Chapter 9 - Straight Lines

Given, intercept c = – 3 and slope m = tan(θ) = 3/5. The equation of the line is y = mx + c, which is y = (3/5)x - 3. This simplifies to 5y = 3x - 15, or 5y - 3x + 15 = 0.

The intercept form of a line is x/a + y/b = 1. If intercepts are of equal length, a = b (or a = -b). If a = b, x + y = a, slope is -1. If a = -b, x - y = a, slope is 1. Usually, 'equal lengths' implies the numerical value, but the most common interpretation leading to one of the options is that the intercepts themselves are equal or opposite. If the intercepts are non-zero and equal, say 'k', the line passes through (k, 0) and (0, k). The slope is (k-0)/(0-k) = -1.

The slope of the line y = x is 1. The slope of a line perpendicular to it is -1/1 = -1. The equation of the required line passing through (3, 2) is y - 2 = -1(x - 3), which simplifies to y - 2 = -x + 3, or x + y = 5.

The slope of the line x + y + 7 = 0 is -1. The slope of the perpendicular line is -1/(-1) = 1. The equation of the line passing through (1, 2) with slope 1 is y - 2 = 1(x - 1), which simplifies to y - 2 = x - 1, or y - x - 1 = 0.

The line is 3x - 2y + 4 = 0. The distance from (1, -3) is |3(1) - 2(-3) + 4| / √(3² + (-2)²) = |3 + 6 + 4| / √(9 + 4) = 13 / √13 = √13. None of the options are correct. The correct distance is √13.

Let the foot of the perpendicular be (h, k). It lies on the line, so h + k - 11 = 0. The slope of the given line is -1. The slope of the perpendicular line joining (2, 3) and (h, k) is (k-3)/(h-2). So, (k-3)/(h-2) = 1, which means k - 3 = h - 2, or h - k = -1. Solving h + k = 11 and h - k = -1, we get 2h = 10, so h = 5. Then k = 6. The coordinates are (5, 6).

Let the x-intercept be 'a'. Then the y-intercept is '2a'. The equation is x/a + y/(2a) = 1. It passes through (1, 2), so 1/a + 2/(2a) = 1, which gives 1/a + 1/a = 1, so 2/a = 1, and a = 2. The equation is x/2 + y/4 = 1. Multiplying by 4 gives 2x + y = 4.

Let the intercepts be (a, 0) and (0, b). P(1, 2) is the midpoint. So, (a+0)/2 = 1 and (0+b)/2 = 2. This gives a = 2 and b = 4. The equation of the line in intercept form is x/a + y/b = 1, so x/2 + y/4 = 1. This simplifies to 2x + y = 4.

Using the point-slope form y - y1 = m(x - x1), we have y - 3 = 2(x - 2). This simplifies to y - 3 = 2x - 4, or 2x - y - 1 = 0.

The lines are mx - y + c1 = 0 and mx - y + c2 = 0. The distance between two parallel lines Ax + By + C1 = 0 and Ax + By + C2 = 0 is |C1 - C2| / √(A² + B²). Here A = m, B = -1. So the distance is |c1 - c2| / √(m² + (-1)²) = |c1 - c2| / √(1 + m²).

The line is 3x - y + 4 = 0. Let the foot be (h, k). The slope of the given line is 3. The slope of the perpendicular is -1/3. So, (k-3)/(h-2) = -1/3 => 3k - 9 = -h + 2 => h + 3k = 11. Also, (h, k) lies on the line, so 3h - k + 4 = 0 => k = 3h + 4. Substituting this into the first equation: h + 3(3h + 4) = 11 => h + 9h + 12 = 11 => 10h = -1 => h = -1/10. Then k = 3(-1/10) + 4 = -3/10 + 40/10 = 37/10. The point is (-1/10, 37/10).

Let the intercepts be (a, 0) and (0, b). The midpoint is (a/2, b/2) = (3, 2). So a = 6 and b = 4. The equation is x/6 + y/4 = 1. Multiplying by 12 (LCM of 6 and 4) gives 2x + 3y = 12.

The slope of the given line is 3. A parallel line will have the same slope. Using the point-slope form with point (1, 2) and slope 3, we get y - 2 = 3(x - 1).

The vertices of the square are (0, 0), (1, 0), (0, 1), and (1, 1). The first diagonal connects (0, 0) and (1, 1), its equation is y = x. The second diagonal connects (1, 0) and (0, 1), its equation is (y-0) = (1-0)/(0-1) * (x-1), which is y = -1(x-1), or x + y = 1.

To uniquely define a straight line, two independent parameters are required. For example: two points, or a point and a slope, or a slope and an intercept.

(i) Reflection of (4, 1) about y = x gives (1, 4). (ii) Translation of (1, 4) through 2 units along the positive x-axis means adding 2 to the x-coordinate. The new point is (1+2, 4) = (3, 4).

The locus of points equidistant from two lines is the angle bisector. We need to find a point that lies on the bisectors of all three pairs of lines. A simpler way is to test the given points. For (0, 0): d1 = |10|/√(16+9) = 10/5 = 2. d2 = |26|/√(25+144) = 26/13 = 2. d3 = |-50|/√(49+576) = 50/25 = 2. Since the distances are equal, (0, 0) is the point.

The slope of 3x + y = 3 is -3. The slope of the perpendicular line is 1/3. The equation of the line is y - 2 = (1/3)(x - 2). To find the y-intercept, set x = 0: y - 2 = (1/3)(-2) => y = 2 - 2/3 = 4/3.

Let the lines be L1: 3x+4y+5=0, L2: 3x+4y-5=0, and L: 3x+4y+2=0. The distance between L1 and L is |5-2|/√(9+16) = 3/5. The distance between L and L2 is |2-(-5)|/√(9+16) = 7/5. The ratio of the distances is (3/5) : (7/5), which is 3:7.

Let the vertex be (h, k). The centroid is the origin (0, 0). The centroid lies between the vertex and the opposite side. Thus, the vertex and the centroid must be on the same side of the line x + y - 2 = 0. For the centroid (0,0), the value of the expression is 0 + 0 - 2 = -2 (negative). For the vertex (-2, -2), the value is -2 - 2 - 2 = -6 (negative). Since both are negative, they are on the same side. The other possible vertex (2,2) gives 2+2-2=2 (positive), which is on the opposite side. Therefore, (-2, -2) is the correct vertex.