Question 1: Reduce the following equations into slopeintercept form and find their slopes and the yintercepts.
(i) x + 7y = 0
(ii) 6x + 3y – 5 = 0
(iii) y = 0
ANSWER :  (i) The given equation is x + 7y = 0.
It can be written as
This equation is of the form y = mx + c, where .
Therefore, equation (1) is in the slopeintercept form, where the slope and the yintercept are and 0 respectively.
(ii) The given equation is 6x + 3y – 5 = 0.
It can be written as
Therefore, equation (2) is in the slopeintercept form, where the slope and the yintercept are –2 and respectively.
(iii) The given equation is y = 0.
It can be written as
y = 0.x + 0 … (3)
This equation is of the form y = mx + c, where m = 0 and c = 0.
Therefore, equation (3) is in the slopeintercept form, where the slope and the yintercept are 0 and 0 respectively.
Question 2: Reduce the following equations into intercept form and find their intercepts on the axes.
(i) 3x + 2y – 12 = 0
(ii) 4x – 3y = 6
(iii) 3y + 2 = 0.
ANSWER :  (i) The given equation is 3x + 2y – 12 = 0.
It can be written as
This equation is of the form , where a = 4 and b = 6.
Therefore, equation (1) is in the intercept form, where the intercepts on the x and y axes are 4 and 6 respectively.
(ii) The given equation is 4x – 3y = 6.
It can be written as
This equation is of the form , where a = and b = –2.
Therefore, equation (2) is in the intercept form, where the intercepts on the x and y axes are and –2 respectively.
(iii) The given equation is 3y + 2 = 0.
It can be written as
This equation is of the form , where a = 0 and b = .
Therefore, equation (3) is in the intercept form, where the intercept on the yaxis is and it has no intercept on the xaxis.
Question 3: Find the distance of the point (–1, 1) from the line 12(x + 6) = 5(y – 2).
ANSWER :  The given equation of the line is 12(x + 6) = 5(y – 2).
⇒ 12x + 72 = 5y – 10
⇒12x – 5y + 82 = 0 … (1)
On comparing equation (1) with general equation of line Ax + By + C = 0, we obtain A = 12, B = –5, and C = 82.
It is known that the perpendicular distance (d) of a line Ax + By + C = 0 from a point (x_{1}, y_{1}) is given by .
The given point is (x_{1}, y_{1}) = (–1, 1).
Therefore, the distance of point (–1, 1) from the given line
Question 4: Find the points on the xaxis, whose distances from the line are 4 units.
ANSWER :  The given equation of line is
On comparing equation (1) with general equation of line Ax + By + C = 0, we obtain A = 4, B = 3, and C = –12.
Let (a, 0) be the point on the xaxis whose distance from the given line is 4 units.
It is known that the perpendicular distance (d) of a line Ax + By + C = 0 from a point (x_{1}, y_{1}) is given by .
Therefore,
Thus, the required points on the xaxis are (–2, 0) and (8, 0).
Question 5: Find the distance between parallel lines
(i) 15x + 8y – 34 = 0 and 15x + 8y + 31 = 0
(ii) l (x + y) + p = 0 and l (x + y) – r = 0
ANSWER :  It is known that the distance (d) between parallel lines Ax + By + C_{1} = 0 and Ax + By + C_{2} = 0 is given by .
(i) The given parallel lines are 15x + 8y – 34 = 0 and 15x + 8y +31 = 0.
Here, A = 15, B = 8, C_{1} = –34, and C_{2} = 31.
Therefore, the distance between the parallel lines is
(ii) The given parallel lines are l (x + y) + p = 0 and l (x + y) – r = 0.
lx + ly + p = 0 and lx + ly – r = 0
Here, A = l, B = l, C_{1} = p, and C_{2} = –r.
Therefore, the distance between the parallel lines is
Question 6: Find equation of the line parallel to the line 3x – 4y + 2 = 0 and passing through the point (–2, 3).
ANSWER :  The equation of the given line is
, which is of the form y = mx + c
∴ Slope of the given line
It is known that parallel lines have the same slope.
∴ Slope of the other line =
Now, the equation of the line that has a slope of and passes through the point (–2, 3) is
Question 7: Find equation of the line perpendicular to the line x – 7y + 5 = 0 and having x intercept 3.
ANSWER :  The given equation of line is .
, which is of the form y = mx + c
∴Slope of the given line
The slope of the line perpendicular to the line having a slope of is
The equation of the line with slope –7 and xintercept 3 is given by
y = m (x – d)
⇒ y = –7 (x – 3)
⇒ y = –7x 21
⇒ 7x + y = 21
Question 8: Find angles between the lines
ANSWER :  The given lines are .
The slope of line (1) is , while the slope of line (2) is .
The acute angle i.e., θ between the two lines is given by
Thus, the angle between the given lines is either 30° or 180° – 30° = 150°.
Question 9: The line through the points (h, 3) and (4, 1) intersects the line 7x – 9y – 19 = 0. at right angle. Find the value of h.
ANSWER :  The slope of the line passing through points (h, 3) and (4, 1) is
The slope of line 7x – 9y – 19 = 0 or is .
It is given that the two lines are perpendicular.
Thus, the value of h is .
Question 10: Prove that the line through the point (x_{1}, y_{1}) and parallel to the line Ax + By + C = 0 is A (x –x_{1}) + B (y – y_{1}) = 0.
