Chapter 9 - Straight Lines

Let the equation of the line be x/a + y/b = 1. Given that intercepts are equal, so a = b. The equation becomes x/a + y/a = 1 or x + y = a. The line passes through (1, -2), so 1 + (-2) = a, which gives a = -1. Therefore, the equation of the line is x + y = -1, or x + y + 1 = 0.

The slope of the line joining (2, 3) and (3, -1) is m1 = (-1 - 3) / (3 - 2) = -4. The slope of the required line, which is perpendicular to it, is m2 = -1/m1 = 1/4. The equation of the line passing through (5, 2) with slope 1/4 is y - 2 = (1/4)(x - 5). This simplifies to 4y - 8 = x - 5, or x - 4y + 3 = 0.

The slope of the first line is m1 = 2 - √3. The slope of the second line is m2 = 2 + √3. The angle θ between them is given by tan(θ) = |(m2 - m1) / (1 + m1*m2)|. m2 - m1 = (2 + √3) - (2 - √3) = 2√3. 1 + m1*m2 = 1 + (2 - √3)(2 + √3) = 1 + (4 - 3) = 1 + 1 = 2. So, tan(θ) = |2√3 / 2| = √3. Therefore, θ = 60° or π/3.

Let the intercepts be a and b. So, a + b = 14, or b = 14 - a. The equation is x/a + y/(14-a) = 1. It passes through (3, 4), so 3/a + 4/(14-a) = 1. This gives 3(14-a) + 4a = a(14-a) => 42 - 3a + 4a = 14a - a² => 42 + a = 14a - a² => a² - 13a + 42 = 0. Factoring gives (a-6)(a-7) = 0. So a = 6 or a = 7. If a = 6, b = 8. Eq: x/6 + y/8 = 1 => 4x + 3y = 24. If a = 7, b = 7. Eq: x/7 + y/7 = 1 => x + y = 7. The point (3,4) satisfies both equations.

Let the point be (h, k). It lies on x + y = 4, so h + k = 4, or k = 4 - h. The distance from (h, k) to 4x + 3y - 10 = 0 is 1. So, |4h + 3k - 10| / √(4² + 3²) = 1. |4h + 3(4-h) - 10| / 5 = 1. |4h + 12 - 3h - 10| = 5. |h + 2| = 5. So, h + 2 = 5 or h + 2 = -5. If h + 2 = 5, h = 3. Then k = 4 - 3 = 1. Point is (3, 1). If h + 2 = -5, h = -7. Then k = 4 - (-7) = 11. Point is (-7, 11).

Using the distance formula d = |Ax1 + By1 + C| / √(A² + B²). Here (x1, y1) = (3, -5) and the line is 3x - 4y - 26 = 0. d = |3(3) - 4(-5) - 26| / √(3² + (-4)²) = |9 + 20 - 26| / √(9 + 16) = |3| / √25 = 3/5.

First, find the intersection point. From x - y + 2 = 0, we get x = y - 2. Substitute into the first equation: 3(y - 2) + 4y = 7 => 3y - 6 + 4y = 7 => 7y = 13 => y = 13/7. Then x = 13/7 - 2 = (13 - 14)/7 = -1/7. The point is (-1/7, 13/7). The slope is 5. The equation is y - 13/7 = 5(x + 1/7). 7y - 13 = 35(x + 1/7) = 35x + 5. So, 35x - 7y + 18 = 0.

The family of lines passing through the intersection is (2x+y-5) + k(x+3y+8) = 0. This is (2+k)x + (1+3k)y + (-5+8k) = 0. The slope is -(2+k)/(1+3k). The slope of the parallel line 3x+4y=7 is -3/4. So, -(2+k)/(1+3k) = -3/4. 4(2+k) = 3(1+3k) => 8+4k = 3+9k => 5k=5 => k=1. The equation is (2x+y-5) + 1(x+3y+8) = 0, which simplifies to 3x+4y+3=0.

Let intercepts be a and b. a + b = 9, so b = 9 - a. Equation: x/a + y/(9-a) = 1. It passes through (2, 2). 2/a + 2/(9-a) = 1. 2(9-a) + 2a = a(9-a) => 18 - 2a + 2a = 9a - a² => a² - 9a + 18 = 0. (a-3)(a-6) = 0. So a = 3 or a = 6. If a = 3, b = 6. Eq: x/3 + y/6 = 1 => 2x + y = 6. If a = 6, b = 3. Eq: x/6 + y/3 = 1 => x + 2y = 6.

