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NCERT Class 11 Mathematics Chapter 9 Straight Liners
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Straight Liners
Chapter – 9
| Exercise 9.1 |
1. Draw a quadrilateral in the Cartesian plane, whose vertices are (-4, 5), (0, 7), (5,-5) and (-4, -2). Also, find its area.
Ans: Let ABCD be the given quadrilateral with vertices A (-4, 5), B (0, 7), C (5, -5), and D (-4, -2).
Then, by plotting A, B, C, and D on the Cartesian plane and joining AB, BC, CD, and DA, the given quadrilateral can be drawn as
To find the area of quadrilateral ABCD, we draw one diagonal, say AC. Accordingly, area (ABCD) = area (ΔABC) + area (AACD)
We know that the area of a triangle whose vertices are (x₁, y₁) (x₂, y₂) and (x₃, y₃) is
Therefore, area of ∆ABC
Area of triangle ∆ACD
Thus, area (ABCD)
2. The base of an equilateral triangle with side 2a lies along the y-axis such that the mid-point of the base is at the origin. Find vertices of the triangle.
Ans: Let ABC be the given equilateral triangle with side 2a.
Accordingly, AB = BC = CA = 2a
Assume that base BC lies along the y-axis such that the mid-point of BC is at the origin.
i.e., BO = OC = a, where O is the origin.
Now, it is clear that the coordinates of point C are (0, a), while the coordinates of point B are (0, -a).
It is known that the line joining a vertex of an equilateral triangle with the mid-point of its opposite side is perpendicular.
Hence, vertex A lies on the y-axis.
On applying Pythagoras theorem to ∆AOC we obtain
(AC)² = (OA)² + (OC)²
⇒ (2a)² = (OA)² + a²
⇒4a² – a² = (OA)²
⇒ (OA)² = 3a²
⇒ OA = √3a
∴ Coordinates of point A = (±√3a,0)
Thus, the vertices of the given equilateral triangle are (0, a) ( 0 ,- a) and (√3a,0)or (0, a) (0, -a), and (-√3a,0)
3. Find the distance between P (x₁, y₁ and Q(x₂, y₂) when:
(i) PQ is parallel to the y-axis.
Ans: When PQ is parallel to the y-axis, X₁ = X₂.
In this case, distance between P and Q =√(x₂-x₁)² + (y₂-y₁)²
= √(y₂ – y₁)²
= |y₂-y₁|
(ii) PQ is parallel to the x-axis.
Ans: When PQ is parallel to the x-axis, y₁ = y₂
In this case, distance between P and Q =
= √(x₂ – x₁)² + (y₂ – y₁)²
= √(x₂ – x₁)²
= |x₂ – x₁|
4. Find a point on the x-axis, which is equidistant from the points (7, 6) and (3, 4).
Ans: Let (a, 0) be the point on the x axis that is equidistant from the points (7, 6) and (3, 4).
On squaring both sides, we
obtain a² – 14a + 85 = a² – 6a
+25 ⇒- 14a + 6a =25 – 85
⇒ – 8a = – 60
Thus, the required point on the x-axis is
5. Find the slope of a line, which passes through the origin, and the mid-point of the line segment joining the points P(0,- 4) and B (8, 0).
Ans: The coordinates of the mid-point of the line segment joining the points
P ( 0 ,- 4) and B (8, 0) are
It is known that the slope (m) of a non-vertical line passing through the points (x₁, y₁) and (x₂, y₂) is given by
Therefore, the slope of the line passing through (0, 0) and ( 4 ,- 2) is
Hence, the required slope of the line is – ½.
6. Without using the Pythagoras theorem, show that the points (4, 4), (3, 5) and (-1,-1) are the vertices of a right angled triangle.
Ans: The vertices of the given triangle are A (4, 4), B (3, 5), and C( – 1, – 1)
It is known that the slope (m) of a non-vertical line passing through the points (x₁, y₁) and (x₂, y₂) is given by
It is observed that m₁ m₃ = – 1
This shows that line segments AB and CA are perpendicular to each other
i.e., the given triangle is right-angled at A (4, 4).
Thus, the points (4, 4), (3, 5), and ( – 1,- 1) are the vertices of a right-angled triangle.
7. Find the slope of the line, which makes an angle of 30° with the positive direction of y-axis measured anticlockwise.
Ans: If a line makes an angle of 30° with the positive direction of the y-axis measured anticlockwise, then the angle made
by the line with the positive direction of the x-axis measured anticlockwise is 90°+ 30°= 120°
Thus, the slope of the given line is tan 120° =tan( 180° – 60° )= – tan 60°
8. Without using distance formula, show that points (-2,-1), (4, 0), (3, 3) and (-3, 2) are the vertices of a parallelogram.
Ans: Let points (- 2, -1), (4, 0), (3, 3), and (-3, 2) be respectively denoted by A, B, C, and D.
⇒ Slope of AB = Slope of CD
⇒ AB and CD are parallel to each other.
⇒ Slope of BC = Slope of AD
⇒ BC and AD are parallel to each other.
