Class 11 - Mathematics
Straight Lines - Exercise 10.2
Top Block 1
Question 1:
Write the equations for the x and y-axes.
Answer:
The y-coordinate of every point on the x-axis is 0.
Therefore, the equation of the x-axis is y = 0.
The x-coordinate of every point on the y-axis is 0.
Therefore, the equation of the y-axis is y = 0.
Question 2:
Find the equation of the line which passes through the point (–4, 3) with slope 1/2.
Answer:
We know that the equation of the line passing through point (x0, y0), whose slope is m, is
y – y0 = m(x – x0)
Thus, the equation of the line passing through point (–4, 3), whose slope is 1/2, is
y – 3 = (x + 4)/2
=> 2y – 6 = x + 4
=> x + 4 – 2y + 6 = 0
=> x – 2y + 10 = 0
Question 3:
Find the equation of the line which passes though (0, 0) with slope m.
Answer:
We know that the equation of the line passing through point (x0, y0), whose slope is m, is
y – y0 = m(x – x0)
Thus, the equation of the line passing through point (0, 0), whose slope is m, is
y – 0 = m(x – 0)
=> y = mx
Question 4:
Find the equation of the line which passes though (2, 2√3) and is inclined with the x-axis at an angle of 75°.
Answer:
The slope of the line that inclines with the x-axis at an angle of 75° is m = tan 75°
=> m = tan(45° + 30°)
=> m = (tan 45° + tan 30°)/(1 – tan 45° * tan 30°)
=> m = (1 + 1/√3)/(1 – 1 * 1/√3)
=> m = (1 + 1/√3)/(1 – 1/√3)
=> m = (√3 + 1)/(√3 – 1)
We know that the equation of the line passing through point (x0, y0), whose slope is m, is
y – y0 = m(x – x0)
Thus, if a line passes though (2, 2√3) and inclines with the x-axis at an angle of 75°, then the
equation of the line is given as
(y – 2√3) = {(√3 + 1)/(√3 – 1)}(x – 2)
=> (y – 2√3)(√3 – 1) = (√3 + 1)(x – 2)
=> y(√3 – 1) – 2√3(√3 – 1) = x(√3 + 1) – 2(√3 + 1)
=> y(√3 – 1) – 6 + 2√3 = x(√3 + 1) – 2√3 – 2
=> x(√3 + 1) – 2√3 – 2 – y(√3 – 1) + 6 – 2√3 = 0
=> x(√3 + 1) – y(√3 – 1) + 4 – 4√3 = 0
=> x(√3 + 1) – y(√3 – 1) + 4(1 – √3) = 0
Question 5:
Find the equation of the line which intersects the x-axis at a distance of 3 units to the left of origin with slope –2.
Answer:
It is known that if a line with slope m makes x-intercept d, then the equation of the line is
given as y = m(x – d)
For the line intersecting the x-axis at a distance of 3 units to the left of the origin, d = –3.
The slope of the line is given as m = –2
Thus, the required equation of the given line is
=> y = –2[x – (–3)]
=> y = –2[x + 3]
=> y = –2x – 6
=> 2x + y + 6 = 0
Question 6:
Find the equation of the line which intersects the y-axis at a distance of 2 units above the origin and makes an angle of 30°
with the positive direction of the x-axis.
Answer:
It is known that if a line with slope m makes y-intercept c, then the equation of the line
is given as y = mx + c
Here, c = 2 and m = tan 30° = 1/√3
Thus, the required equation of the given line is
=> y = x/√3 + 2
=> y = (x + 2√3)/ √3
=> y√3 = x + 2√3 => x – √3y – 2√3 = 0
Question 7:
Find the equation of the line which passes through the points (–1, 1) and (2, –4).
Answer:
It is known that the equation of the line passing through points (x1, y1) and (x2, y2) is
(y – y1) = {(y2 – y1)/(x2 – x1)}(x – x1)
Therefore, the equation of the line passing through the points (–1, 1) and (2, –4) is
(y – 1) = {(-4 – 1)/(2 + 1)}(x + 1)
=> (y – 1) = (-5/3)(x + 1)
=> 3(y – 1) = -5(x + 1)
=> 3y – 3 = -5x – 5
=> 5x + 3y + 2 = 0
Question 8:
Find the equation of the line which is at a perpendicular distance of 5 units from the origin and the angle
made by the perpendicular with the positive x-axis is 30°
Answer:
If p is the length of the normal from the origin to a line and ω is the angle made by the normal
with the positive direction of the x-axis, then the equation of the line is given by
X cos ω + y sin ω = p.