The slope of the line joining (5, -1) and (2, 3) is m1 = (3 - (-1)) / (2 - 5) = 4 / -3 = -4/3. The slope of the required perpendicular line is m2 = -1/m1 = 3/4. The equation of the line passing through (6, -2) with slope 3/4 is y - (-2) = (3/4)(x - 6). 4(y + 2) = 3(x - 6) => 4y + 8 = 3x - 18 => 3x - 4y - 26 = 0.

The slope of the line is m = tan(120°) = -√3. The equation is y = -√3x + c, or √3x + y - c = 0. The distance from the origin (0, 0) is 4. So, |√3(0) + 0 - c| / √( (√3)² + 1²) = 4. |-c| / √4 = 4. |c| / 2 = 4 => |c| = 8. So c = ±8. The equations are √3x + y - 8 = 0 and √3x + y + 8 = 0.

Let the sides through (2, 2) be AB and AC. The slope of the hypotenuse is -3/4. The sides of an isosceles right-angled triangle make an angle of 45° with the hypotenuse. Let the slope of a side be m. tan(45°) = |(m - (-3/4)) / (1 + m(-3/4))|. 1 = |(4m + 3) / (4 - 3m)|. So, 4 - 3m = 4m + 3 or 4 - 3m = -(4m + 3). Case 1: 7m = 1 => m = 1/7. Eq: y - 2 = 1/7(x - 2) => 7y - 14 = x - 2 => x - 7y + 12 = 0. Case 2: 4 - 3m = -4m - 3 => m = -7. Eq: y - 2 = -7(x - 2) => y - 2 = -7x + 14 => 7x + y - 16 = 0.

(a) Let point be (h,k). h+5k=13. Distance from 12x-5y+26=0 is 2. |12h-5k+26|/13=2. |12(13-5k)-5k+26|=26. |156-60k-5k+26|=26. |182-65k|=26. 182-65k=26 or 182-65k=-26. 65k=156 => k=12/5, h=1. Or 65k=208 => k=16/5, h=-3. Matches (iii). (b) Solved in Q5. Matches (i). (c) P divides AB in ratio 1:2, Q divides in 2:1. P = ((1*3+2*(-2))/(1+2), (1*1+2*5)/(1+2)) = (-1/3, 11/3). Q = ((2*3+1*(-2))/(2+1), (2*1+1*5)/(2+1)) = (4/3, 7/3). Matches (ii).

Line: (2+6λ)x + (3-λ)y + (4+12λ) = 0. (a) Parallel to x-axis means coeff of x is 0. 2+6λ=0 => λ=-1/3. This matches (a)-(ii). (d) Parallel to y-axis means coeff of y is 0. 3-λ=0 => λ=3. This matches (d)-(iii). (c) Passes through (1,2). (2+6λ)(1) + (3-λ)(2) + (4+12λ) = 0. 2+6λ+6-2λ+4+12λ=0. 16λ+12=0 => λ=-12/16=-3/4. This matches (c)-(i). (b) Perpendicular to 7x+y-4=0. Slope of our line is -(2+6λ)/(3-λ). Slope of other line is -7. Perpendicular slope is 1/7. So -(2+6λ)/(3-λ) = 1/7. -7(2+6λ) = 3-λ. -14-42λ=3-λ. -17=41λ. λ=-17/41. This matches (b)-(iv). So: (a)-(ii), (b)-(iv), (c)-(i), (d)-(iii).

Intersection: 4x-6y=0 and 4x-5y=2. Subtracting: -y=-2 => y=2. Then 2x-3(2)=0 => 2x=6 => x=3. Point is (3,2). (a) Line through (3,2) and (2,1). Slope = (2-1)/(3-2)=1. Eq: y-1=1(x-2) => x-y-1=0. Matches (a)-(iii). (b) Perpendicular to x+2y+1=0 (slope -1/2). Required slope is 2. Eq: y-2=2(x-3) => y-2=2x-6 => 2x-y-4=0. Matches (b)-(i). (c) Parallel to 3x-4y+5=0 (slope 3/4). Required slope is 3/4. Eq: y-2=3/4(x-3) => 4y-8=3x-9 => 3x-4y-1=0. Matches (c)-(ii).