Therefore, both pairs of opposite sides of quadrilateral ABCD are parallel. Hence, ABCD is a parallelogram.
Thus, points ( – 2, – 1) (4, 0), (3, 3), and ( – 3,2) are the vertices of a parallelogram.
9. Find the angle between the x-axis and the line joining the points (3,-1) and (4,-2).
Ans: The slope of the line joining the points ( 3 , — 1) and ( 4 , — 2) is
Now, the inclination (θ) of the line joining the points ( 3, – 1) and ( 4 , – 2) is given by
tan θ = – 1 ⇒ θ =(90° + 45°)=135°
Thus, the angle between the x-axis and the line joining the points ( 3 , – 1) and ( 4 , – 2) is 135°.
10. The slope of a line is double of the slope of another line. If tangent of the angle between them is 1/3, find the slopes of the lines.
Ans: Let m₁ and m be the slopes of the two given lines such that m₁ = 2m
We know that if θisthe angle between the lines/₁ and /₂ with slopes m₁and m₂ then
It is given that the tangent of the angle between the two lines is ⅓
Case I
If m = – 1 , then the slopes of the lines are – 1 and – 2 .
If m = – 1/2 then the slopes of the lines are – 1/2 and – 1.
Case II
⇒ 2m² + 1 = 3m
⇒ 2m² – 3m + 1 = 0
⇒ 2m² – 2m – m + 1 = 0
⇒ 2m(m – 1) – 1(m – 1) = 0
⇒ (m – 1)(2m – 1) = 0
⇒ m = 1 or m = 1/2
If m = 1 , then the slopes of the lines are 1 and 2.
If m = 1/2 , then the slopes of the lines are 1/2 and 1
Hen – 1/2 ce, the slopes of the lines are -1 and – 2 or
and – 1 or 1 and 2 or 1/2 and 1.
11. A line passes through (x₁, y₁) and (h, k) If slope of the line is m, show that k – y₁ = m(h – x₁).
Ans: The slope of the line passing through (x¹, y₁) and (h, k) is
It is given that the slope of the line is m.
⇒ k – y₁ = m(h – x₁)
Hence, k – y₁ = m(h – x₁)
| Exercise 9.2 |
In Exercises 1 to 8, find the equation of the line which satisfy the given conditions:
1. Write the equations for the x-and y-axes.
Ans: 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 x = 0 .
2. Passing through the point (-4, 3) with slope ½.
Ans: 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) = 1/2 (x + 4)
2(y – 3) = x + 4
2y – 6 = x + 4
i.e., x – 2y + 10 = 0
3. Passing through (0, 0) with slope m.
Ans: 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)
i.e., y = mx
4. Passing through (2, 2√3) and inclined with the x-axis at an angle of 75°.
Ans: The slope of the line that inclines with the x-axis at an angle of 75° is
m = tan 75°
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(√3 – 1) – 2√3 (√3 – 1) = x(√3 + 1) – 2(√3 + 1)
(√3 + 1) x – (√3 – 1) y = 2√3 + 2 – 6 + 2√3
(√3 + 1) x – (√3 – 1) y = 4√3 – 4
i.e.., (√3 + 1) x – (√3 – 1) y = 4(√3 – 1)
5. Intersecting the x-axis at a distance of 3 units to the left of origin with slope -2.
Ans: 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 = – 2x – 6
i.e., 2x + y + 6 = 0
6. Intersecting the y-axis at a distance of 2 units above the origin and making an angle of 30° with positive direction of the x-axis.
Ans: 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
√3 y = x + 2√3
i.e., x – √3y + 2√3 = 0.
7. Passing through the points (-1, 1) and (2,-4).
Ans: It is known that the equation of the line passing through points (x₁, y₁) and (x₂, y₂) is
Therefore, the equation of the line passing through the points ( – 1,1) and ( 2 , – 4) is
3(y – 1) = – 5(x + 1)
3y – 3 = – 5x – 5
i.e., 5x + 3y + 2 = 0
8. The vertices of ∆PQR are P (2, 1), Q(- 2, 3) and R (4, 5). Find equation of the median through the vertex R.
Ans: It is given that the vertices of the triangle ∆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
It is known that the equation of the line passing through points (x₁, y₁) and (x₂, y₂) is
Therefore, the equation of RL can be determined by substituting (x₁, y₁) = (4, 5) and (x₂, y₂) = (0, 2) .
⇒ 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.
9. Find the equation of the line passing through (-3, 5) and perpendicular to the line through the points (2, 5) and (-3, 6).
Ans: The slope of the line joining the points (2, 5) and ( – 3,6) is
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)
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
i.e., 5x – y + 20 = 0
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.
Ans: 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
The slope of the line joining the points (1, 0) and (2, 3) is
We know that two non – vertical lines are perpendicular to each other if and only if their slopes are negative reciprocals of each other.
⇒ 3[(n + 1) y – 3] = – [x(n + 1) – (n + 2)]
⇒ 3(n + 1) y – 9 = – (n + 1) x + n + 2
⇒ (1 + n) x + 3(1 + n) y = n + 11
11. Find the equation of a line that cuts off equal intercepts on the coordinate axes and passes through the point (2, 3).
