2 2q = k 2 cos θ sin θ 2q = k sin 2θ Square both sides
Add Equation (I) and (II) p p
Hence, proved.
Let AD be the altitude of triangle ABC from vertex A. D lies in BC and AD ⊥ BC. Given vertices of the triangle are A (2, 3), B (4, -1) and C (1, 2) Let the slope of the line BC = m m m Let the slope of the line AD be m AD is perpendicular to BC. m -1 × m m The equation of the line passing through the point (2, 3) and having a slope of 1 will be y - 3 = 1 × (x - 2) y - 3 = x - 2 y - x = 1 Therefore, equation of the altitude of ∆ ABC from vertex A is y - x = 1 The equation of BC will be y + 1 = -1 × (x - 4) y + 1 = -x + 4 x + y - 3 = 0 Perpendicular distance (d) of a line Ax + By + C = 0 from a point (x d = |Ax On comparing x + y - 3 = 0 to Ax + By + C = 0, we get A = 1, B = 1, C = -3 Length of AD = Perpendicular distance from A (2, 3) to BC = d = |1(2) + 1(3) + (-3)|/√(1 = |2 + 3 - 3|/√(1 + 1) = |2|/√2 = √2 units Hence, the equation and length of altitude from the vertex A is y - x = 1 and √2 units.
The equation of a line whose intercepts on the axes are a and b will be x/a + y/b = 1 (bx + ay)/ab = 1 bx + ay = ab bx + ay - ab = 0 Perpendicular distance (d) of a line Ax + By + C = 0 from a point (x d = |Ax On comparing to bx + ay - ab = 0 to Ax + By + C = 0, we get A = b, B = a, C = -ab It is given that p is the length of perpendicular from (0, 0) to line bx + ay - ab = 0. p = |b(0) + a(0) + (-ab)|/√(b p = |-ab|/√(a Square both sides p 1/p 1/p 1/p Hence, proved. ## Miscellaneous Exercise
(k - 3) x - (4 - k (4 - k y = [(k - 3) x + k y = (k - 3)x/(4 - k The above equation is of the form y = mx + c, where m is the slope of the line. So, m = (k - 3)/(4 - k
Slope of the line = Slope of the x-axis (k - 3)/(4 - k k - 3 = 0
Slope of the line = Slope of the y-axis (k - 3)/(4 - k (k - 3)/(4 - k 4 - k k
(k - 3)(0) - (4 - k k k k(k - 1) - 6(k - 1) = 0 (k - 1)(k - 6) = 0 (k - 6) = 0 ⇒ OR (k - 1) = 0 ⇒
√3x + y + 2 = 0 -√3x - y = 2 Divide both sides by √((√-3) -√3x/2 - y/2 = 2/2 x(-√3/2) + y(-1/2) = 1 On comparing the above equation to x cos θ + sin θ = p, we get cos θ = -√3/2 cos θ = -cos 30° sin θ = -1/2 sin θ = -sin 30° and p = 1 Since, cos θ and sin θ are negative. Therefore, θ = 180° + 30° = 210° Hence, value of θ = 210° and value of p = 1
Let the cut-off intercepts on the axes be a and b respectively. a + b = 1 a = 1 - b ab = -6 (1 - b)b = -6 b - b b b b(b - 3) + 2(b - 3) = 0 (b - 3)(b + 2) = 0 (b - 3) = 0 ⇒ OR (b + 2) = 0 ⇒ When b = 3: a = 1 - 3
When b = -2: a = 1 + 2
We know that the equation of the line whose intercepts on the a and b axes is given by x/a + y/b = 1 (bx + ay)/ba = 1 bx + ay = ba bx + ay - ba = 0 Thus, CASE I: a = -2 and b = 3 Equation of the line will be 3x + (-2)y - (3)(-2) = 0 3x - 2y + 6 = 0 3x - 2y = -6 CASE II: a = 3 and b = -2 Equation of the line will be 3x + (-2)y - (3)(-2) = 0 3x - 2y + 6 = 0 3x - 2y = -6-2x + 3y - (-2)(3) = 0 -2x + 3y + 6 = 0 2x - 3y = 6 Hence, the required equation of the lines are 2x - 3y = 6 and -3x + 2y = 6.
x/3 + y/4 = 1 (4x + 3y)/12 = 1 4x + 3y = 12 4x + 3y - 12 = 0 Let the point on y-axis whose distance from the line 4x + 3y - 12 = 0 is 4 be (0, b). On comparing the above equation with the general equation of line Ax + By + C = 0, we get A = 4, B = 3, and C = -12 We know that the perpendicular distance (d) of a line Ax + By + C = 0 from (x d = |Ax The point is (0, b) and distance from the line is 4 units. 4 = |4(0) + 3(b) - 12|/√(4 4 = |3b - 12|/√(16 + 9) 4 = |3b - 12|/5 20 = ± (3b - 12) So, 20 = 3b - 12 3b = 32
OR 20 = -3b + 12 3b = -8
Hence, the required points are (0, 32/3) and (0, -8/3).
