Question 11
Question: 
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
Question: 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
Question: 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
Question: 
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
Question: 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
Question: 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
Question: 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
Question: 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
Question: 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
Question: 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.