Ans: The equation of a line in the intercept form is
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.
Accordingly, equation (i) reduces to
Since the given line passes through point (2, 3), equation (ii) reduces to
2 + 3 =a ⇒ a=5
On substituting the value of a in equation (ii), we obtain
x + y = 5 which is the required equation of the line
12. Find equation of the line passing through the point (2, 2) and cutting off intercepts on the axes whose sum is 9.
Ans: The equation of a line in the intercept form is
Here, a and b are the intercepts on x and y axes respectively.
It is given that + b =9 ⇒ b=9 – a…(ii)
From equations (i) and (ii), we obtain
It is given that the line passes through point (2, 2). Therefore, equation (iii) reduces to
⇒ 18 = 9a – a²
⇒ a² – 9a + 18 = 0
⇒ a² – 6a – 3a + 18 = 0
⇒ a(a – 6) – 3(a – 6) = 0
⇒ (a – 6)(a – 3) = 0
⇒ a = 6 or a = 3
If a = 6 and b = 9 – 6 = 3, then the equation of the line is
If a = 3 and b = 9 – 3 = 6, then the equation of the line is
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.
Ans:
Now, the equation of the line passing through point (0, 2) and having a slope -√3 is (y-2)= -√3(x-0)
y – 2 = – √3
i.e., √3x + y – 2 = 0
The slope of line parallel to line √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 -√3 is
y – (- 2) = – √3(x – 0)
y + 2 = – √3x
√3x + y + 2 = 0
14. The perpendicular from the origin to a line meets it at the point (-2, 9), find the equation of the line.
Ans: The slope of the line joining the origin (0, 0) and point ( – 2,9) is
Accordingly, the slope of the line perpendicular to the line joining the origin and point (- 2,9) is
Now, the equation of the line passing through point ( – 2,9) and having a slope m₂ is
(y – 9) = 2/9 (x + 2)
9y – 81 = 2x + 4
i.e., 2x – 9y + 85 = 0
15. The length L (in centimetre) 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.
Ans: It is given that when C = 20 the value of L is 124.942, whereas when C = 110 the value of L is 125.134.
Accordingly, points (20, 124.942) and (110, 125.134) satisfy the linear relation between L and C.
Now, 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)
16. 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?
Ans: 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).
i.e., y = 120(x – 14) + 980
When x = Rs 17/litre,
y = 120(17 – 14) + 980
⇒ y = 120 × 3 + 980 = 360 + 980 = 1340
Thus, the owner of the milk store could sell 1340 litres of milk weekly at Rs 17/litre.
17. P (a, b) is the mid – point of a line segment between axes. Show that equation of the line is
Ans: 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.
Since P(a,b) is the mid – point of AB,
∴ 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) is
a(y – 2b) = – bx
ay – 2ab = – bx
i.e., bx + ay = 2ab
On dividing both sides by ab, we obtain
18. Point R (h, k) divides a line segment between the axes in the ratio 1: 2. Find equation of the line.
Ans: Let AB be the line segment between the axes such that point R (h, k) divides AB in the ratio 1: 2.
Let the respective coordinates of A and B be (x, 0) and (0, y) .
Since point R (h, k) divides AB in the ratio 1: 2, according to the section formula,
hy = – 2kx + 3hk
i.e., 2kx + hy = 3hk
Thus, the required equation of the line is2kx + hy = 3hk
19. By using the concept of equation of a line, prove that the three points (3, 0), (-2,-2) and (8, 2) are collinear.
Ans: 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
5y = 2x – 6
i.e., 2x – 5y = 6
It is observed that at x = 8 and y = 2,
L.H.S. = 2 x 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, points (3, 0), ( – 2, – 2) and (8, 2) are collinear.
| Exercise 9.3 |
1. Reduce the following equations into slope – intercept form and find their slopes and the y intercepts.
(i) x + 7y = 0.
Ans: 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 slope-intercept form, where the slope and the y-intercept are –1/7 and 0 respectively.
(ii) 6x + 3y – 5 = 0.
Ans: The given equation is 6x + 3y – 5 = 0.
It can be written as
Therefore, equation (2) is in the slope-intercept form, where the slope and the y-intercept are – 2 and 5/3 respectively.
(iii) y = 0
Ans: 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 slope-intercept form, where the slope and the y-intercept are 0 and 0 respectively.
2. Reduce the following equations into intercept form and find their intercepts on the axes.
(i) 3x + 2y – 12 = 0.
Ans: The given equation is 3x + 2y — 12 = 0.
It can be written as
3x + 2y = 12
Therefore, equation (1) is in the intercept form, where the intercepts on the x and y axes are 4 and 6 respectively.
(ii) 4x – 3y = 6
Ans: The given equation is 4x – 3y = 6
It can be written as
and b = – 2 3/2
Therefore, equation (2) is in the intercept form, where the intercepts on the x and y axes are 3/2 and – 2 respectively.