Equation of the line joining the points (cos θ, sin θ) and (cos φ, sin φ) will be y - sin θ = (sin φ - sin θ)/(cos φ - cos θ) × (x - cos θ) (y - sin θ)(cos φ - cos θ) = (sin φ - sin θ)(x - cos θ) y(cos φ - cos θ) - sin θ (cos φ - cos θ) = x(sin φ - sin θ) - cos θ (sin φ - sin θ) y(cos φ - cos θ) + x(sin θ - sin φ) + cos θ sin φ - cos θ sin θ + sin θ cos θ - sin θ cos φ = 0 x(sin θ - sin φ) + y(cos φ - cos θ) + sin(φ - θ) = 0 On compare the above equation to Ax + By + C = 0, we get A = sin θ - sin φ, B = cos φ - cos θ, and C = sin(φ - θ) We know that the perpendicular distance (d) of a line Ax + By + C = 0 from (x d = |Ax The point is (0, 0), so distance from the line will be d = |(sin θ - sin φ)(0) + (cos φ - cos θ)(0) + sin(φ - θ)|/√((sin θ - sin φ) = |sin(φ - θ)|/√(sin = |sin(φ - θ)|/√(sin = |sin(φ - θ)|/√(1 + 1 - 2(cos (φ - θ))) = |sin(φ - θ)|/√2(1 - cos (φ - θ)) = |sin(φ - θ)|/√2(2 sin = |(sin(φ - θ))/2 sin
We know that the equation of any line parallel to the y-axis is of the form x = a Given lines are x - 7y + 5 = 0 3x + y = 0 ⇒ y = -3x Substitute y = -3x in x - 7y + 5 = 0 x - 7(-3x) + 5 = 0 x + 21x + 5 = 0 22x + 5 = 0 22x = -5
So, y = -3(-5/22)
So, (-5/22, 15/22) is the point of intersection of the two given lines. If x = a passes through this point then -5/22 = a. Therefore, required equation will be x = -5/22
x/4 + y/6 = 1 (6x + 4y)/24 = 1 6x + 4y = 24 2(3x + 2y) = 24 3x + 2y = 12 2y = 12 - 3x y = 6 - 3x/2 y = x(-3/2) + 6 The above equation is of the form y = mx + c, where m is the slope of the line. So, m = -3/2 Therefore, slope of the line perpendicular to the given line = -1/m = -1/(-3/2) = 2/3 Let the point at which the given line meets the y-axis be (0, y). 3x + 2y = 12 3(0) + 2y = 12 2y = 12 y = 6 Therefore, the given line intersects the y-axis at the point (0, 6). The equation of the line passing through the point (0, 6) and having a slope of 2/3 will be y - 6 = 2/3 × (x - 0) 3(y - 18) = 2x 3y - 18 = 2x 2x - 3y + 18 = 0 Hence, the required equation of the line is 2x - 3y + 18 = 0.
y - x = 0 … (I) x + y = 0 … (II) x - k = 0 … (III) Point of intersection of lines (I) and (II): y - x = 0 y = x Substitute y = x in (II) x + x = 0 2x = 0
So, Thus, lines (I) and (II) intersect at (0, 0) Point of intersection of lines (II) and (III): x + y = 0 x = -y Substitute x = -y in (III) -y - k = 0
So, Thus, lines (II) and (III) intersect at (k, -k). Point of intersection of lines (I) and (III): y - x = 0 y = x Substitute y = x in (III) y - k = 0
So, Thus, lines (I) and (III) intersect at (k, k). Therefore, the triangle formed by the three lines will have the vertices (0, 0), (k, -k), and (k, k). We know that the area of triangle whose vertices are (x 1/2 × |x Area of the given triangle = 1/2 × |0(-k - k) + k(k - 0) + k(0 - (-k))| = 1/2 × |0 + k = 1/2 × |2k = k
3x + y - 2 = 0 … (I) px + 2y - 3 = 0 … (II) 2x - y - 3 = 0 … (III) Point of intersection of lines (I) and (III): 3x + y = 2 y = 2 - 3x Substitute y = 2 - 3x in (III) 2x - 2 + 3x = 3 5x = 5
So, y = 2 - 3
Thus, lines (I) and (III) intersect at (1, -1). Since, all three lines intersect at one point. Therefore, the point of intersection of line (I) and (III) will satisfy the equation line (II). p(1) + 2(-1) - 3 = 0 p - 5 = 0
y = m y = m y = m Subtract (I) from (II) y - y = (m 0 = m 0 = x(m x(m
Substitute x = (c y = m y = (m y = (m
So, the point ((c Since, the three given lines are concurrent. Therefore, the point of intersection of line (I) and (II) will satisfy the equation line (III). (m (m m m
Hence, proved.
