In Exercises 1 to 8, find the equation of the line which satisfy the given conditions:
Question 1. Write the equations for the x-and y-axes.
Solution:
The y-coordinate of every point on x-axis is 0 and the x-coordinate of every point on y-axis is 0
So, the equation of x-axis is y = 0 and the equation of y-axis is y = 0.
Question 2. Passing through the point (– 4, 3) with slope 1/2.
Solution:
Given that point p(x1, y1) is (-4, 3) and slope m = 1/2
Equation of Line can be derived by formula y – y1 = m (x – x1),
Where m is slope of line and (x1, y1) is co-ordinate of p from which line passes.
y – 3 = 1/2(x – (-4))
y – 3 = 1/2 (x + 4)
2(y – 3) = x + 4
2y – 6 = x + 4
x + 4 – (2y – 6) = 0
x + 4 – 2y + 6 = 0
x – 2y + 10 = 0
So, the equation of the line is x – 2y + 10 = 0.
Question 3. Passing through (0, 0) with slope m.
Solution:
Given that Point p(x1, y1) is (0, 0) and slope is m.
So, the equation of Line can be derived by formula y – y1 = m (x – x1),
Where m is slope of line and (x1, y1) is co-ordinate from which line passes.
So, y – 0 = m (x – 0)
y = mx
y – mx = 0
So, the equation of the line is y – mx = 0.
Question 4. Passing through (2, 2√3) and inclined with the x-axis at an angle of 75o.
Solution:
Given that point p(x1, y1) is (2, 2√3) and θ = 75°
So, the equation of line is (y – y1) = m (x – x1)
where, m = slope of line = tan θ and (x1, y1) are the points through which line passes
So, m = tan 75°
Now, finding tan 75° using below formula:
tan(A + B) = (tanA + tanB)/(1 – tanA.tanB)
tan 75° = tan(45° + 30°)
tan 75° = (tan 45° + tan 30°)/(1 – tan 30°.tan 45°)
tan 75° = (1 + 1/√3)/(1 – 1/√3)
tan 75° = (√3 + 1)/(√3 – 1)
By rationalizing, we get
tan 75° = 2 + √3
Equation of Line will be,
(y – 2√3) = (2 + √3)(x – 2)
y – 2√3 = 2 x – 4 + √3x – 2 √3
y = 2 x – 4 + √3x
(2 + √3)x – y – 4 = 0
So, the equation of the line is (2 + √3)x – y – 4 = 0.
Question 5. Intersecting the x-axis at a distance of 3 units to the left of origin with slope –2.
Solution:
Given that Slope m = –2, x-intercept = 3,
that means Line is passing from point p(x1, y1) that is (3, 0)
Equation of line will be
y – 0 = (–2) × (x − 3).
y = (–2) × (x + 3)
y = –2x – 6
2x + y + 6 = 0
So, the equation of the line is 2x + y + 6 = 0.
Question 6. Intersecting the y-axis at a distance of 2 units above the origin and making an angle of 30° with the positive direction of the x-axis.
Solution:
Given that θ = 30° then slope(m) will be = tan θ
m = tan30° = (1/√3)
As Y intercept is 2 that means line is passing from (0, 2) point p(x1, y1) = (0, 2)
Equation of Line will be
y – 2 = (1/√3)x
y = (1/√3)x + 2
y = (x + 2√3) / √3
√3 y = x + 2√3
x – √3 y + 2√3 = 0
So, the equation of the line is x – √3 y + 2√3 = 0.
Question 7. Passing through the points (–1, 1) and (2, – 4).
Solution:
Given that point p1(x1, y1) is (–1, 1) and point p2(x2, y2) is (2, –4),
It is mention in question that line passes from p1 and p2.
That means, the Slope(m) of line will be (y2 – y1)/(x2 – x1)
m = (–4 – 1)/(2 – (–1))
m = –5/3
Equation of Line will be
(y – y1) = m(x – x1)
y – 1 = –5/3 (x + 1)
3 (y – 1) = (–5)(x + 1)
3y – 3 = –5x – 5
3y – 3 + 5x + 5 = 0
5x + 3y + 2 = 0
So, the equation of the line is 5x + 3y + 2 = 0.
Question 8. The vertices of ΔPQR are P (2, 1), Q (–2, 3), and R (4, 5). Find the equation of the median through the vertex R.
Solution:
Given Vertices of ΔPQR that are P (2, 1), Q (–2, 3) and R (4, 5)
Let RS be the median of vertex R => S is a midpoint of PQ.
As S is midpoint of PQ => S = (P + Q)/2
S = (2 – 2, 1 + 3)/2
S = (0, 2)
Equation of the line passing through the points (x1, y1) and (x2, y2) is given by
y – 5 = –3/ –4(x – 4)
(–4)(y – 5) = (–3)(x – 4)
–4y + 20 = –3x + 12
–4y + 20 + 3x – 12 = 0
3x – 4y + 8 = 0
So, the equation of median through the vertex R is 3x – 4y + 8 = 0.