(iii) 3y + 2 = 0.
Ans: The given equation is 3y + 2 = 0.
It can be written as
3y = – 2
Therefore, equation (3) is in the intercept form, where the intercept on the y-axis is – 2/3 and it has no intercept on the x-axis.
3. Find the distance of the point (-1, 1) from the line 12(x + 6) = 5(y – 2).
Ans: 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 (x₁, y₁) is given by
The given point is (x₁, y₁) =( – 1,1)
Therefore, the distance of point ( – 1,1) from the given line
Ans: The given equation of line is
or, 4x + 3y – 12 = 0 …(1)
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 x-axis 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 (x₁, y₁) is given by
⇒ |4a – 12| = 20
⇒ ± (4a – 12) = 20
⇒ (4a – 12) = 20 or – (4a – 12) = 20
⇒ 4a = 20 + 12 or 4a = – 20 + 12
⇒a = 8 or -2
Thus, the required points on the x-axis are (- 2,0) and (8, 0).
5. Find the distance between parallel lines
(i) 15x + 8y – 34 = 0 and 15x + 8y + 31 = 0.
Ans: The given parallel lines are 15x + 8y – 34 = 0 and 15x + 8y + 31 = 0.
Here, A = 15, B = 8, C₁ = – 34 and C₂ = 31
Therefore, the distance between the parallel lines is
(ii) l(x + y) + p = 0 and l(x + y) – r = 0.
Ans: The given parallel lines are / (x + y) + p = 0 and / (x + y) – r = 0 lx
+ ly + p = 0 and lx + ly – r = 0
Here, A = I, B = l, C₁ = p, and C₂ = – r
Therefore, the distance between the parallel lines is
6. Find equation of the line parallel to the line 3x – 4y + 2 = 0 and passing through the point (-2, 3).
Ans: The equation of the given line is
3x – 4y + 2 = 0
∴ Slope of the other line =
Now, the equation of the line that has a slope of 3/4 and passes through the point ( – 2,3) is
4y – 12 = 3x + 6
i.e., 3x – 4y + 18 = 0
7. Find the equation of the line perpendicular to the line x – 7y + 5 = 0 and having x intercept 3.
Ans: The given equation of line is x – 7y + 5 = 0
The equation of the line with slope – 7 and x-intercept 3 is given
by y=m(x – d) ⇒ y = – 7(x – 3) ⇒ y = – 7x +21 ⇒7x+y= 21
8. Find angles between the lines √3 x + y = 1 and x + √3y = 1.
Ans:
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° – 30⁰ = 150°.
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.
Ans: The slope of the line passing through points (h, 3) and (4, 1) is
It is given that the two lines are perpendicular.
∴ m₁ × m₂ = – 1
10. Prove that the line through the point (x₁, y₁) and parallel to the line Ax + By + C = 0 is A(x – x₁) + B(y – y₁) = 0.
Ans:
y – y₁ = m(x – x₁)
B(y – y₁) = – A (x – x₁)
A(x – x₁) + B(y – y₁) = 0
Hence, the line through point (x₁, y₁) and parallel to line A x + By + C = 0 is
A(x – x₁) + B (y – y₁)=0
11. Two lines passing through the point (2, 3) intersects each other at an angle of 60°. If the slope of one line is 2, find the equation of the other line.
Ans: It is given that the slope of the first line, m₁ = 2 .
Let the slope of the other line be m₂
The angle between the two lines is 60°.
(2√3 – 1)y – 3(2√3 – 1) = – (2 + √3) x + 2(2 +√3)
(2√3 – 1) y + (2 + √3) x = 4 + 2√3 + 6√3 – 3
(2 + √3) x + (2√3 – 1) y = 1 + 8√3
In this case, the equation of the other line is
(2 + √3) x + (2√3 – 1) y = 1 + 8√3
Thus, the required equation of the other line is
(√3 – 2) x + (2√3 + 1) y = – 1 + 8√3
or (2 + √3) x + (2√3 – 1) y = 1 + 8√3
12. Find the equation of the right bisector of the line segment joining the points (3, 4) and (-1, 2).
Ans: The right bisector of a line segment bisects the line segment at 90°.
The end-points of the line segment are given as A (3, 4) and B ( – 1,2)
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.
13. Find the coordinates of the foot of perpendicular from the point (-1, 3) to the line 3x – 4y – 16 = 0.
Ans: Let (a, b) be the coordinates of the foot of the perpendicular from the point (- 1, 3) to the line 3x – 4y – 16 = 0.
Since these two lines are perpendicular, m₁ m₂ = – 1
⇒ 3b – 9 = – 4a – 4
⇒ 4a + 3b = 5 …(1)
Point (a, b) lies on line 3x – 4y = 16
∴ 3a – 4b = 16 … (2)
On solving equations (1) and (2), we obtain
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.
Ans: 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.
∴ m × – 2 = – 1 [The two lines are perpendicular]
⇒ m = 1/2
Since point ( – 1,2) lies on the given line, it satisfies the equation y = mx + c .