Let the slope of required line be m x - 2y = 3 2y = x - 3 y = x/2 - 3/2 The above equation is of the form y = mx + c, where m is the slope of the line. Therefore, slope of the given line m We know that if θ is the acute angle between lines l tan θ = |(m tan 45° = |(1/2 - m 1 = |{(1 - 2m 1 = |(1 - 2m 1 = ± (1 - 2m CASE I: 1 = (1 - 2m 2 + m 3m
The equation of the line passing through (3, 2) and having a slope -1/3 will be y - 2 = (-1/3) (x - 3) 3(y - 2) = 3 - x 3y - 6 = 3 - x
CASE II: 1 = -(1 - 2m 2 + m
The equation of the line passing through (3, 2) and having a slope 3 will be y - 2 = (3) (x - 3) y - 2 = 3x - 9
Hence, the equation of the lines are x + 3y = 9 and 3x - y = 7.
Equation of line having equal intercepts on the axes will be x/a + y/b = 1 a = b (x + y)/a = 1 x + y = a Given lines are 4x + 7y - 3 = 0 2x - 3y + 1 = 0 Point of intersection of given two lines: 4x + 7y - 3 = 0 4x = 3 - 7y x = (3 - 7y)/4 Substitute the value of x in 2x - 3y + 1 = 0 2(3 - 7y)/4 - 3y + 1 = 0 3/2 - 7y/2 - 3y + 1 = 0 -(7y + 6y)/2 = -(3 + 2)/2 13y/2 = 5/2 13y = 5
So, x = (3 - 7(5/13))/4 = (39 - 35)/13(4)
Therefore, the point of intersection of the given two lines is (1/13, 5/13). The equation x + y = a also passes through this point. Thus, 1/13 + 5/13 = a
Therefore, x + y = 6/13 13x + 13y = 6 Hence, the required equation of the line is 13x + 13y = 6.
Let y = m m It is given that the line makes an angle θ with the line y = mx + c. Therefore, tan θ = |(m tan θ = |(y/x - m)/(1 + ym/x)| tan θ = ± (y/x - m)/(1 + ym/x) CASE I: tan θ = (y/x - m)/(1 + ym/x) tan θ + (ym tan θ)/x = y/x - m m + tan θ = y/x × (1 - m tan θ)
CASE II: tan θ = -(y/x - m)/(1 + ym/x) tan θ + (ym tan θ)/x = m - y/x m - tan θ = y/x × (1 + m tan θ)
Hence, y/x = (m ± tan θ)/(1 ∓ m tan θ).
Equation of the line joining the points (-1, 1) and (5, 7) will be y - 1 = (7 - 1)/(5 + 1) × (x + 1) y - 1 = x + 1 x - y + 2 = 0 x = y - 2 Substitute x = y - 2 in x + y = 4 y - 2 + y = 4 2y = 6
So, Therefore, the line x - y + 2 = 0 intersects the line x + y = 4 at (1, 3). Let the ratio in which the point (1, 3) divides the line x - y + 2 = 0 be 1 : k. Using section formula, we get: 1 = (k(-1) + 1(5))/(1 + k) 1 + k = -k + 5 2k = 4
AND 3 = (k(1) + 1(7))/(1 + k) 3 + 3k = k + 7 2k = 4
Hence, the line joining the points (-1, 1) and (5, 7) is divided by the line x + y = 4 in the ratio 1: 2.
