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Given: equation is 3x − 2y + 6 = 0 

Concept Used: 

Line in intercepts form is \(\frac{x}{a}+\frac{y}{b}=1\) ( a and b are x and y intercepts resp.)

Explanation:

3x − 2y = - 6

 \(\frac{3}{-6}x+\frac{2y}{6}=1\) [Dividing both sides by - 6]

\(\frac{x}{-2}+\frac{y}{3}=1\)

Thus, the intercept form of the given line 

∴ x - intercept = −2 and y - intercept = 3

The given lines can be written as 

5x + 3y − 7 = 0 … (1) 

5x + 3y + \(\frac{14}{3}\) = 0 ……(2) 

Let d be the distance between the lines 5x + 3y − 7 = 0 and 15x + 9y + 14 = 0 

Then, d = \(\Big|\frac{(-7-\frac{14}{2})}{\sqrt{5^2+3^2}}\Big|\) 

⇒ d = \(\frac{35}{3\sqrt{34}}\)

The area of a triangle with vertex (x₁, y₁), (x₂, y₂) and (x₃, y₃) is

∆ = \(\frac{1}{2}\)|x₁(y₂ − y₃) + x₂(y₃ − y₁) + x₃(y₁ − y₂)|

Here, x₁ = 2, x₂ = 3, x₃ = x

y₁ = 1, y₂ = – 2, y₃ = x + 3.

∆ = \(\frac{1}{2}\)|2(−2 − x − 3) + 3(x + 3 − 1) + x(1 + 2)|

⇒ 5 = \(\frac{1}{2}\)|2(−5 − x) + 3 (x + 2) + 3x|

⇒ 10 = |−10 − 2x + 3x + 6 + 3x|

⇒ |4x − 4| = 10

⇒ 4x − 4 = 10 or 4x – 4 = – 10

⇒ x = \(\frac{7}{2}\) or = − \(\frac{3}{2}\)

⇒ y = \(\frac{13}{2}\) ⇒ y = \(\frac{3}{2}\)

∴ (x, y.) = (\(\frac{7}{2}\), \(\frac{13}{2}\)) or (−\(\frac{3}{2}\), \(\frac{3}{2}\)).

x – 2y + 1 = 0 ……. (1) 

2x – 5y + 3 = 0 ….. (2) 

5x – 9y + k = 0 …….. (3) 

Solve (1) and (2) x/(-6 + 5) = - y/(3 - 2) = 1

x/-1 = - y/-1 = 1/-1

⇒ x = 1, y = 1 

point of intersection of (1) and (2) is (1,1), 

⇒ (1, 1) lies on (3) = 5 – 9 + k = 0 = k = 4

Let the intercepts cut by the given lines on the axes be a and b. It is given that 

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.

Equation of a line in intercept form is 

\(\frac{x}{a}+\frac{y}{b}=1\) 

Given, a + b = 1 

and a × b = – 6 

⇒ a −\(\frac{6}{a}\) = 1 

⇒ a² − a − 6 = 0 

⇒ (a − 3) (a + 2) = 0 

⇒ a = 3, – 2. 

If a = 3, b = – 2 and if a = – 2, b = 3. 

∴ Equation of the line is 

\(\frac{x}{3}-\frac{y}{2}=1\) or  \(\frac{x}{-2}-\frac{y}{3}=1\)  

⇒ 2x – 3y – 6 = 0 or 3x – 2y + 6 = 0, 

is the required equation of the line

Equation of the line in two point form is \(\frac {y-y_1}{y_2-y_1} = \frac {x-x_1}{x_2-x_1}\)

Equation of side AB is \(\frac {y-4}{3-4} = \frac {x-1}{2-1}\)

y – 4 = -1(x – 1) 

y – 4 = -x + 1 

x + y = 5

Equation of side BC is \(\frac {y-3}{6-3} = \frac {x-2}{1-2}\)

-1(y – 3) = 3(x – 2) 

-y + 3 = 3x – 6 

∴ 3x + y = 9

Since both the points A and C have same x co-ordinates i.e. 1 

the points A and C lie on a line parallel to Y-axis. 

∴ The equation of side AC is x = 1.