Question 9. Find the equation of the line passing through (–3, 5) and perpendicular to the line through the points (2, 5) and (–3, 6).
Solution:
Given that Points are (2, 5) and (-3, 6).
So, the slope, m1 = (y2 – y1)/(x2 – x1)
= (6 – 5)/(–3 – 2)
= 1/–5 = –1/5
As we know that two non-vertical lines are perpendicular to each other
if their slopes are negative reciprocals of each other.
Then, m = (–1/m1)
= –1/(–1/5)
= 5
As we know that the point p(x, y) lies on the line with
slope m through the fixed point (x1, y1),
If its coordinates satisfy the equation y – y1 = m (x – x1)
Then, y – 5 = 5(x – (–3))
y – 5 = 5x + 15
5x + 15 – y + 5 = 0
5x – y + 20 = 0
So, the equation of the line is 5x – y + 20 = 0
Question 10. 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.
Solution:
Slope of line segment will be m1 = (y2 – y1)/(x2 – x1)
= (3 – 0)/(2 – 1)
= 3/1
= 3
If two line are perpendicular to each other than slope of line1 X slope of line2 = -1
So, the slope of line will be m2 = (-1/m1)
m2 = -1/3
As we know that the coordinates of a point (p, q) dividing the line
segment joining the points (x1, y1) and (x2, y2) internally
In the ratio m : n are:
(p, q) = ((mx2 + nx1)/(m + n), (my2 + ny1)/(m + n))
(p, q) = ((1 × 2 + n × 1)/(1 + n), (1 × 3 + n × 0)/(1 + n))
p = (2 + n)/(1 + n)
q = 3/(1 + n)
We know that the point (p, q) lies on the line with slope m2,
Equation of line will be y – q = m2(x – p)
y – 3/(1 + n) = (–1/3)(x – (2 + n)/(1 + n))
3((1 + n) y – 3) = (–(1 + n) x + 2 + n)
3(1 + n) y – 9 = – (1 + n) x + 2 + n
(1 + n) x + 3(1 + n) y – n – 9 – 2 = 0
(1 + n) x + 3(1 + n) y – n – 11 = 0
So, the equation of the line is (1 + n) x + 3(1 + n) y – n – 11 = 0
Question 11. Find the equation of a line that cuts off equal intercepts on the coordinate axes and passes through the point (2, 3).
Solution:
Given that the line cuts off equal intercepts on the coordinate axes i.e. a = b.
So, the equation of the line intercepts a and b on x-and y-axis, respectively, which is
x/a + y/b = 1
So, x/a + y/a = 1
x + y = a -(1)
Given point is (2, 3)
2 + 3 = a
a = 5
On substituting the value of ‘a’ in eq(1), we get
x + y = 5
x + y – 5 = 0
So, the equation of the line is x + y – 5 = 0.
Question 12. Find equation of the line passing through the point (2, 2) and cutting off intercepts on the axes whose sum is 9.
Solution:
Equation of the line making intercepts a and b on x-and y-axis, respectively, is
x/a + y/b = 1 -(1)
Given that sum of intercepts = 9
a + b = 9
b = 9 – a
On substituting the value of b in the eq(1), we get
x/a + y/(9 – a) = 1
Given that the line passes through the point (2, 2),
So, 2/a + 2/(9 – a) = 1
[2(9 – a) + 2a] / a(9 – a) = 1
[18 – 2a + 2a] / a(9 – a) = 1
18/a(9 – a) = 1
18 = a (9 – a)
18 = 9a – a2
a2 – 9a + 18 = 0
By using factorization method for quadratic equation solving, we get
a2 – 3a – 6a + 18 = 0
a (a – 3) – 6 (a – 3) = 0
(a – 3) (a – 6) = 0
a = 3 or a = 6
Let us substitute in eq(1),
Case 1 (a = 3):
Then b = 9 – 3 = 6
x/3 + y/6 = 1
2x + y = 6
2x + y – 6 = 0
Case 2 (a = 6):
Then b = 9 – 6 = 3
x/6 + y/3 = 1
x + 2y = 6
x + 2y – 6 = 0
So, the equation of the line is 2x + y – 6 = 0 or x + 2y – 6 = 0
Question 13. 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.
Solution:
Given that
Point (0, 2) and θ = 2π/3
slope (m)= tan θ
m = tan (2π/3) = -√3
Now, the equation of line passing through point(p,q) with slope m will be:
y – q = m (x – p)
y – 2 = -√3 (x – 0)
y – 2 = -√3 x
√3 x + y – 2 = 0
Now we need to find equation of line parallel to above obtained equation
crosses the y-axis at a distance of 2 units below the origin.
So, the point = (0, -2) and for parallel line slope will be same that is m = -√3
From point slope form equation,
y – (–2) = –√3 (x – 0)
y + 2 = -√3 x
√3 x + y + 2 = 0
So, the equation of line is √3 x + y – 2 = 0 and the line parallel to it is √3 x + y + 2 = 0.
