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Straightlines and Family of Straightlines



Question (1): Find the equation of the line which makes intercepts of 3 and 5 with x-axis and y-axis respectively.
Answer: Here, a = 3, b = 5.

Question (2): Find the equation of the line through (4,6) and the sum of the intercepts on the coordinate axis is 20.
Answer: Let a be the intercept on the x-axis, then 20 - a is the intercept on the y-axis. The equation of the line is This line passes through (4,6). When a = 8, b = 12 and when a = 10, b = 10. The equation of the line is

Question (3): A straight line moves so that the sum of reciprocals of intercepts on the axes of coordinates is constant. Show that it will pass through a fixed point.
Answer: Let the equation of the line in the intercept form be From the problem, we have

Question (4): The part of line intercepted between the axes is divided by the point (-5,2) in the ratio 2:3. Find the equation of the line.
Answer: Then, the intercepts made by the line on the x-axis is a and on y-axis is b. P divides the line joining the point A(a,0) and B(0,b) in the ratio AP : PB::2 : 3.

Question (5): Find the equation of the straight line passing through the point (3,-4) and cutting off equal intercepts but of opposite signs from the axis.
Answer: Let the equation of the line in the intercept form be From the problem, (a,-a) are the intercepts. This line passes through (3,-4).

Question (6): Find the equation of the line passing through (-2,4) and whose intercepts on x-axis is thrice as that on the y-axis.
Answer: Let the equation of the line in the intercept form be The intercepts on the axes are a and b. But a = 3b. This line passes through (-2,4).

Question (7): Find the equation of the straight line which passes through (-3,5) and cuts the axes at A and B, so that OA + OB = 4, where O is the origin.
Answer: Let the equation of the line in the intercept form be From the problem a + b = 4, b = 4 - a This line passes through (-3,5). \ a - 6 or a 2 When a = - 6, b = 4 - (-6) = 10 When a = 2, b = 4 - 2 = 2 Line is x + y = 2.

Question (8): Find the equation of the straight line whose intercepts on x and y axes are respectively twice and thrice of those made by the line 3x - 4y = 12.
Answer: Given line is 3x - 4y = 12. Now put y = 0 to get x-intercept. 3x = 12y 3x = 12(0) x = 4 \ a = 4 Next, put x = 0 to obtain y-intercept. The intercepts made by the new required line are a = 2(4), b = 3(-3) = -9.

Question (9): Find the ratio in which the straight line 2x + 3y - 5 = 0 divides the line joining two points (-4,1) and (2,7).
Answer: Let m:n be the ratio in which the line divides the line joining the points (-4,1) and (2,7).

Question (10): Find the equation of the straight line passing through the origin and the middle point of the intercept of the line ax + by + c = 0 between the axes.
Answer: Given line is ax + by + c = 0. The equation of line is

Question (11): distance from the place to the canal is exactly north-east. A village is 3 miles north and 4 miles east from the place. Does it lie by the nearer edge of the canal?
Answer: Let O be the position of the given place. Choose O as the origin and the east and the north lines as x and y axes respectively, let AB represent the nearest edge of the canal. From the problem, Now, the equation of the canal line is The position of the village is (4,3). The position of the village will be on Hence, the village does not lie by the nearer edge of the canal.

Question (12): Find the equations of straight lines which pass through the origin and trisect the portion of the line 3x + 4y = 36 between the axes.
Answer: Let the line AB represent 3x + 4y = 36. This line makes intercepts OA and OB on the axes respectively. Let C and D be the points of trisection on AB. The intercept made by the line on the x-axis is obtained by putting y = 0 in 3x + 4y = 36. 3x + 4(0) = 36 x = 12 \ a = 12 The intercept made by the line on the y-axis is obtained by putting x = 0 in the equation 3x + 4y = 36. 3(0) + 4y = 36 y = 9 \ b = 12 The coordinates of A are (12,0) and the coordinates of B are (0,9). The point C divides AB in the ratio AC:CB::1:2. The coordinates of C are The point D divides AB in the ratio AD:DB::2:1. The coordinates of D are Equation of the line OC is Equation of the line OD is

Question (13): Prove that every equation of the first degree in x and y represents a straight line.
Answer: The general equation of the first degree in x and y is Ax + By + C = 0 ...(i) Let (x₁,y₁) and (x₂,y₂) be any two points on the line (locus) represented by (i). Every point on the locus satisfies the equation of the locus. (x₁, y₁) lies on (i), (x₂, y₂) lies on (i), Divide by (1 + k), straight line joining the points (x₁, y₁) and (x₂, y₂) in the ratio k:1 lies on the locus represented by (i). Now k is any arbitrary constant. Every point on the straight line joining (x₁, y₁) and (x₂, y₂) lies on the locus represented by (i). Hence, Ax + By + C = 0 represents a straight line.

Question (14): i) The slope-intercept form ii) The intercept form iii) The normal form
Answer: Comparing this with y = mx + c, we get Dividing by -8 throughout, we get Changing the sign throughout to make RHS positive, we get Comparing this with x cos a + y sin a = P. Since both cosa and sina are negative, a is in the third quadrant. \ Equation of the line in the normal form is

Question (15): Reduce the equation 5x + 12y + 39 = 0 to Normal form.
Answer: Given equation is 5x + 12y + 39 = 0 5x + 12y = -39 Changing the sign throughout, we have -5x - 12y = 39.