Comparing the equation 4x² + 4xy + y² = 0 with ax² + 2hxy + by² = 0, we get,

a = 4, 2h = 4, i.e. h = 2 and b = 1

∴ h² – ab = (2)² – 4(1) = 4 – 4 = 0

Since the equation 4x² + 4xy + y² = 0 is a homogeneous equation of second degree and h² – ab = 0, the given equation represents a pair of lines which are real and coincident.

Given: p = 4 and ∝ = 1800

Here p is the perpendicular that makes an angle ∝ with positive direction of x-axis, hence the equation of the straight line is given by:

Formula used:

x cos ∝ + y sin ∝ = p

x cos 1800 + y sin 1800 = 4

i.e; cos 1800 = cos (5 × 360) = cos(5 × 2π )= cos 360 = 1

similarly, sin 1800 = sin (5 × 360) = sin(5 × 2π )= sin 360 = 0

hence, x × 1+ y ×0 = 4

Hence The required equation of the line is x = 4.

Given: p = 2 and ∝ = 3000

Here p is the perpendicular that makes an angle ∝ with positive direction of x-axis, hence the equation of the straight line is given by:

Formula used:

X cos ∝ + y sin ∝ = p

X cos 3000 + y sin 3000 = 2

i.e; cos 3000 = cos ((8×360) +120) = cos((8×2π )+120) = cos 120 = cos(180-60) = cos60

similarly, sin 3000 = sin ((8×360) +120) = sin((8×2π )+120) = sin 120

= sin(180 - 60) = - sin 60

hence, x cos 60 + y (–sin 60) = 2

x × (1/2) - y × (√3/2) = 2

Hence The required equation of the line is x - √3y = 4

Given: p = 3 and ∝ = 2250

Here p is the perpendicular that makes an angle ∝ with positive direction of x-axis, hence the equation of the straight line is given by:

Formula used:

x cos ∝ + y sin ∝ = p

x cos 2250 + y sin 2250 = 3

i.e; cos 2250 = cos ((6×360) +90) = cos((6×2π )+ 90) = cos 90

similarly, sin 2250 = sin ((6×60) + 90) = sin((6×2π ) +90) = sin 90

hence, x cos 90 + y sin 90 = 3

x × (0)+y ×1 = 3

Hence The required equation of the line is y = 3

Given: p = 8 and ∝ = 1500

Here p is the perpendicular that makes an angle ∝ with positive direction of x-axis , hence the equation of the straight line is given by:

Formula used:

x cos ∝ + y sin ∝ = p

x cos 1500 + y sin 1500 = 8

i.e; cos 1500 = cos ((4×360) +60) = cos((4×2π )+60) = cos 60

similarly, sin 1500 = sin ((4×360) +60) = sin((4×2π )+60) = sin 60

x × (1/2)+ y × (√3/2) = 8

Hence The Required equation of the line is x + √3 y = 16.

Given: p = 5 and ∝ = 1350

Here p is the perpendicular that makes an angle ∝ with positive direction of x-axis, hence the equation of the straight line is given by:

Formula used:

x cos ∝ + y sin ∝ = p

x cos 1350 + y sin 1350 = 5

i.e; cos 1350 = cos ((4 360) - 90) = cos((4×2π) - 90)= cos 90

similarly, sin 1350 = sin ((4 360) - 90) = sin((4×2π ) - 90) = -sin 90

hence, x cos 90 + y (–sin 90) = 5

x × (0) - y ×1 = 5

Hence The required equation of the line is y = -5.

To Find: The equation of the line.

Given: p = 3 and ∝ = 450

Here p is the perpendicular that makes an angle ∝ with positive direction of x-axis, hence the equation of the straight line is given by:

Formula used:

X cos ∝ + y sin ∝ = p

X cos 450 + y sin 450 = 3

i.e; cos 450 = cos (360 +90) = cos 90 [∵ cos(360+ x) = cos x]

similarly, sin 450 = sin (360 +90) = sin 90 [∵ sin(360+ x) = sin x]

hence, x cos 90 + y sin 90 = 3

x × (0)+ y ×1 = 3

Hence the required equation of the line is y = 3

Answer is (C)

Line is parallel to the line y = 3x – 1.