Here, p = 5 units and ω = 30°
Thus, the required equation of the given line is x
x cos 30° + y sin 30° = 5
√3x/2 + y/2 = 5
=> √3x + y = 5 * 2
=> √3x + y = 10
Question 9:
The vertices of ∆PQR are P (2, 1), Q (–2, 3) and R (4, 5). Find equation of the median through the vertex R.
Answer:
It is given that the vertices of ∆PQR are P (2, 1), Q (–2, 3), and R (4, 5).
Let RL be the median through vertex R.
Accordingly, L is the mid-point of PQ.
By mid-point formula, the coordinates of point L are given by {(2 – 2)/2, (1 + 3)/2} = (0, 2)
(y – y1) = {(y2 – y1)/(x2 – x1)}(x – x1)
Therefore, the equation of RL can be determined by substituting
(x1, y1) = (4, 5) and (x2, y2) = (0, 2)
(y – 5) = {(2 – 5)/(0 – 4)}(x – 4)
=> (y – 5) = (-3/-4)(x – 4)
=> 4(y – 5) = 3(x – 4)
=> 4y – 20 = 3x – 12
=> 3x – 4y + 8 = 0
Thus, the required equation of the median through vertex R is 3x – 4y + 8 = 0
Question 10:
Find the equation of the line passing through (–3, 5) and perpendicular to the line through the points (2, 5) and (–3, 6).
Answer:
The slope of the line joining the points (2, 5) and (–3, 6) is
M = (6 – 5)/(-3 – 2) = -1/5
We know that two non-vertical lines are perpendicular to each other if and only if their slopes
are negative reciprocals of each other.
Therefore, slope of the line perpendicular to the line through the points (2, 5) and (–3, 6)
= -1/m = -1/(-1/5) = 5
Now, the equation of the line passing through point (–3, 5), whose slope is 5, is
y – 5 = 5(x + 3)
=> y – 5 = 5x + 15
=> 5x – y + 20 = 0
Question 11:
A line perpendicular to the line segment joining the points (1, 0) and (2, 3) divides it in the ratio 1 : n. Find the equation of the line.
Answer:
According to the section formula, the coordinates of the point that divides the line segment
joining the points (1, 0) and (2, 3) in the ratio 1 : n is given by
[(n * 1 + 1 * 2)/(1 + n), (n * 0 + 1 * 3)/(1 + n)] = [(n + 2)/(n + 1), 3/(n + 1)]
The slope of the line joining the points (1, 0) and (2, 3) is
m = (3 – 0)/(2 – 1) = 3
We know that two non-vertical lines are perpendicular to each other if and only if their slopes
are negative reciprocals of each other.
Therefore, slope of the line that is perpendicular to the line joining the points (1, 0) and (2, 3)
= -1/m = -1/3
Now, the equation of the line passing through [(n + 2)/(n + 1), 3/(n + 1)] and whose slope is
-1/3 given by
y – 3/(n + 1) = (-1/3)[x – (n + 2)/(n + 1)]
=> y(n + 1) – 3 = (-1/3)[x(n + 1) – (n + 2)]
=> 3y(n + 1) – 9 = -[x(n + 1) – (n + 2)]
=> 3y(n + 1) – 9 + x(n + 1) – (n + 2) = 0
=> (n + 1)x – 3(n + 1)y – 9 – n – 2 = 0
=> (n + 1)x – 3(n + 1)y – n – 11 = 0
=> (n + 1)x – 3(n + 1)y = n + 11
Question 12:
Find the equation of a line that cuts off equal intercepts on the coordinate axes and passes through the point (2, 3).
Answer:
The equation of a line in the intercept form is
x/a + y/b = 1 …………..1
Here, a and b are the intercepts on x and y axes respectively.
