Chapter 10 Straight lines Ex 10.3 Solutions
Question - 11 : - Prove that the line through thepoint (x1, y1) and parallel to the line Ax + By + C = 0is A (x – x1) + B (y – y1) = 0.
Answer - 11 : -
Let the slope of line Ax + By + C = 0 be m
Ax + By + C = 0
So, y = -A/Bx – C/B
m = -A/B
By using the formula,
Equation of the line passing through point (x1, y1)and having slope m = -A/B is
y – y1 = m (x – x1)
y – y1= -A/B (x – x1)
B (y – y1) = -A (x – x1)
∴ A(x – x1) + B(y – y1) = 0
So, the line through point (x1, y1)and parallel to the line Ax + By + C = 0 is A (x – x1) + B (y – y1)= 0
Hence proved.
Question - 12 : - Two lines passing through the point (2, 3) intersects each other at an angle of 60o. If slope of one line is 2, find equation of the other line.
Answer - 12 : -
Given: m1 = 2
Let the slope of the first line be m1
And let the slope of the other line be m2.
Angle between the two lines is 60°.
So,
Question - 13 : - Find the equation of the right bisector of the line segment joining the points (3, 4) and (–1, 2).
Answer - 13 : -
Given:
The right bisector of a line segment bisects the linesegment at 90°.
End-points of the line segment AB are given as A (3, 4) andB (–1, 2).
Let mid-point of AB be (x, y)
x = (3-1)/2= 2/2 = 1
y = (4+2)/2= 6/2 = 3
(x, y) = (1, 3)
Let the slope of line AB be m1
m1 = (2 – 4)/(-1 – 3)
= -2/(-4)
= 1/2
And let the slope of the line perpendicular to AB be m2
m2 = -1/(1/2)
= -2
The equation of the line passing through (1, 3) and having aslope of –2 is
(y – 3) = -2 (x – 1)
y – 3 = – 2x + 2
2x + y = 5
∴ The required equation of the line is 2x + y = 5
Question - 14 : - Find the coordinates of the foot of perpendicular from the point (–1, 3) to the line 3x – 4y – 16 = 0.
Answer - 14 : -
Let us consider the co-ordinates of the foot of the perpendicular from (-1, 3) to the line 3x – 4y – 16 = 0 be (a, b)
So, let the slope of the line joining (-1, 3) and (a, b) be m1
m1 = (b-3)/(a+1)
And let the slope of the line 3x – 4y – 16 = 0 be m2
y = 3/4x – 4
m2 = 3/4
Since these two lines are perpendicular, m1 × m2 = -1
(b-3)/(a+1) × (3/4) = -1
(3b-9)/(4a+4) = -1
3b – 9 = -4a – 4
4a + 3b = 5 …….(1)
Point (a, b) lies on the line 3x – 4y = 16
3a – 4b = 16 ……..(2)
Solving equations (1) and (2), we get
a = 68/25 and b = -49/25
∴ The co-ordinates of the foot of perpendicular is (68/25, -49/25)
Question - 15 : - 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 - 15 : -
Given:
The perpendicular from the origin meets the given line at(–1, 2).
The equation of line is y = mx + c
The line joining the points (0, 0) and (–1, 2) isperpendicular to the given line.
So, the slope of the line joining (0, 0) and (–1, 2) =2/(-1) = -2
Slope of the given line is m.
m × (-2) = -1
m = 1/2
Since, point (-1, 2) lies on the given line,
y = mx + c
2 = 1/2 × (-1) + c
c = 2 + 1/2 = 5/2
∴ The values of m and c are 1/2 and 5/2 respectively.
Question - 16 : - If p and q are the lengths ofperpendiculars from the origin to the lines x cos θ − y sin θ = k cos 2θ and xsec θ + y cosec θ = k, respectively, prove that p2 + 4q2 =k2
Answer - 16 : -
Given:
The equations of given lines are
x cos θ – y sin θ = k cos 2θ …………………… (1)
x sec θ + y cosec θ = k ……………….… (2)
Perpendicular distance (d) of a line Ax + By + C = 0 from a point (x1, y1) is given by
q = k cos θ sin θ
Multiply both sides by 2, we get
2q = 2k cos θ sin θ = k × 2sin θ cos θ
2q = k sin 2θ
Squaring both sides, we get
4q2 = k2 sin22θ…………………(4)
Now add (3) and (4) we get
p2 + 4q2 = k2 cos2 2θ+ k2 sin2 2θ
p2 + 4q2 = k2 (cos2 2θ+ sin2 2θ) [Since, cos2 2θ + sin2 2θ= 1]
∴ p2 + 4q2 = k2
Hence proved.
Answer - 17 : -
Let AD be the altitude of triangle ABC from vertex A.
So, AD is perpendicular to BC
Given:
Vertices A (2, 3), B (4, –1) and C (1, 2)
Let slope of line BC = m1
m1 = (- 1 – 2)/(4 – 1)
m1 = -1
Let slope of line AD be m2
AD is perpendicular to BC
m1 × m2 = -1
-1 × m2 = -1
m2 = 1
The equation of the line passing through point (2, 3) andhaving a slope of 1 is
y – 3 = 1 × (x – 2)
y – 3 = x – 2
y – x = 1
Equation of the altitude from vertex A = y – x = 1
Length of AD = Length of the perpendicular from A (2, 3) toBC
Equation of BC is
y + 1 = -1 × (x – 4)
y + 1 = -x + 4
x + y – 3 = 0 …………………(1)
Perpendicular distance (d) of a line Ax + By + C = 0 from apoint (x1, y1) is given by
Now compare equation (1) to the general equation of linei.e., Ax + By + C = 0, we get
Lengthof AD =
[where, A = 1, B = 1 and C = -3]
∴ The equation and the length of the altitude from vertex Aare y – x = 1 and
√2 units respectively.
If p is the length ofperpendicular from the origin to the line whose intercepts on the axes are aand b, then show that 1/p2 = 1/a2 + 1/b2
Answer - 18 : -
Equation of a line whose intercepts on the axes are a and bis x/a + y/b = 1
bx + ay = ab
bx + ay – ab = 0 ………………..(1)
Perpendicular distance (d) of a line Ax + By + C = 0 from apoint (x1, y1) is given by
Now square on both the sides we get
∴ 1/p2 = 1/a2 + 1/b2
Hence proved.