So, slope of the line is‘3’.

Also, line passes through the point (1,2).

So, equation of the line is: y – 2 = 3(x – 1)

To Find: The values of a and b when the line x/a+y/b = 1 passes through the points (8, -9) and (12, -15).

Given: the equation of the line: x/a+y/b = 1 equation 1

Also (8, -9) passes through equation 1

8/a - 9/b = 1

8b - 9a = ab equation 2

And (12, -15) passes through equation 1

12/a − 15/b = 1

12b - 15a = ab equation 3

Solving equation 2 and 3

a = 2.

Put a = 2 in equation 2

8b - 9a = ab

8b -18 = 2b

6b = 18

⟹ b = 3

Hence the values of a and b are 2 and 3 respectively.

Answer is (B)

Reflection of A (4, 1) in y = x is 5(1,4).

Now translation of point B through a distance ‘2’ units along the positive x-axis shifts B to C( 1 + 2,4) or C(3,4).

The given line is

x – 2y = 3

⇒ y = \(\frac{x}{2}\) − \(\frac{3}{2}\)

∴ Slope, m₁ = \(\frac{1}{2}\)

Let ‘m’ be the slope of line AB which passes through (3, 2)

Since the angle between the two lines is 60°.

∴ tan 45° = ± \(\frac{m_2 − m_1}{1 + m_1m_2}\)

⇒ 1 = ± \(\frac{m_2 − \frac{1}{2}}{1 + \frac{1}{2}m_2}\)

⇒ = ± \(\frac{2m_2 − 1}{ m_2 + 2}\)

∴ \(\frac{2m_2 − 1}{ m_2 + 2}\) = 1 or \(\frac{2m_2 − 1}{ m_2 + 2}\) = −1

⇒ m₂ = 3 or m₂ = − \(\frac{1}{3}\).

∴ Equation of AB is

y – 2 = 3 (x – 3)

⇒ 3x – y = 7   (m₂ = 3)

Or,

y − 2 = − \(\frac{1}{3}\)(x − 3) (m₂ = − \(\frac{1}{3}\))

⇒ x + 3y = 9

Using intercept form, required equations are of the form y = mx + c and y = m(x - d)

i.e, y = 1/2x - 3/2 and y = 1/2(x - 4)

Let (h, k) be the point to which the origin is shifted. Then,

x = - 4, y = 2, X = 3 and Y = -2

∴ x = X + h and y = Y + k

⇒ - 4 = 3 + h and 2 = - 2 + k

⇒ h = -7 and k = 4

Hence, the origin must be shifted to (-7, 4)

Given lines are 4x + 1y-3 = 0  …………….(1) 2x – 3y + 1 = 0 ………………. (2) First, find the point of intersection of (1) and (2), ∴ 3 x (1) + 7 x (2) gives 12 x + 14x – 9 + 7 = 0

⇒ 26x - 2 = 0 ⇒ x = 1/13

Put x = 1/13 in (2), we get

 2/13 - 3y + 1 = 0 ⇒ 3y = 15/13 ⇒ y = 5/13

∴ (1) and (2) meet at (1/13, 5/13)

Since the required line has equal intercepts on the axes then its equation is of the form,

x/a + y/a = 1 (∵a = b)

i.e., x + y = a  ...(3)

the line (3) passes through (1/13, 5/13) then

1/13 + 5/13 = a  ⇒ a = 6/13

Putting the value of 'a' in(3), we get

x + y = 6/13

i.e., 13x + 13y = 6

Given: 

4x− 7y − 3 = 0 … (1) 

2x − 3y + 1 = 0 … (2) 

To find: 

Equation of line passing through the point of intersection of lines. 

Concept Used: 

Point of intersection of two lines. 

Explanation: 

Solving (1) and (2) using cross - multiplication method:

\(\frac{x}{-7-9}=\frac{y}{-6-4}=\frac{1}{-12+14}\)

⇒ x = - 8 , y = - 5 

Thus, the point of intersection of the given lines is (- 8, - 5). 