2x - y = 0 … (I) 4x + 7y + 5 = 0 … (II) From (I), we have 2x = y Substitute y = 2x in (II) 4x + 7(2x) + 5 = 0 4x + 14x + 5 = 0 18x + 5 = 0 18x = -5
So, y = 2(-5/18)
So, the given two lines intersect at the point (-5/18, -5/9). Distance between (1, 2) and (-5/18, -5/9) = √{(1 + 5/18) = √(23 = √(23/9) = 23/9 × √(5/4) = 23√5/18 units Hence, the required distance is 23√5/18 units.
Let the line passing through the point (-1, 2) be y = mx + c So, 2 = m(-1) + c 2 = -m + c c = m + 2 Substitute the value of c in the equation y = mx + m + 2 Given equation is x + y = 4 Substitute the value of y in the given equation. x + mx + m + 2 = 4 x(1 + m) = 2 - m
Thus, y = m(2 - m)/(m + 1) + m + 2 y = m × [(2 - m)/(m + 1) + 1] + 2 y = m[2 - m + m + 1]/(m + 1) + 2 y = (3m + 2m + 2)/(m + 1)
Therefore, ((2 - m)/(m + 1), (5m + 2)/(m + 1)) is the point of intersection of the two lines. This point is at a distance of 3 units from (-1, 2). Therefore, 3 = √(((2 - m)/(m + 1) + 1) Square both sides 9 = (2 - m + m + 1) 9 = 1/(m + 1) 9(m + 1) 9(m 9(m 18m = 0
Thus, the slope of the required line must be zero. Hence, the line must be parallel to the x-axis.
Let the right angled triangle be ABC, with ∠C = 90° m is the slope of the line AC then slope of BC = -1/m Equation of AC where A is (1, 3) will be
Equation of BC where B is (-4, 1) will be y - 1 = (-1/m)(x + 4)
If m = 0, Equations of legs will be y - 3 = 0 AND x + 4 = 0 If m = ∞, Equations of legs will be (y - 3)/m = x - 1 x - 1 = 0 AND 1 - y = (x + 4)/m 1 - y = 0
Let the given point be A (3, 8) and its image be B (a, b). x + 3y = 7 is the perpendicular bisector of the line AB. Slope of AB = m = (b - 8)/(a - 3) Slope of the given line = -1/3 m × (-1/3) = -1 (b - 8)/(a - 3) × (-1/3) = -1 (b - 8) = 3(a - 3) b - 8 = 3a - 9
b = 3a - 1 Mid-point of the line AB will be ((a + 3)/2, (b + 8)/2) This point will satisfy the perpendicular bisector x + 3y = 7. So, (a + 3)/2 + 3(b + 8)/2 = 7 (3b + 24 + a + 3) = 14 a + 3b = -13 Substitute b = 3a - 1 a + 3(3a - 1) = -13 a + 9a - 3 = -13 10a = -10
Thus, b = -3 - 1
Hence, the image of the given point with respect to the line x + 3y = 7 is (-1, -4).
y = 3x + 1 Slope of the line = m 2y = x + 3 y = (1/2)x + 3/2 Slope of the line = m y = mx + 4 Slope of the line = m It is given that y = 3x +1 and 2y = x + 3 are equally inclined to the line y = mx + 4. Therefore, |(m |(3 - m)/(1 + 3m)| = |(1/2 - m)/(1 + m/2)| |(3 - m)/(1 + 3m)| = |{(1 - 2m)/2}/{(2 + m)/2}| |(3 - m)/(1 + 3m)| = |(1 - 2m)/(2 + m)| (3 - m)/(1 + 3m) = ± (1 - 2m)/(2 + m) CASE I: (3 - m)/(1 + 3m) = (1 - 2m)/(2 + m) (3 - m)(2 + m) = (1 - 2m)(1 + 3m) 6 - 2m + 3m - m 6 + m - m 5m 5(m m m = √-1 This value of m is not real. Thus, (3 - m)/(1 + 3m) = (1 - 2m)/(2 + m) is rejected. CASE II: (3 - m)/(1 + 3m) = -(1 - 2m)/(2 + m) (3 - m)(2 + m) = -(1 - 2m)(1 + 3m) 6 - 2m + 3m - m 6 + m - m 7m Using the quadratic formula, m = (2 ± √(4 - 4(-49)))/2(7) m = (2 ± √4(1 + 49))/14 m = 2[1 ± 5√2]/14
Perpendicular distance of P (x, y) from the line x + y - 5 = 0 will be d d Perpendicular distance of P (x, y) from the line 3x - 2y + 7 = 0 will be d d It is given that 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. Therefore, d |x + y - 5|/√2 + |3x - 2y + 7|/√13 = 10 {√13|x + y - 5| + √2|3x - 2y + 7|}/√26 = 10 √13|x + y - 5| + √2|3x - 2y + 7| = 10√26 Taking x + y - 5 and 3x - 2y + 7 as positive, √13(x + y - 5) + √2(3x - 2y + 7) = 10√26 x√13 + y√13 - 5√5 + 3x√2 - 2y√2 + 7√2 = 10√26
This is equation of a line. For any signs of (x + y - 5) and (3x - 2y + 7), an equation can be obtained. Hence, the point P must move on a line.
