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 150owith 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.
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