Now, the equation of a line having equal intercept as a is \(\frac{x}{a}+\frac{y}{a}=1\)

This line passes through ( - 8, - 5)

\(-\frac{8}{a}-\frac{5}{a}=1\)

⇒ - 8 - 5 = a 

⇒ a = - 13 

Hence, the equation of the required line is \(\frac{x}{-13}=\frac{y}{-13}=1\)  or x + y + 13 = 0

Let the new origin be (h, k) = (2, -1) and (x, y) = (-3, 5) be the given point.

Let the new coordinates be (X, Y)

We use the transformation formula:

x = X + h and y = Y + k

⇒ -3 = X + 2 and 5 = Y + (-1)

⇒ X = -5 and Y = 6

Thus, the new coordinates are (-5, 6)

Let the new origin be (h, k) = (0, -2) and (x, y) = (3, 2) be the given point.

Let the new coordinates be (X, Y)

We use the transformation formula:

x = X + h and y = Y + k

⇒ 3 = X + 0 and 2 = Y + (-2)

⇒ X = 3 and Y = 4

Thus, the new coordinates are (3, 4)

Given:

The equations, 2x – 7y + 11 = 0 and x + 3y – 8 = 0

The equation of the straight line passing through the points of intersection of 2x − 7y + 11 = 0 and x + 3y − 8 = 0 is given below:

2x − 7y + 11 + λ(x + 3y − 8) = 0

(2 + λ)x + (− 7 + 3λ)y + 11 − 8λ = 0

(i) The required line is parallel to the x-axis. So, the coefficient of x should be zero.

2 + λ = 0

λ = -2

Now, substitute the value of λ back in equation, we get

0 + (− 7 − 6)y + 11 + 16 = 0

13y − 27 = 0

∴ The equation of the required line is 13y − 27 = 0

(ii) The required line is parallel to the y-axis. So, the coefficient of y should be zero.

-7 + 3λ = 0

λ = 7/3
Now, substitute the value of λ back in equation, we get

(2 + 7/3)x + 0 + 11 – 8(7/3) = 0

13x – 23 = 0

∴ The equation of the required line is 13x – 23 = 0

Let the new origin be (h, k) = (1, -2)

Then, the transformation formula become:

x = X + 1 and y = Y + (- 2) = Y – 2

Substituting the value of x and y in the given equation, we get

2x² + y² – 4x + 4y = 0

Thus,

2(X + 1)² + (Y – 2)² – 4(X + 1) + 4(Y – 2) = 0

⇒ 2(X² + 1 + 2X) + (Y² + 4 – 4Y) – 4X – 4 + 4Y – 8 = 0

⇒ 2X² + 2 + 4X + Y² + 4 – 4Y – 4X + 4Y – 12 = 0

⇒ 2X² + Y² – 6 = 0

⇒ 2X² + Y² = 6

Hence, the transformed equation is 2X² + Y² = 6

Given:

The equations, 2x + 3y + 1 = 0 and 3x – 5y – 5 = 0

The equation of the straight line passing through the points of intersection of 2x + 3y + 1 = 0 and 3x − 5y − 5 = 0 is

2x + 3y + 1 + λ(3x − 5y − 5) = 0

(2 + 3λ)x + (3 − 5λ)y + 1 − 5λ = 0

y = – [(2 + 3λ) / (3 – 5λ)] – [(1 – 5λ) / (3 – 5λ)]

The required line is equally inclined to the axes. So, the slope of the required line is either 1 or − 1.

So,

– [(2 + 3λ) / (3 – 5λ)] = 1 and – [(2 + 3λ) / (3 – 5λ)] = -1

-2 – 3λ = 3 – 5λ and 2 + 3λ = 3 – 5λ

λ = 5/2 and 1/8

Now, substitute the values of λ in (2 + 3λ)x + (3 − 5λ)y + 1 − 5λ = 0, we get the equations of the required lines as:

(2 + 15/2)x + (3 – 25/2)y + 1 – 25/2 = 0 and (2 + 3/8)x + (3 – 5/8)y + 1 – 5/8 = 0