Let the point on the line which is equidistant from the given lines 9x + 6y - 7 = 0 and 3x + 2y + 6 = 0 be P (a, b). Perpendicular distance of P (a, b) from the line 9x + 6y - 7 = 0 will be d d d Perpendicular distance of P (a, b) from the line 3x + 2y + 6 = 0 will be d d Since, P (a, b) is equidistant from the given two lines. Therefore, d |9a + 6b - 7|/3√13 = |3a + 2b + 6|/√13 |9a + 6b - 7| = 3|3a + 2b + 6| 9a + 6b - 7 = ± 3(3a + 2b + 6) CASE I: 9a + 6b - 7 = 3(3a + 2b + 6) 9a + 6b - 7 = 9a + 6b + 18 -7 = 18 which is not possible. So, 9a + 6b - 7 = 3(3a + 2b + 6) is rejected. CASE II: 9a + 6b - 7 = -3(3a + 2b + 6) 9a + 6b - 7 = -9a - 6b - 18 18a + 12b + 11 = 0 Hence, the required equation of the line equidistant from the given two lines is 18a + 12b + 11 = 0.
Let the coordinates of point A be (a, 0). Draw the figure. Construct a line AL perpendicular to the x-axis. We know that the angle of incidence = angle of reflection ∠BAL = ∠CAL = ϕ ∠CAX = ϕ Now, ∠OAB = 180° - (θ + 2ϕ) = 180° - [θ + 2(90° - θ)] ∠OAB = 180° - θ - 180° + 2θ ∠OAB = θ So, ∠BAX = 180° - θ Slope of the line AC = (3 - 0)/(5 - a)
Slope of the line AB = (2 - 0)/(1 - a) tan (180° - θ) = 2/(1 - a) -tan θ = 2/(1 - a)
Therefore, 3/(5 - a) = 2/(a - 1) 3a - 3 = 10 - 2a 5a = 13
Hence, coordinates of the point A are (13/5, 0).
Given line is (x/a) cos θ + (y/b) sin θ = 1 (bx cos θ + ay sin θ)/ab = 1 bx cos θ + ay sin θ = ab bx cos θ + ay sin θ - ab = 0 Length of the perpendicular drawn from the point (√(a d d Length of the perpendicular drawn from the point (√-(a d d d Now, product of the perpendicular distances is d = |{b cos θ √(a = |(b cos θ √(a = |b = |a = b = b = b = b = b = b = b = b Hence, proved.
2x - 3y + 4 = 0 … (I) 3x + 4y - 5 = 0 … (II) 6x - 7y + 8 = 0 … (III) It is given that the person is standing at the point of intersection of line (I) and (II). Therefore, 2x - 3y + 4 = 0 2x = 3y - 4 x = (3y - 4)/2 Substitute the value of x in equation (II) 3(3y - 4)/2 + 4y - 5 = 0 (9y - 12)/2 + 4y = 5 (9y - 12 + 8y)/2 = 5 (17y - 12)/2 = 5 17y - 12 = 10 17y = 22
So, x = (3(22/17) - 4)/2 = (66 - 68)/17/2 = -2/17(2)
Thus, the person is standing at (-1/17, 22/17). In order to reach the path of equation (III), the person must follow a perpendicular path from his current position. 6x - 7y + 8 = 0 7y = 6x + 8 y = (6/7)x + 8/7 Slope of the line = m = 6/7 Slope of the line perpendicular to this line = -1/m = -7/6 Equation of the line passing through (-1/17, 22/17) and having slope -7/6 will be y - 22/17 = (-7/6)(x + 1/17) (17y - 22)/17 = (-7/6) (17x + 1)/17 6(17y - 22) = -7(17x + 1) 102y - 132 = -119x - 7 119x + 102y - 125 = 0 Hence, the person should follow the path 119x + 102y - 125 = 0. |