15. If p and q are the lengths of perpendiculars from the origin to the lines x cos θ – y sin θ = k cos 2 θ and x sec θ + y cosec θ = k respectively, prove that p² + 4q² = k².
Ans: The equations of given lines are x
cos θ – y sin θ = k cos 2 θ…(1)x
secθ + y cosec θ =k…(2)
The perpendicular distance (d) of a line Ax + By + C = 0 from a point (x₁, y₁) is given by
On comparing equation (1) to the general equation of line i.e., Ax + By + C = 0 we obtain A = cos θ B = – sin θ, 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
= k² cos² 2θ + 4k² sin² θ cos² θ
= k² cos² 2θ + k² (2sin θ cos θ)²
= k² cos² 2θ + k² sin² 2θ
= k² (cos² 2θ + sin² 2θ)
= k²
Hence, we proved that p² + 4q² = k².
16. In the triangle ABC with vertices A (2, 3), (4, – 1) and C(1,2) find the equation and length of altitude from the vertex A.
Ans:
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
⇒ (y + 1) = – 1 (x – 4)
⇒ y + 1 = – x + 4
⇒ x + y – 3 = 0 ..(1)
The perpendicular distance (d) of a line Ax + By + C = 0 from a point (x₁, y₁) 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 √2 units respectively.
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
Ans: It is known that the equation of a line whose intercepts on the axes are a and b is
or bx + ay = ab
or bx + ay – ab = 0 …(1)
The perpendicular distance (d) of a line Ax + By + C = 0 from a point (x₁, y₁) is given by
On comparing equation (1) to the general equation of line A x + By + C = 0 we obtain A = b B = a and C = – ab.
Therefore, if p is the length of the perpendicular from point (x₁, y₁) = (0, 0) to line (1), we obtain
| Miscellaneous Exercise on Chapter 9 |
1. Find the values of k for which the line (k-3) x-(4-k²) y+k²-7k+ 6 = 0 is
(a) Parallel to the x-axis,
Ans: If the given line is parallel to the x-axis, then
Slope of the given line = Slope of the x-axis
The given line can be written as
(4 – 2 )y=(k – 3)x+k² – 7k + 6 = 0
Slope of the x-axis = 0
, which is of the form y = mx + c
⇒ k – 3 = 0
⇒ k = 3
Thus, if the given line is parallel to the x-axis, then the value of k is 3.
(b) Parallel to the y-axis,
Ans: If the given line is parallel to the y-axis, it is vertical. Hence, its slope will be undefined.
The slope of the given line is
Now
is undefined at k² = 4
k² =4 ⇒
k = ± 2
Thus, if the given line is parallel to the y-axis, then the value of k is ±2.
(c) Passing through the origin.
Ans: If the given line is passing through the origin, then point (0, 0) satisfies the given equation of line.
(k – 3)(0) – (4 – k²)(0) + k² – 7k + 6 = 0
k² – 7k + 6 = 0
k² – 6k – k + 6 = 0
(k – 6)(k – 1) = 0
k = 1 or 6
Thus, if the given line is passing through the origin, then the value of k is either 1 or 6.
2. Find the equations of the lines, which cut-off intercepts on the axes whose sum and product are 1 and – 6, respectively.
Ans: Let the intercepts cut by the given lines on the axes be a and b.
It is given that a + b = 1 … (1) ab = – 6 … (2)
On solving equations (1) and (2), we obtain
a = 3 and b =- 2 or a = – 2 and b = 3
It is known that the equation of the line whose intercepts on the axes are a and b is
Case I: a = 3 and b = – 2
In this case, the equation of the line is – 2x + 3y + 6 = 0 i.e., 2x – 3y = 6 .
Case II: a= – 2 and b = 3
In this case, the equation of the line is 3x – 2y + 6 = 0 i.e., – 3x + 2y = 6
Thus, the required equation of the lines are 2x – 3y = 6 and – 3x + 2y = 6
3. What are the points on the y-axis whose distance from the line
is 4 units.
Ans: Let (0, b) be the point on the y-axis whose distance from line
is 4 units.
The given line can be written as 4x + 3y – 12 = 0 … (1)
On comparing equation (1) to the general equation of line Ax + By + C = 0 we obtain A = 4, B = 3, and C = – 12.
It is known that the perpendicular distance (d) of a line Ax + By + C = 0 from a point (x₁, y₁) is given by
Therefore, if (0, b) is the point on the y-axis whose distance from line
is 4 units,then:
⇒ 20 = |3b – 12|
⇒ 20 = ± (3b – 12)
⇒ 20 = (3b – 12) or 20 = – (3b – 12)
