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Straightlines and Family of Straightlines



Question (1): If p is the length of perpendicular drawn from the origin on the line AB whose intercepts on the axes are a and b, then show
Answer: Aliter: Equation of the line AB in the intercept form is ...(i) Since (i) and (ii) represent the same line, we have Squaring and adding, we get

Question (2): A straight line passes through the point (a, b) and this bisects the part of the line intercepted between the axes. Show that
Answer: Let the straight line passing through p(a, b) cut the x-axis at A(a,0) and B(0,b) on the x-axis and the y-axis respectively.

Question (3): Find the equation of the straight line which passes through the point (2,3) and whose intercepts on the y-axis is thrice that on the x-axis.
Answer: Let the equation of the line in the intercept form be From the problem, b = 3a. (ii) passes through (2,3). b = 3a = 3 x 3 = 9

Question (4): Find the equation of the straight line which passes through the point (3,2) and cuts off intercepts a and b respectively on the x and y axis such that a - b = 2.
Answer: Let the equation of the line be ...(i) From the problem, a - b = 2 a = b+2 This line passes through (3,2). The two lines are

Question (5): Find the equation of the line which makes thrice the intercepts made by the line 3x + 4y = 12 on the coordinate axes.
Answer: The given line is 3x+4y = 12. ......(i) The intercept made by the line on the x-axis is obtained by substituting y = 0 in (i). The intercept made by the line on the y-axis is obtained by substituting x = 0 in (i). The intercepts made by the required line are 3a and 3b. i.e. 12 and 9 respectively. Equation of the required line is Note:

Question (6): Find the equation of the straight line, which passes through p(1,-7) and meets the axes at A and B respectively, so that 4AP - 3BP = 0, where O is the origin.
Answer: Now, the point p(1,-7) divides AB in the ratio AP:PB::3:4. Let the coordinates of A be (a,0) and that of B be (0,b). The equation of the line is

Question (7):
Answer: Let C divide AB in the ratio m:n. The coordinates of C are Since lies on ax + by + c = 0, we have which is the required ratio.

Question (8): Find the equations of the diagonals of the quadrilateral whose ratio in which each diagonal divides the other.
Answer: Equation of diagonal AC is or Let this line divide DB in the ratio m:n. Equation of the diagonal BD is Let this line divide the diagonal AC in the ratio m:n.

Question (9): Find the equation of the line passing through (2,-1) and making an angle of 45^o with the positive direction of the x-aixs. Also find the points on the line at a distance from the point (2,-1).
Answer: Here, Equation of the line is y + 1 = 1(x - 2) x - y = 3

Question (10): Find the equation of the line which passes through the point points on the line that is 5 units away from the point (3,2).
Answer: The equation of the line in the point-slope form is Here, The points on the line at a distance 5 units from (3,2). The points are (7,5) and (-1,-1).

Question (11): Find the equation of the straight line which passes through the origin and trisect the intercepts of the line 2x + 3y = 18 between the axes.
Answer: The given line can be written as Intercepts on y-axis = b = 6 = OB The points C and D are the points of trisection. C = (6, 2) D = (3, 4)

Question (12): route from the place to canal is exactly northeast. A village is 5 miles north and 12 miles east from the place. Does the village lie by the nearer edge of the canal.
Answer: x + y = 17 The position of the village is (12, 5). The position of the village on the edge of the canal is 12 + 5 = 17, which is true.

Question (13): i) Slope-intercept form ii) Intercept form iii) Normal form
Answer: Given the line Ax + By + C = 0 i) Slope intercept form (y = mx + c) Comparing this equation with y = mx+c, we have ...(i) Multiplying by k throughout, we have kAx + kBy = -kC Substituting the value of k in (ii), we obtain ...(iii) Note: p is always positive. When C > 0, equation (iii) will take the form Again, when C < 0 Procedure to reduce the general equation to the normal form: i) Put the constant term to the RHS and make it positive if not so, by changing the sign of every term.

Question (14): i) Slope intercept form ii) Intercept form iii) Normal form
Answer: (iii) Normal form Comparing with xcosa + ysina = p, we have The equation in the normal form is

Question (15): Find the length of perpendicular drawn from (2,-1) on the line 3x + 4y = 12.
Answer: The length of the perpendicular drawn from (2,-1) on the line 3x + 4y = 12 is

Question (16): If P and P' are the length of the perpendiculars from the origin respectively, prove that 4p² + p² = a².
Answer: The two lines are 4p²= a²sin²2q ....(iv) p'² = a²cos²2q ....(v) Adding (iv) and (v), we get 4p²+ p'²= a²

Question (17): Find the distance between the pair of parallel lines 3x + 4y - 15 = 0 and 6x + 8y = 25.
Answer: The given lines are 3x + 4y - 15 = 0 ... (i) 6x + 8y - 25 = 0 ... (ii) The perpendicular distance of 3x + 4y - 15 = 0 from the origin is The perpendicular distance of 6x + 8y - 25 = 0 from the origin is

Question (18): Find the point of intersection of the lines.
Answer: a) b)

Question (19): Find the coordinates of the foot of the perpendicular from (4, - 4) on the line 5x + 7y - 12 = 0.
Answer: The equation of the line perpendicular to 5x + 7y - 12 = 0 is 7x - 5y + k = 0. The line passes through (4,-4). The line is 7x - 5y - 48 = 0. To find the foot of the perpendicular, solve the equations

Question (20):
Answer: The given lines are The coordinates of the point of intersection of the lines (i) and (ii) are The coordinates of the point of intersection of (i) and (ii) are (1,1). Hence the three lines are concurrent. Aliter: The determinant \The three lines are concurrent.

Question (21): and its slope is -1. Find the equation of the line.
Answer: Let the equation of the line be y = mx + c. mx - y + c =0 ...(i)

Question (22): Find the length of perpendicular drawn from (-3,-4) on the line 3x + 4y = 12.
Answer: The length of the perpendicular from (-3,-4) on the line 3x+4y=12 is

Question (23): Find the distance of the line y = x from the point (3,-4).
Answer: The given line is y = x. The given point is (3,-4). The distance of the line y = x from the point (3,-4) is

Question (24): Determine the value of 'a' so that the distance 'd' from the line 12x + ay + 10 = 0 to the point (4,-4) is numerically equal to 6 units.
Answer: Squaring both sides, we get

Question (25):
Answer: Let the sides AB, BC CA of triangle ABC be represented by Solving (i) and (ii), we get the coordinates of B. Solving (ii) and (iii), we get the coordinates of C. Solving (i) and (iii), we get