It is given that the line cuts off equal intercepts on both the axes.
This means that a = b.
Now equation 1 becomes
x/a + y/a = 1
=> x + y = a ……..2
Since the given line passes through point (2, 3), equation 2 reduces to
2 + 3 = a
=> a = 5
On substituting the value of a in equation 2, we get
x + y = 5
which is the required equation of the line
Question 13:
Find equation of the line passing through the point (2, 2) and cutting off intercepts on the axes whose sum is 9.
Answer:
The equation of a line in the intercept form is
x/a + y/b = 1 ………1
Here, a and b are the intercepts on x and y axes respectively.
It is given that
a + b = 9
=> b = 9 – a …………2
From equations 1 and 2, we obtain
x/a + y/(9 – a) = 1 ………..3
It is given that the line passes through point (2, 2). Therefore, equation 3 becomes
2/a + 2/(9 – a) = 1
=> 2{1/a + 1/(9 – a)} = 1
=> 2[(9 – a + a)/{a(9 – a)}] = 1
=> 18/{9a – a2} = 1
=> 18 = 9a – a2
=> a2 – 9a + 18 = 0
=> (a – 6)(a – 3) = 0
=> a = 3, 6
If a = 6 and b = 9 – 6 = 3, then the equation of the line is
x/6 + y/3 = 1
=> (x + 2y)/6 = 1
=> x + 2y = 6
If a = 3 and b = 9 – 3 = 6, then the equation of the line is
x/3 + y/6 = 1
=> (2x + y)/6 = 1
=> 2x + y = 6
Question 14:
Find equation of the line through the point (0, 2) making an angle 2π/3 with the positive x-axis.
Also, find the equation of line parallel to it and crossing the y-axis at a distance of 2 units below the origin.
Answer:
The slope of the line making an angle 2π/3 with the positive x-axis is
m = tan 2π/3 = -√3
Now, the equation of the line passing through point (0, 2) and having a slope -√3 is
(y – 2) = -√3(x – 0)
=> y – 2 = -√3x
=> √3x + y – 2 = 0
The slope of line parallel to √3x + y – 2 = 0 is -√3
It is given that the line parallel to line √3x + y – 2 = 0 crosses the y-axis 2 units below the origin
i.e., it passes through point (0, –2).
Hence, the equation of the line passing through point (0, –2) and having a slope is
y – (-2) = -√3(x – 2)
=> y + 2 = -√3x
=> √3x + y + 2 = 0
Question 15:
The perpendicular from the origin to a line meets it at the point (– 2, 9), find the equation of the line.
Answer:
The slope of the line joining the origin (0, 0) and point (–2, 9) is
m1 = (9 – 0)/(-2 – 0) = -9/2
Again, the slope of the line perpendicular to the line joining the origin and point (– 2, 9) is
m2 = -1/ m1 = -1/(-9/2) = 2/9
Now, the equation of the line passing through point (–2, 9) and having a slope m2 is
y – 9 = (2/9)(x + 2)
=> 9(y – 9) = 2(x + 2)
=> 9y – 81 = 2x + 4
=> 2x – 9y + 85 = 0
Question 16:
The length L (in centimeter) of a copper rod is a linear function of its Celsius temperature C.
In an experiment, if L = 124.942 when C = 20 and L = 125.134 when C = 110, express L in terms of C.
Answer:
It is given that when C = 20, the value of L is 124.942
and when C = 110, the value of L is 125.134
Now, points (20, 124.942) and (110, 125.134) satisfy the linear relation between L and C.
Assuming C along the x-axis and L along the y-axis, we have two points i.e., (20, 124.942) and
(110, 125.134) in the XY plane.
Therefore, the linear relation between L and C is the equation of the line passing through
points (20, 124.942) and (110, 125.134)
L – 124.942 = {(125.134 – 124.942)/(110 – 20)}(C – 20)
=> L – 124.942 = (0.192/90)(C – 20)
=> L = (0.192/90)(C – 20) + 124.942
Which is the required linear relation.
Question 17:
The owner of a milk store finds that, he can sell 980 litres of milk each week at Rs 14/litre and 1220 litres of milk each week at Rs 16/litre.
