Subject > Math > Geometry > Straightlines > Question and Answers 4

Straightlines and Family of Straightlines



Question (1):
Answer: Line through the origin and perpendicular to x + y - 4 = 0 is x - y = 0 ...(iv) \ The foot of the perpendicular is given by solving (i) and (iv), x = y = 2. (2,2) are the coordinates of the foot of the perpendicular. Line through the origin perpendicular to x + 5y - 26 =0 is 5x - y = 0 .......... (v) Solving (ii) and (v) gives the coordinates of the foot of the perpendicular. Coordinates of the foot of the perpendicular are (1,5). Equation of the line perpendicular to 3x - y + 10 = 0 and through the origin is 3y + x = 0. The foot of the perpendicular is obtained by solving the two equations 3x - y - 10 = 0 and x + 3y = 0. We get x = 3, y = -1. \ The coordinates of the foot of the perpendicular are (3,-1). The three points are A(2,2), B(1, 5) and C(3, -1). A,B and C are collinear. Equation of the line joining the feet of the perpendicular is

Question (2): Show that the lines 3x - y - 6 = 0, 4x - y - 7 = 0 and 2x - y - 5 = 0 are concurrent.
Answer: Given lines are Solving (i) and (ii) for x and y, we get The point of intersection is (1,-3). Now substituting x = 1 and y = -3 in the LHS of (iii), we get \ The three given lines are concurrent.

Question (3): Show that the point (3,4) and (-2,-5) lie on the opposite sides of the line 3x + 4y + 16 = 0.
Answer: Let d = distance between (3,4) and the line 3x + 4y + 16 = 0. d'= distance between (-2,-5) and the line 3x + 4y + 16 = 0. Since d and d^\| are of opposite signs, the points lie on either side of the line.

Question (4): Show that the origin and the point (2,-3) lie on the opposite sides of the line 6x - 5y + 8 = 0.
Answer: Let d = distance between the origin and the line. Let d' = distance between the point and the line. Since d and d' are of opposite signs the origin and the point lie on opposite sides of the line.

Question (5): Find the equations of the bisectors of the angles between the lines 5x + 12y + 8 = 0 and 6x + 8y - 9 = 0.
Answer: The given equations can be written as 5x + 12y + 8 = 0 and 6x + 8y - 9 = 0 Equation of the angle bisectors is Taking only the positive sign, we get Taking only the negative sign, we get

Question (6):
Answer: From (1) and (2) and by rule of cross multiplication, we get Eliminating, we have

Question (7): Write down the equation of line with slope 4 and y-intercept 2.
Answer: Since the slope and the intercept are given, the equation is of the form y = mx+c. Here m = 4 and c = 2 \The equation of the line is y = 4x + 2.

Question (8): units on the negative direction of y-axis and makes an angle of 150^owith the positive direction of x-axis.
Answer: \ Required equation of the line is or

Question (9):
Answer: Let the equation of the line be Aliter Let P (x, y) be any point on the line, then

Question (10): Find the equation of the line passing through the points A (2,4) and B (-3,1).
Answer: Let P (x, y) be any point on the line joining A and B. Then, Aliter: Let equation of the line be y = mx + c ..........(i) (i) passes through (2,4), \ 4 = 2m + c ......... (a) (i) Passes through (-3,1), \ 1 = -3m + c ......... (b) Subtracting (b) from (a), we have Now substituting these values in (i), we get

Question (11): Find the equation of the line through the points
Answer: Let R (x, y) be any point on the line joining the point P and Q. \The equation of the line is

Question (12): The vertices of the triangle are (0,0), (4,6) and (8,4). Find the equation of its sides.
Answer: Let O (0,0), A (4,6) and B (8,4) be the vertices of D OAB. Equation of the side OA is Equation of the side OB is Equation of the side AB is

Question (13): Reduce each of the following straight line equation to the slope intercept form. Find m and c
Answer: The slope-intercept form of the equation of line is y = mx + c. i) Divide by the coefficient of y i.e., by 3. Then, Comparing with y = mx+c, we have m = 3 and c = 9. Dividing the coefficient of y i.e., by 3, we get

Question (14): Write the equation of the line passing through the origin with slope m.
Answer: Let the equation of the line be y = mx + c. This must pass through (0,0). \The required equation of the line is y = mx.