⇒ 3b = 20 + 12 or 3b = – 20 + 12
Thus, the required points are
4. Find perpendicular distance from the origin to the line joining the points (cos θ, sin θ) and (cos Φ, sin Φ).
Ans: The equation of the line joining the points (cos θ, sin θ) and(cos phiΦ, sin Φ) is given by
y(cos Φ – cos θ) – sin θ (cos Φi – cos θ) = x(sin Φ – sin θ) – cos θ (sin Φ – sin θ)
x(sin θ – sin Φ) + y(cos Φ – cos θ) + cos θ sin Φ – cos θ sin θ – sin θ cos Φ + sin θ cos θ = 0
x(sin θ – sin Φ) + y(cos Φ – cos θ) + sin(Φ – θ) = 0
Ax + By + C = 0 where A = sin θ – sin Φ , B = cos Φ – cos θ and C = sin(Φ – θ)
It is known that the perpendicular distance (d) of a line Ax + By + C = 0 from a point (x₁, y₁) is given by
Therefore, the perpendicular distance (d) of the given line from point (x₁, y₁) = (0, 0) is
5. Find the equation of the line parallel to y-axis and drawn through the point of intersection of the lines x – 7y + 5 = 0 and 3x + y = 0.
Ans: The equation of any line parallel to the y-axis is of the form
x=a… (1)
The two given lines are
x – 7y + 5 = 0 … (2)
3x + y = 0 (3)
On solving equations (2) and (3), we obtain
Therefore,
is the point of intersection of lines (2) and (3).
Since line x = a passes through
Thus, the required equation of the
line is
6. Find the equation of a line drawn perpendicular to the line x/4 + y/6 = 1 through the point, where it meets the y-axis.
Ans: The equation of the given line is x/4 + y/6 = 1
This equation can also be written as 3x + 2ya – 12 = 0
which is of the form y = mx + c
∴ Slope of the given line
∴ Slope of line perpendicular to the given line
Let the given line intersect the y-axis at (0, y)
On substituting x with 0 in the equation of the given line, we obtain
∴ The given line intersects the y-axis at (0, 6).
The equation of the line that has a slope of 2/3 and passes through point (0, 6) is
(y – 6) = 2/3 (x + 0)
3y – 18 = 2x
2x – 3y + 18 = 0
Thus, the required equation of the line is 2x – 3y + 18 = 0.
7. Find the area of the triangle formed by the lines y – x = 0 x + y = 0 and x – k = 0.
Ans: The equations of the given lines are
y – x=0… (1) x + y =0…(2)x
– k = 0 … (3)
The point of intersection of lines (1) and (2) is given by x = 0 and y = 0
The point of intersection of lines (2) and (3) is given by x = k and y = – k
The point of intersection of lines (3) and (1) is given by x = k and y = k
Thus, the vertices of the triangle formed by the three given lines are (0, 0), (k, – k ), and (k, k)
We know that the area of a triangle whose vertices are (x₁, y₁) , (x₂, y₂) and (x₃, y₃) is
Therefore, area of the triangle formed by the three given lines
8. Find the value of p so that the three lines 3x + y – 2 = 0 , px + 2y – 3 = 0 and 2x – y – 3 = 0 may intersect at one point.
Ans: The equations of the given lines are
3x + y – 2 =0…(1)
px + 2y – 3 =0…(2)
2x-y-3= 0 … (3)
On solving equations (1) and (3), we obtain
x = 1 and y = – 1
Since these three lines may intersect at one point, the point of intersection of lines (1) and (3) will also satisfy line (2).
p(1) + 2(- 1) – 3 = 0p – 2 – 3 = 0p = 5
Thus, the required value of p is 5.
9. If three lines whose y = m₁ x + c₁ , y = m₂x + c₂ and y = m₃x + c₃ are equations are concurrent, then show that m₁(c₂ – c₃) + m₂(c₃- c₁) + m₃ (c₁- c₂) = 0.
Ans: The equations of the given lines are
y = m₁ x + c₁… (1) y = m₂x + c₂…
(2) y = m₃x + C₃….(3)
On subtracting equation (1) from (2), we obtain
0 = (m₂ – m₁) x + (c₂ – c₁)
⇒ (m₁ – m₂) x = c₂ – c₁
On substituting this value of x in (1), we obtain
is the point of intersection of lines (1) and (2).
It is given that lines (1), (2), and (3) are concurrent. Hence, the point of intersection of lines (1) and (2) will also satisfy equation (3).
m₁c₂ – m₂c₁ – m₃c₂ + m₃c₁ – c₃m₁ + c₃m₂= 0
m₁(c₂ – c₃) + m₂(c₃- c₁) + m₃ (c₁ – c₂) = 0
Hence, m₁ (c₂ – c₃) + m₂ (c₃ – c₁) + m₃(c₁ – c₂) = 0.
10. Find the equation of the lines through the point (3, 2) which make an angle of 45° with the line x – 2y = 3.
Ans: Let the slope of the required line be m₁.
The given line can be
written as ,which is of the form y = mx + c
∴ Slope of the given line
It is given that the angle between the required line and line x – 2y = 3 is 45°.
We know that if θisthe acute angle between lines /₁ and /₂ with slopes m₁ and m₂ respectively, then
Case I: m₁ = 3
The equation of the line passing through (3, 2) and having a slope of 3 is:
y – 2 = 3(x – 3)
y – 2 = 3x – 9
3x – y = 7
Case II: m₁=- 1/3
The equation of the line passing through (3, 2) and having a slope of – 1/3 is:
y – 2 = – 1/3 (x – 3)
3y – 6 = – x + 3
x + 3y = 9
Thus, the equations of the lines are 3x – y = 7 and x + 3y = 9.