Assuming a linear relationship between selling price and demand, how many litres could he sell weekly at Rs 17/litre?
Answer:
The relationship between selling price and demand is linear.
Assuming selling price per litre along the x-axis and demand along the y-axis, we have
two points i.e., (14, 980) and (16, 1220) in the XY plane that satisfy the linear relationship
between selling price and demand.
Therefore, the linear relationship between selling price per litre and demand is the equation
of the line passing through points (14, 980) and (16, 1220).
y – 980 = {(1220 – 980)/(16 – 14)}(x – 14)
=> y – 980 = (240/2)(x – 14)
=> y – 980 = 120(x – 14)
=> y = 120(x – 14) + 980
When x = Rs 17/litre,
=> y = 120(17 – 14) + 980
=> y = 120 * 3 + 980
=> y = 360 + 980
=> y = 1340
Thus, the owner of the milk store could sell 1340 litres of milk weekly at Rs 17/litre.
Question 18:
P (a, b) is the mid-point of a line segment between axes. Show that equation of the line is
x/a + y/b = 2
Answer:
Let AB be the line segment between the axes and let P (a, b) be its mid-point.
Let the coordinates of A and B be (0, y) and (x, 0) respectively.
Mddle block 1
{(0 + x)/2, (0 + y)/2} = (a, b)
=> (x/2, y/2) = (a, b)
=> x/2 = a and y/2 = b
=> x = 2a and y = 2b
Thus, the respective coordinates of A and B are (0, 2b) and (2a, 0).
The equation of the line passing through points (0, 2b) and (2a, 0)
y – 2b = {(0 – 2b)/(2a – 0)}(x – 0)
=> y – 2b = (-2b/2a)(x)
=> y – 2b = (-b/a)(x)
=> a(y – 2b) = -bx
=> ay – 2ab = -bx
=> bx + ay = 2ab
On dividing both sides by ab, we obtain
=> bx/ab + ay/ab = 2ab/ab
=> x/a + y/b = 2
Thus, the equation of the line is x/a + y/b = 2
Question 19:
Point R (h, k) divides a line segment between the axes in the ratio 1 : 2. Find equation of the line.
Answer:
Let AB be the line segment between the axes such that point R (h, k) divides AB in the ratio
1: 2.
Since point R (h, k) divides AB in the ratio 1: 2, according to the section formula,
(h, k) = {(1 * 0 + 2 * x)/(1 + 2), (1 * y + 2 * 0)/(1 + 2)}
=> (h, k) = (2x/3, y/3)
=> h = 2x/3 and k = y/3
=> x = 3h/2 and y = 3k
Therefore, the respective coordinates of A and B are (3h/2, 0) and (0, 3k).
Now, the equation of line AB passing through points (3h/2, 0) and (0, 3k) is
y – 0 = {(3k – 0)/(0 – 3h/2)}(x – 3h/2)
=> y = (-2k/h)(x – 3h/2)
=> hy = (-2k)(x – 3h/2)
=> hy = (-2k)(x – 3h/2)
=> hy = -2kx + 3hk
=> 2kx + hy = 3kh
Thus, the required equation of the line is 2kx + hy = 3hk
Question 20:
By using the concept of equation of a line, prove that the three points (3, 0), (–2, –2) and (8, 2) are collinear.
Answer:
In order to show that points (3, 0), (–2, –2), and (8, 2) are collinear, it suffices to show that the
line passing through points (3, 0) and (–2, –2) also passes through point (8, 2).
The equation of the line passing through points (3, 0) and (–2, –2) is
y – 0 = {(-2 – 0)/(-2 – 3)}(x – 3)
=> y = (-2/-5)(x – 3)
=> y = (2/5)(x – 3)
=> 5y = 2(x – 3)
=> 5y = 2x – 6
=> 2x – 5y – 6 = 0
It is observed that at x = 8 and y = 2,
L.H.S. = 2 * 8 – 5 * 2 = 16 – 10 = 6 = R.H.S.
Therefore, the line passing through points (3, 0) and (–2, –2) also passes through point (8, 2).
Hence, the points (3, 0), (–2, –2), and (8, 2) are collinear.
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