19x – 19y – 23 = 0 and 19x + 19y + 3 = 0

∴ The required equation is 19x – 19y – 23 = 0 and 19x + 19y + 3 = 0

Given:

The lines x + y = 4 and 2x – 3y = 1

The equation of the straight line passing through the point of intersection of x + y = 4 and 2x − 3y = 1 is

x + y − 4 + λ(2x − 3y − 1) = 0

(1 + 2λ)x + (1 − 3λ)y − 4 − λ = 0 … (1)

y = – [(1 + 2λ) / (1 – 3λ)]x + [(4 + λ) / (1 – 3λ)]

The equation of the line with intercepts 5 and 6 on the axis is

x/5 + y/6 = 1 …. (2)

So, the slope of this line is -6/5

The lines (1) and (2) are perpendicular.

∴ -6/5 × [(-1 + 2λ) / (1 – 3λ)] = -1

λ = 11/3

Now, substitute the values of λ in (1), we get the equation of the required line.

(1 + 2(11/3))x + (1 – 3(11/3))y − 4 – 11/3 = 0

(1 + 22/3)x + (1 – 11)y – 4 – 11/3 = 0

25x – 30y – 23 = 0

∴ The required equation is 25x – 30y – 23 = 0

A line which cut off an intercept 5 from the y – axis are equally to axes. 

To Find: Find the equation 

The formula used: The equation of the line is y = mx + c 

Explanation: If a line is equally inclined to the axis, then 

Angle θ = 45° and θ = (180 – 45) = 135° 

Then, The slope of the line, m = tan θ 

m = 1 

Since, the intercept is 5, C = 5 

Now, The equation of the line is y = mx + c

y = 1x + 5 

y – x = 5 

Hence, The equation of the line is y – x = 5.

5x² – 9y² = 0

∴ (√5x)² – (3y)² = 0

∴ (√5x + 3y)(√5x – 3y) = 0

∴ the separate equations of the lines are

√5x + 3y = 0 and √5x – 3y = 0.

Given   \(-\frac{\pi}{4}\)

To Find: Slope of the line 

Angle made with the positive x - axis is  \(-\frac{\pi}{4}\)

The Slope of the line is m 

Formula Used: m = tanθ So, 

The slope of Line is m = tan \(\Big(-\frac{\pi}{4}\Big)\) = -1

Hence, The slope of the line is – 1

Here the slope of the line is m = tanθ = tan 30° =1/√3 and the given point is (2, 3). Therefore, using point slope formula of the equation of a line, we have

y – 3 = (1/√3)(x-2)

or, x – √(3y) + (3√3 – 2) = 0.

Let the coordinate of the foot of perpendicular from the point (2, 3) on the line x + y – 11 = 0 be (x, y) 

Now, the slope of the perpendicular = -1 

The equation of perpendicular is given by 

y – 3 = 1(x – 2) 

= > x – y + 1 = 0 

Solving x + y – 11 = 0 and x – y + 1 = 0, we get 

x = 5 and y = 6

Given, inclination of the line, θ = 30°.

Slope of the line, m = tan 30°

= \(\frac{1}{\sqrt{3}}\)

Equation of a line in one-point form is

y – y₀ = m(x – x₀)

Since m = \(\frac{1}{\sqrt{3}}\) and (x₀, y₀) = (2, 3)

∴ y − 3 = \(\frac{1}{\sqrt{3}}\)(x − 2)

⇒ \(\sqrt{3}\)y − 3\(\sqrt{3}\) = x – 2

⇒ x − \(\sqrt{3}\)y + (2 − 3\(\sqrt{3}\)) = 0, is the required equation of the line.

Let ‘a’ be the acute angle made by the line with positive x-axis.

∴ Equation of the line is x cos a + y sin a = 3 …(1)

Since the line passes through (– 5, 0), so

⇒ – 5 cos a + 0. Sin a = 3

⇒ cos a = – \(\frac{3}{5}\)

⇒ sin a = \(\sqrt{1 − (-\frac{3}{5})^2}\) = \(\sqrt{1 − \frac{9}{25}}\) = \(\frac{4}{5}\)

(1) ⇒ x(− \(\frac{3}{5}\)) + y(\(\frac{4}{3}\)) = 3

⇒ 3x – 4y + 15 = 0, is the required of line.