11. Find the equation of the line passing through the point of intersection of the lines
4x + 7y – 3 = 0 and 2x – 3y + 1 = 0 that has equal intercepts on the axes.
Ans: Let the equation of the line having equal intercepts on the axes be
On solving equations 4x + 7y – 3 = 0 and 2x – 3y + 1 = 0 we obtain
is the point of intersection of the two given lines.
Since equation (1) passes through point
∴ Equation (1)
becomes
Thus, the required 13x + 13y = 6 equation of the line is
12. Show that the equation of the line passing through the origin and making an angle θ with the line
Ans: Let the equation of the line passing through the origin be y = m₁ x .
If this line makes an angle of θ with line y = mx + c then angle θ is given by
13. In what ratio, the line joining (-1, 1) and (5, 7) is divided by the line x + y = 4?
Ans: The equation of the line joining the points ( – 1,1) and (5, 7) is given by
x – y + 2 = 0 …. (1)
The equation of the given line is
x + y – 4 =0…(2)
The point of intersection of lines (1) and (2) is given by
x = 1 and y = 3
Let point (1, 3) divide the line segment joining ( – 1,1) and (5, 7) in the ratio 1:k. Accordingly, by section formula,
⇒ – k + 5 = 1 + k
⇒ 2k = 4
⇒ k = 2
Thus, the line joining the points ( – 1,1) and (5, 7) is divided by line x + y = 4 in the ratio 1:2.
14. Find the distance of the line 4x + 7y + 5 = 0 from the point (1, 2) along the line 2x – y = 0.
Ans: The given lines are
2x – y = 0…(1)
4x + 7y + 5 =0…(2)
A (1, 2) is a point on line (1).
Let B be the point of intersection of lines (1) and (2).
On solving equations (1)
and (2), we obtain the
∴ Coordinates of point B
are
By using distance formula, the distance between points A and B can be obtained as
Thus, the required distance is
units
15. Find the direction in which a straight line must be drawn through the point (-1, 2) so that its point of intersection with the line x + y = 4 may be at a distance of 3 units from this point.
Ans: Let y = mx + c be the line through point ( – 1 ,2).
Accordingly, 2 = m(- 1) + c
⇒ 2 = – m +c ⇒ c=
m+2 :. y = mx + m + 2
… (1) The given line is
x + y =4…(2)
On solving equations (1) and (2), we obtain
is the point of intersection of lines (1) and (2).
Since this point is at a distance of 3 units from point ( – 1 ,2) according to distance formula
⇒ 1 + m² = m² + 1 + 2m
⇒ 2m = 0
⇒ m = 0
Thus, the slope of the required line must be zero i.e., the line must be parallel to the x-axis.
16. The hypotenuse of a right angled triangle has its ends at the points (1, 3) and (-4, 1). Find an equation of the legs (perpendicular sides) of the triangle which are parallel to the axes.
Ans: Let A(1,3) and B(-4,1) be the coordinates of the end points of the hypotenuse.
Now, plotting the line segment joining the points A(1,3) and B(-4,1) on the coordinate plane, we will get two right triangles with AB as the hypotenuse. Now from the diagram, it is clear that the point of intersection of the other two legs of the right triangle having AB as the hypotenuse can be either P or Q.
CASE 1: When / ” APB is taken.
The perpendicular sides in ∠”APB are AP and PB.
Now, side PB is parallel to x-axis and at a distance of 1 units above x-axis.
So, equation of PB is, y = 1 or y – 1 = 0
The side AP is parallel to y – axis and at a distance of 1 units on the right of y-axis.
So, equation of AP is x = 1 or x – 1 = 0
CASE 2: When ∠” AQB is taken.
The perpendicular sides in ∠” AQB are AQ and QB.
Now, side AQ is parallel to x-axis and at a distance of 3 units above x-axis.
So, equation of AQ is, y = 3 or y – 3 = 0
The side QB is parallel to y-axis and at a distance of 4 units on the left of y-axis.
So, equation of QB is x = – 4 or x + 4 = 0
Hence, the equation of the legs are:
x = 1 y = 1 or x = – 4 y = 3
17. Find the image of the point (3, 8) with respect to the line x + 3y = 7 assuming the line to be a plane mirror.
Ans: The equation of the given line is
x + 3y =7…(1)
Let point B (a, b) be the image of point A (3, 8).
Accordingly, line (1) is the perpendicular bisector of AB.
Since line (1) is perpendicular to AB,
⇒ b – 8 = 3a – 9
⇒ 3a – b = 1 …(2)
The mid – point of line segment AB will also satisfy line (1).
Hence, from equation (1), we have
⇒ a + 3+ 3b + 24 = 14
⇒a + 3b = -13 …..(3)
On solving equations (2) and (3). we obtain a = – 1 and b = – 4
Thus, the image of the given point with respect to the given line is (- 1, – 4).