Question (15): Find the equation of the line passing through (5,1) and parallel to 7x - 2y = 10.
Answer: The line parallel to 7x - 2y = 10 is 7x - 2y = k. This line must pass through (5,1). \The equation of the required line is 7x - 2y = 33.

Question (16): If the lines ax + by + c = 0 and a'x + b'y + c' = 0 are parallel then show that ab' = a'b.
Answer: Reduce both line equations to slope-intercept form, then we get ii) Intercept form Since the lines are parallel, their slopes must be identical.

Question (17): Find the equation of the perpendicular bisector of the line segment joining the points (-4,5) and (2,9).
Answer: Let P (x, y) be any point on the perpendicular bisector. \Equation of the perpendicular bisector through (-1,7) is

Question (18): Find the equation of the straight line which bisects the distance between the two points (3, -4) and (5,6) and also bisects the distance between the points (-6,3) and (-10,7).
Answer: Co-ordinates of the mid-point of the line joining (3, -4) and (2,6) is Co-ordinates of the mid-point of the line joining (-6,3) and (-10,7) We have to find the equation of the line joining the points P and Q.

Question (19):
Answer:

Question (20): Find the image of the point (4,3) with respect to the mirror line x + y - 5 = 0.
Answer: Let the image of the point P (4,3) be Q (h, k) in the mirror line AB, whose equation is x + y - 5 = 0 ......(1) PQ is perpendicular to AB and is bisected at C. Since AB and PQ are perpendicular to each other, we have Solving for h and k from the equation (ii) and (iii), we have h = 2 and k = 1 The image of P (4,3) is Q (2,1) w.r.t line x + y - 5 = 0.

Question (21): Find the parametric form of a straight line.
Answer: In the figure, let AP = r This is the parametric form of the straight line.

Question (22): Find the co-ordinates of the foot of the perpendicular from the point (3,-3) on the line 4x - 7y + 25 = 0.
Answer: This line passes through (3, -3). The line is 7x + 4y = 9 -(ii) Solving (i) and (ii), we get

Question (23):
Answer: Equation of the line through (2,-1) with slope m is Slope of given line is -1.

Question (24): Find the coordinates of the points A (7,4), B (-3,6) which divides the line segment joining the points A and B in the ratio 3:2 i) internally and ii) externally.
Answer: x₁ = 7, x₂ = -3 y₁ = 4, y₂ = 6 m = 3, n = 2 i) Internal division: ii) External division: x = -23, y = 10 \ The required point is (-23,10).

Question (25): Find the coordinates of the points A(-3,-4), B(-8,7) which divides the line segment joining the points A and B in the ratio 7:5 i) internally and ii) externally.
Answer: x₁ = -3, x₂ = -8 y₁ = -4, y₂ = 7 m = 7, n = 5 i)Internal division: ii) External division:

Question (26): What are the co-ordinates of the point which trisect the line segment joining the points (1,2) and (11,9)?
Answer: Let C and D trisect the line segment AB. C divides AB in the ratio AC:CB::1:2. Co-ordinates of C are Co-ordinates of D are

Question (27): A (4,5) and B (1, -7) are two points and C is a point on AB produced such that AC = 4AB. Find the coordinates of the point C.
Answer: Let the coordinates of C be (x, y). Let AB = m, then AC = 4m. BC = AC - AB = 4m - m = 3m \ B divides AC in the ratio 1:3. x + 12 = 4 15 + y = -28 x = -18 y = -43 \Coordinates of C are (-8, -43).

Question (28): Find the slope and the angle of inclination q of the lines through each of the following pairs of points. a) (8,5), (3,5) b) (9,12), (11,14)
Answer: a) b) c) d)

Question (29): Find the value of y, so that the line through A (-3,y) and B (5,9) is parallel to the line through P (9,6) and Q (7,8).
Answer: Since AB \|\|PQ, we have Slope of AB = Slope of PQ

Question (30): Show that the line through (8, -5) and (4, 7) is perpendicular to the line through (10, 8) and (7, 7).
Answer: Since the product of the slopes = -1, the two lines are perpendicular to each other.