Let the image of point P(3, 8) be Q (a, b) with respect to line AB which is given by

x + 3y – 7 = 0 …(1)

The mid-point of P and Q is \(\big(\frac{3 + a}{2} , \frac{8 + b}{2}\big)\), which lies on line (1).

∴ (1) ⇒ \(\frac{3 + a}{2} +3. \frac{8 + b}{2}\) − 7 = 0

⇒ 3 + a + 24 + 3b – 14 = 0

⇒ a + 3b + 13 = 0 …(2)

Now, slope of AB = – \(\frac{1}{3}\)

and slope of PQ = \(\frac{b−8}{a−3}\)

∴ − \(\frac{1}{3}\) × \(\frac{b−8}{a−3}\) = −1

⇒ 3a – b – 1 = 0

Solving (2) and (3), we arrive at a = – 1, b = – 4

∴ Image of (3, 8) is (– 1, – 4).

Assuming: 

x denotes the price per liter, and y denote the quality of the milk sold at this price. 

Since there is a linear relationship between the price and the quality, the line representing this 

Given: Relationship passes through (14, 980) and (16, 1220). 

To find: How many liters could he sell weekly at Rs. 17 per liter. 

Explanation: 

So, the equation of the line passing through these points is

y-y₁ = \(\Big(\frac{y_2-y_1}{x_2-x_1}\Big)(x-x_1)\) 

y - 980 = \(\frac{1220-980}{16-14}\) (x - 14)

⇒ y – 980 = 120(x - 14) 

⇒ 120 x - y - 700 = 0 

When x = 17 then we have, 

120(17) – y – 700 = 0 

⇒ y = 1340 

Hence, the owner of the milk store can shell 1340 litres of milk at Rs. 17 per litre.

Given line: 3x – 4y – 8 = 0

and given points are (3, 4) and (2, -6)

For point (3, 4)

3(3) – 4(4) – 8

= 9 – 16 – 8

= 9 – 24

= - 15 < 0

For point (2, -6)

3(2) – 4(-6) – 8

= 6 + 24 – 8

= 30 – 8

= 22 > 0

Ans. So, the points (3,4) and (2, -6) are situated on the opposite sides of 3x – 4y – 8 = 0.

Given: 

Line x - √3y + 4 = 0 

To find: 

The perpendicular distance from the origin of the perpendicular from the point (1, 2) upon the straight line x - √3y + 4 = 0 .

Concept Used: 

Distance of a point from a line. 

Explanation: 

The equation of the line perpendicular to x - √3y + 4 = 0 is x - √3y + λ = 0 This line passes through (1,2)

∴  \(\sqrt{-3}\)  + 2 + λ = 0

⇒ λ = \(\sqrt{-3}\) - 2

Substituting the value of λ, we get \(\sqrt{3}x+y-\sqrt{3}-2=0\) 

Let d be the perpendicular distance from the origin to the line  \(\sqrt{3}x+y-\sqrt{3}-2=0\)  

d = \(\frac{0-0-\sqrt{3}-2}{\sqrt{1+3}}=\frac{\sqrt{3}+2}{2}\) 

Hence, the required perpendicular distance is \(\frac{\sqrt{3}+2}{2}\) 

Given: equation is perpendicular to 2x − 3y = 5 and cutting off an intercept 1 on the positive direction of the x-axis. 

Explanation: 

The line perpendicular to 2x − 3y = 5 is 

3x + 2y + λ = 0 

It is given that the line 3x + 2y + λ = 0 cuts off an intercept of 1 on the positive direction of the x axis. 

This means that the line 3x + 2y + λ = 0 passes through the point (1, 0). 

∴ 3 + 0 + λ = 0 

⇒ λ = -3 

Substituting the value of λ, we get 3x + 2y – 3 = 0, 

Hence, equation of the required line.

Given: equation is perpendicular to lx + my + n = 0 and passing through (α, β) 

To find: 

The equation of the straight line through the point (α, β) and perpendicular to the line lx + my + n = 0. 