ANSWER :  The slope of line Ax + By C = 0 or is
It is known that parallel lines have the same slope.
∴ Slope of the other line =
The equation of the line passing through point (x_{1}, y_{1}) and having a slope is
Hence, the line through point (x_{1}, y_{1}) and parallel to line Ax + By + C = 0 is
A (x –x_{1}) + B (y – y_{1}) = 0
Question 11: Two lines passing through the point (2, 3) intersects each other at an angle of 60°. If slope of one line is 2, find equation of the other line.
ANSWER :  It is given that the slope of the first line, m_{1} = 2.
Let the slope of the other line be m_{2}.
The angle between the two lines is 60°.
The equation of the line passing through point (2, 3) and having a slope of is
In this case, the equation of the other line is .
The equation of the line passing through point (2, 3) and having a slope of is
In this case, the equation of the other line is .
Thus, the required equation of the other line is or .
Question 12: Find the equation of the right bisector of the line segment joining the points (3, 4) and (–1, 2).
ANSWER :  The right bisector of a line segment bisects the line segment at 90°.
The endpoints of the line segment are given as A (3, 4) and B (–1, 2).
Accordingly, midpoint of AB
Slope of AB
∴Slope of the line perpendicular to AB =
The equation of the line passing through (1, 3) and having a slope of –2 is
(y – 3) = –2 (x – 1)
y – 3 = –2x + 2
2x + y = 5
Thus, the required equation of the line is 2x + y = 5.
Question 13: Find the coordinates of the foot of perpendicular from the point (–1, 3) to the line 3x – 4y – 16 = 0.
ANSWER :  Let (a, b) be the coordinates of the foot of the perpendicular from the point (–1, 3) to the line 3x – 4y – 16 = 0.
Slope of the line joining (–1, 3) and (a, b), m_{1}
Slope of the line 3x – 4y – 16 = 0 or
Since these two lines are perpendicular, m_{1}m_{2} = –1
Point (a, b) lies on line 3x – 4y = 16.
∴3a – 4b = 16 … (2)
On solving equations (1) and (2), we obtain
Thus, the required coordinates of the foot of the perpendicular are .
Question 14: The perpendicular from the origin to the line y = mx + c meets it at the point (–1, 2). Find the values of m and c.
ANSWER :  The given equation of line is y = mx + c.
It is given that the perpendicular from the origin meets the given line at (–1, 2).
Therefore, the line joining the points (0, 0) and (–1, 2) is perpendicular to the given line.
∴Slope of the line joining (0, 0) and (–1, 2)
The slope of the given line is m.
Since point (–1, 2) lies on the given line, it satisfies the equation y = mx + c.
Thus, the respective values of m and c are .
Question 15: If p and q are the lengths of perpendiculars from the origin to the lines x cos θ – y sin θ = k cos 2θ and x sec θ + y cosec θ = k, respectively, prove that p^{2} + 4q^{2} = k^{2}
ANSWER :  The equations of given lines are
x cos θ – y sinθ = k cos 2θ … (1)
x secθ + y cosec θ= k … (2)
The perpendicular distance (d) of a line Ax + By + C = 0 from a point (x_{1}, y_{1}) is given by .
On comparing equation (1) to the general equation of line i.e., Ax + By + C = 0, we obtain A = cosθ, B = –sinθ, and C = –k cos 2θ.
It is given that p is the length of the perpendicular from (0, 0) to line (1).
On comparing equation (2) to the general equation of line i.e., Ax + By + C = 0, we obtain A = secθ, B = cosecθ, and C = –k.
It is given that q is the length of the perpendicular from (0, 0) to line (2).
From (3) and (4), we have
Hence, we proved that p^{2} + 4q^{2} = k^{2}.
Question 16: In the triangle ABC with vertices A (2, 3), B (4, –1) and C (1, 2), find the equation and length of altitude from the vertex A.
ANSWER :  Let AD be the altitude of triangle ABC from vertex A.
Accordingly, AD⊥BC
The equation of the line passing through point (2, 3) and having a slope of 1 is
(y – 3) = 1(x – 2)
⇒ x – y + 1 = 0
⇒ y – x = 1
Therefore, equation of the altitude from vertex A = y – x = 1.
Length of AD = Length of the perpendicular from A (2, 3) to BC
The equation of BC is
The perpendicular distance (d) of a line Ax + By + C = 0 from a point (x_{1}, y_{1}) is given by .
On comparing equation (1) to the general equation of line Ax + By + C = 0, we obtain A = 1, B = 1, and C = –3.
∴Length of AD
Thus, the equation and the length of the altitude from vertex A are y – x = 1 and units respectively.
Question 17: If p is the length of perpendicular from the origin to the line whose intercepts on the axes are a and b, then show that .
ANSWER :  It is known that the equation of a line whose intercepts on the axes are a and b is
The perpendicular distance (d) of a line Ax + By + C = 0 from a point (x_{1}, y_{1}) is given by .
On comparing equation (1) to the general equation of line Ax + By + C = 0, we obtain A = b, B = a, and C = –ab.
Therefore, if p is the length of the perpendicular from point (x_{1}, y_{1}) = (0, 0) to line (1), we obtain
On squaring both sides, we obtain
Hence, we showed that .
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