Question 14. The perpendicular from the origin to a line meets it at the point (–2, 9), find the equation of the line.
Solution:
Perpendicular line will pass through point (0,0) and (-2,9) as per given question,
Then slope of perpendicular line will be m1 = (y2 – y1)/(x2 – x1)
= (9 – 0)/(–2 – 0)
= –9/2
As we know that two non-vertical lines are perpendicular to each
other if their slopes are negative reciprocals of each other.
So, we need to find equation of line let say it AB.
Slope of line AB will be m = (–1/m1) = –1/(–9/2) = 2/9
As perpendicular line and line intersecting at (–2, 9) that means (–2, 9) lies on the line AB
By using Slope Point Form of line, we get
y – y1 = m (x – x1)
y – 9 = (2/9) (x – (–2))
9(y – 9) = 2(x + 2)
9y – 81 = 2x + 4
2x + 4 – 9y + 81 = 0
2x – 9y + 85 = 0
So, the equation of line is 2x – 9y + 85 = 0.
Question 15. 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.
Solution:
Let us considered ‘L’ along X-axis and ‘C’ along Y-axis,
So, we have two points (124.942, 20) and (125.134, 110) in XY-plane.
As we know that the equation of the line passing through the points (x1, y1) and (x2, y2) is given by
0.192(C-20) = 90(L – 124.942)
L can be express in Terms of C in following manner
L = (0.192 × (C – 20)/90) + 124.942
Question 16. The owner of a milk store finds that he can sell 980 liters of milk each week at Rs. 14/liter and 1220 liters of milk each week at Rs. 16/liter. Assuming a linear relationship between the selling price and demand, how many liters could he sell weekly at Rs. 17/liter?
Solution:
Let us considered the relationship between selling price and demand is linear.
So, the selling price per liter along X-axis and demand along Y-axis,
We have two points (14, 980) and (16, 1220) in XY-plane
We know that the equation of the line passing through the points (x1, y1) and (x2, y2) is given by
y – 980 = 120 (x – 14)
y = 120 (x – 14) + 980
When x = Rs 17/liter,
y = 120 (17 – 14) + 980
y = 120(3) + 980
y = 360 + 980 = 1340
So, the owner can sell 1340 liters weekly at Rs. 17/liter.
Question 17. 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
Solution:
Let us considered AB be a line segment whose midpoint is P (a, b).
So, the coordinates of A is (0, y) and B is (x, 0)
Mid-Point of AB will be P(a, b) = ((x + 0)/2, (y + 0)/2)
a = x/2
x = 2a
And
b = y/2
y = 2b
That means line AB is passing from (0, 2b) and (2a, 0)
We know that the equation of the line passing through the points (x1, y1) and (x2, y2) is given by
(y – 2b) = ((0 – 2b)/2a) × x
a (y – 2b) = –bx
ay – 2ab = –bx
bx + ay = 2ab
On dividing both the sides with ab, we get
x/a + y/b =2
Question 18. Point R (h, k) divides a line segment between the axes in the ratio 1: 2. Find the equation of the line.
Solution:
Let us consider, PQ be the line segment such that r (h, k) divides it in the ratio 1: 2.
So the coordinates of P and Q be (0, y) and (x, 0) respectively.
We know that the coordinates of a point dividing the line segment
joins the points (x1, y1) and (x2, y2) internally in the ratio m: n is
(h, k) = ((mx2 + nx1)/(m + n), (my2 + ny1)/(m + n))
(h, k) = ((1 × 0 + 2 × x)/(1 + 2), (1 × y + 2 × 0)/(1 + 2))
h = 2x/3 and k = y/3
x = 3h/2 and y = 3k
So, P = (0, 3k) and Q = (3h/2, 0)
We know that the equation of the line passing through the points (x1, y1) and (x2, y2) is given by
y – 3k = ((0 – 3k)/(3h/2))x
3h(y – 3k) = –6kx
3hy – 9hk = –6kx
6kx + 3hy = 9hk
On dividing both the sides by 9hk, we get,
2x/3h + y/3k = 1
So, the equation of the line is given by 2x/3h + y/3k = 1
Question 19. By using the concept of equation of a line, prove that the three points (3, 0), (– 2, – 2), and (8, 2) are collinear.
Solution:
Prove that the given three points (3, 0), (– 2, – 2) and (8, 2) are collinear,
So, we have to also prove that the line passing through the points
(3, 0) and (– 2, – 2) also passes through the point (8, 2).
The equation of the line passing through the points (x1, y1) and (x2, y2) is given by
y – 0 = ((–2 – 0)/(–2 – 3)) × (x – 3)
–5y = –2 (x – 3)
–5y = –2x + 6
2x – 5y = 6
Checking whether (8,2) is on line 2x – 5y = 6 or not,
LHS = 2x – 5y = 2(8) – 5(2)
= 16 – 10
= 6
= RHS
So, the line passing through the points (3, 0) and (– 2, – 2)
also passes through the point (8, 2).
Hence, proved that the points (3, 0), (– 2, – 2) and (8, 2) are collinear,