18. If the lines y = 3x + 1 and 2y = x + 3 are equally inclined to the line y = mx + 4 find the value of m.
Ans: The equations of the given lines are
y=3x+1… (1) 2y =x+3…(2)
y = mx + 4 ….. (3)
Slope of line (1), m₁ = 3
Slope of line (2),
Slope of line (3), m₃ = m
It is given that lines (1) and (2) are equally inclined to line (3). This means that
the angle between lines (1) and (3) equals the angle between lines (2) and (3).
(3 – m)(m + 2) = (1 – 2m)(1 + 3m)
⇒ – m² + m + 6 = 1 + m – 6m²
⇒ 5m² + 5 = 0
⇒ (m² + 1) = 0
⇒m = √- 1 which is not real
Hence, this case is not possible.
Thus, the required value of m is
19. If sum of the perpendicular distances of a variable point P(x,y) from the lines x + y – 5 = 0 and 3x – 2y + 7 = 0 is always 10. Show that P must move on a line.
Ans: The equations of the given lines are
x+y – 5=0… (1)
3x – 2y + 7 = 0 … (2)
The perpendicular distances of P (x, y) from lines (1) and (2) are respectively given by
It is given that d₁ + d₂ = 10
⇒ √13|x + y – 5| + √2|3x – 2y + 7| – 10√26= 0
⇒ √13(x + y – 5) + √2 (3x – 2y + 7) – 10√26= 0
[Assuming (x + y – 5) and (3x – 2y + 7) are positive]
⇒ √13 x + √13y – 5√13 + 3√2x – 2√2 y + 7√2 – 10√26 = 0
⇒ x(√13 + 3√2) + y(√13 – 2√2) +( 7√2 – 5√13 – 10√26) = 0 which is the equation of a line.
Similarly, we can obtain the equation of line for any signs of (x + y – 5) and (3x – 2y + 7)
Thus, point P must move on a line.
20. Find the equation of the line which is equidistant from parallel lines 9x + 6y – 7 = 0 and 3x + 2y + 6 = 0.
Ans: The equations of the given lines are
9x+6y – 7 = 0 … (1)
3x + 2y + 6 = 0 … (2)
Let P (h, k) be the arbitrary point that is equidistant from lines (1) and (2). The perpendicular distance of P (h, k) from line (1) is given by
The perpendicular distance of P (h, k) from line (2) is given by
Since P(h,k) is equidistant from lines (1) and (2), d₁ = d₂
⇒ |9h + 6k – 7| = 3|3h + 2k + 6|
⇒ |9h + 6k – 7| = ± 3 (3h + 2k + 6)
⇒ 9h + 6k – 7 = 3(3h + 2k + 6) or 9h + 6k – 7 = – 3(3h + 2k + 6)
The case 9h + 6k – 7 = 3(3h + 2k + 6) is not possible as
9h+6k-7=3 (3h + 2k + 6) ⇒-7=18 (which is absurd)
9h + 6k – 7 = – 3(3h + 2k + 6)
∴ 9h + 6k – 7 – 9h – 6k – 18
⇒ 18h + 12k + 11 = 0
Thus, the required equation of the line is 18x + 12y + 11 = 0
21. A ray of light passing through the point (1, 2) reflects on the x-axis at point A and the reflected ray passes through the point (5, 3). Find the coordinates of A.
Ans:
Let the coordinates of point A be (a, 0).
Draw a line (AL) perpendicular to the x-axis.
We know that angle of incidence is equal to angle of reflection. Hence, let
∠BAL = ∠CAL = ⍉
Let ∠CAX = θ
∠OAB = 180° – (θ + 2Φ) = 180° – [θ + 2(90° – θ)]
= 180° – Ө – 180° +20
= θ
∴∠BAX = 180° – θ
From equations (1) and (2), we obtain
Thus, the coordinates of point A are
22. Prove that the product of the lengths of the perpendiculars drawn from the points (√a² – b²,0) and (-√a² – b², 0) to to the line
Ans: The equation of the given line is
Or. bx cos θ + ay sin θ – ab = 0 …(1)
Length of the perpendicular from point (√a² – b² ,0) to line (1) is
Length of the perpendicular from point (- √a² – b²), 0) to line (2) is
On multiplying equations (2) and (3), we obtain
Hence proved
23. A person standing at the junction (crossing) of two straight paths represented by the equations 2x – 3y + 4 = 0 and 3x + 4y – 5 = 0 wants to reach the path whose equation is 6x – 7y + 8 = 0 in the least time. Find equation of the path that he should follow.
Ans: The equations of the given lines are
2x – 3y + 4 =0…(1)
3x + 4y – 5 = 0…(2)
6x – 7y + 8 =0…(3)
The person is standing at the junction of the paths represented by lines (1) and (2).
On solving equations (1) and (2),
we obtain.
Thus, the person is standing at point
The person can reach path (3) in the least time if he walks along the perpendicular line to (3) from point
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