Question (16): Find the distance of the line 15x + 8y + 14 = 0 from the point (2,3).
Answer: Line is 15x + 8y + 14 = 0. Point is (2, 3). Distance of a point (x₁, y₁) from a line ax + by + c = 0 is given by 4

Question (17): Find the distance between the lines ax + by + c = 0 and ax + by + d = 0.
Answer: The two given equations are ax + by + c = 0 ...(i) ax + by + d = 0 ...(ii) Take any point (h,k) on the straight line (i). ha + kb + c = 0 ha + kb = -c ...(iii) The distance between two parallel lines = the length of perpendicular drawn from a point on one line to the other line Hence, the distance between (i) and (ii) is = length of perpendicular drawn from (h,k) upon the line (ii) Aliter: The length of the perpendicular drawn from the origin upon the line ax + by + c = 0 is Again, the length of the perpendicular p₂ drawn from the origin upon the line ax + by + d = 0 is Now, the distance between the parallel lines = difference of length of perpendicular drawn from any point upon the given lines. = the difference between the length of perpendicular drawn from the origin upon the given lines. Remark: In this method, first make the sign of the constant terms of both the equations of the given lines same and the signs of the lengths of the perpendiculars should be maintained.

Question (18): Find the equation of the lines parallel to 3x + 4y + 10 = 0 and at a distance equal to 2 units.
Answer: Let P(x', y') be any point on the required line. Then, Simplifying and replacing x' and y' by x and y, we get 3x + 4y = 0 or 3x + 4y + 20 = 0.

Question (19): Find the coordinates of the foot of the perpendicular from (3,-3) on the line 4x - 7y + 32 = 0.
Answer: Given line is 4x - 7y + 32 = 0 ... (i) Equation of the line perpendicular to 4x-7y+32 = 0 is 7x + 4y + k = 0. This line passes through (3,-3). \ 7(3) + 4(-3) + k = 0 k = -9 The line is 7x + 4y - 9 = 0 ... (ii) The point of intersection of the lines (i) and (ii) gives the coordinates of the foot of the perpendicular. Solving, \ The point of intersection is (-1,4) which is the foot of the perpendicular.

Question (20): Find the equation to the straight line through the point of intersection of 2y + x + 3 = 0 and 4y + 3x + 7 = 0 and perpendicular to y - x = 8.
Answer: The given lines are 2y + x + 3 = 0 and 4y + 3x + 7 = 0. \ The line through the intersection is x + 2y + 3 + l (3x + 4y + 7) = 0. (A) is perpendicular to y - x = 8. Slope of y - x = 8 is 1. Slope of (A) is -1. The slope of the line ax + by + c = 0 is given by -a/b \ The line is x + 2y + 3 - 1 (3x + 4y + 7) = 0.

Question (21): Find the equation of the line through the intersection of the lines 6x+5y=11, 8x-5y=3 and parallel to the line 3x - 2y + 12 = 0.
Answer: The given lines are 6x + 5y = 11 ...(i) 8x - 5y = 3 ...(ii) Solving for x and y, the coordinates of the point of intersection of the two lines are (1,1). The slope of the line through (1,1) = Slope of line 3x - 2y + 12 = 0 \ The equation of the line which passes through (1, 1) and whose slope is 3/2 is

Question (22): Show that the points (1,4) and (-1,-2) lie on either side of line 4x + 5y + 10 = 0.
Answer: Let d = distance between the line 4x + 5y + 10 = 0 and the point (1,4). Let d' be the distance between the line 4x + 5y + 10 = 0 and the point (-1,-2). Since d and d' are of the opposite signs, the points lie on the either side of the line.

Question (23): Find the distance of the line 24x + 7y + 12= 0 from the point (4,6).
Answer: Line 24x + 7y + 12 = 0, point (4,6). 6 units (since distance cannot be negative)

Question (24): Find the distance between the following pairs of parallel lines. i) 3x + 4y + 20 = 0, 6x + 8y + 45 =0 ii) 12x + 15y + 41 = 0, 4x + 5y + 41 = 0
Answer: i) 3x + 4y + 20 = 0, 6x + 8y + 45 = 0 Two lines are 3x + 4y + 20 = 0 ...(i) 6x + 8y + 45 = 0 ...(ii) Distance of (i) from the origin (0,0) is Distance of (ii) from the origin (0,0) is Distance between the lines is ii) 12x + 15y + 41 = 0, 4x + 5y + 41 = 0 The lines are 12x + 15y + 41 = 0 ...(i) 4x + 5y + 41 = 0 ...(ii) The distance of line (i) from (0,0) is The distance of line (ii) from (0,0) is Distance between the lines is

Question (25): Find the point of intersection of the lines
Answer: 2x + 3y - 7 = 0 3x - 5y - 1 = 0 The point of intersection is (2,1). ii) mx - y - c = 0 x - my + c = 0 Adding (i) and (ii), we get Subtracting (ii) from (i), we get 2mx = 0 x = 0

Question (26): Find the centroid of the triangle whose sides are x + y - 1 = 0, x - 3y + 3 = 0 and x - y - 1 = 0.
Answer: Let AB represent the side x + y - 1 = 0 ...(i) Let BC represent the side x - 3y + 3 = 0 ... (ii) Let CA represent the side x - y - 1 = 0 ... (iii) Solving (i) and (ii) for x and y, we get the coordinates of B, i.e., B(0,1). Solving (ii) and (iii) , we get the coordinates of C, i.e., C(3,2). Solving (iii) and (i), we get the coordinates of A, i.e., A(1,0). The coordinates of the centroid of the triangle are