Explanation: 

The line perpendicular to lx + my + n = 0 is 

mx – ly + λ = 0 

This line passes through (α, β). 

∴ mα - lβ + λ = 0 

⇒ λ = lβ - mα 

Substituting the value of λ: 

mx – ly + lβ - mα = 0 

⇒ mx – α = ly – β 

Hence, equation of the required line is mx – α = ly – β

Given: 

Point (a, b)

To find: 

Equation representing a pair of lines through the point (a, b) and parallel to the coordinates axes. 

Explanation: 

The lines passing through (a, b) and parallel to the x-axis and y-axis are y = b and x = a, respectively. Therefore, their combined equation is given below: 

(x – a)(y – b) = 0

Given: 

( x₁, y₁) = (0, 0), ( x₂, y₂) = (2, -2) 

Concept Used: 

The equation of the line passing through the two points ( x₁, y₁) and ( x₂, y₂). 

To find: 

The equation of the straight line passing through a pair of points. 

Explanation: 

So, the equation of the line passing through the two points (0, 0) and (2, −2) is 

The formula used: y - y₁ = \(\frac{y_2-y_1}{x_2-x_1} (x-x_1)\) 

⇒ y -0 = \(\frac{-2-0}{2-0}(x-0)\) 

⇒ y = -x Hence, equation of line is y = -x

Let P(x, y) be any point on the required locus. Given, A(1, 3), B(2, 1) and PA = PB 

∴ PA² = PB² 

∴ (x – 1)² + ( y – 3)² = (x – 2)² + (y – 1)² 

∴ x² – 2x + 1 + y² – 6y + 9 = x² – 4x + 4 + y² – 2y + 1 -2x – 6y + 10 = -4x – 2y + 5 

∴ 2x – 4y + 5 = 0 

∴ The required equation of locus is 2x – 4y + 5 = 0.

Let P(x, y) be any point on the required locus. Given, A(4,1), B(5,4) and PA² = 3PB² 

∴ (x – 4)2 + (y – 1)2 = 3[(x – 5)² + (y – 4)² ]

∴ x² – 8x + 16 + y² – 2y + 1 = 3(x² – 10x + 25 + y² – 8y + 16) 

∴ x² – 8x + y² – 2y + 17 = 3x² -30x + 75 + 3y² – 24y + 48

∴ 2x² + 2y² – 22x – 22y + 106 = 0 

∴ x² + y² – 11x – 11y + 53 = 0 

∴ The required equation of locus is x² + y² – 11x – 11y + 53 = 0

Given, The line segment joining the points (1,0) and (2,3) 

To Find: Find the equation of line 

Formula used: The equation of line is (y – y₁) = m(x – x₁) 

Explanation: Here, The right bisector PQ of AB at C and is perpendicular to AB

So, The slope of the line with two points is, m = \(\frac{y_2-y_1}{x_2-x_1}\) 

The slope of the line AB = \(\frac{3-0}{2-1}\) = 3

We know, The product of two slopes of the perpendicular line is always – 1 

Therefore, (slope of AB) × (slope of PQ) = – 1

 Since Slope of PQ = \(-\frac{1}{3}\) 

Now, The coordinate of the mid – points =  \([\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}]\) 

The coordinates of point C are = \([\frac{1+2}{2},\frac{3+0}{2}]\) = \([\frac{3}{2}, \frac{3}{2}]\) 

The required equation of PQ is (y – y₁) = m(x – x₁)

\(\Big(y-\frac{3}{2}\Big)\) = \(-\frac{1}{3}\Big(x-\frac{3}{2}\Big)\) 

6y – 9 = – 2x + 3 

x + 3y = 6 

Hence, The equation of line is x + 3y = 6

Given, The line segment joining the points (3,4) and ( – 1,2) 

To Find: Find the equation of the line 

Formula used: The equation of line is (y – y₁) = m(x – x₁) 

Explanation: Here, The right bisector PQ of AB at C and is perpendicular to AB

Now, The coordinate of the mid – points =  \([\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}]\) 

The coordinates of point C are = \([\frac{3-1}{2},\frac{4+2}{2}]\) = (1,3) 

And, The slope of PQ = \(\frac{1}{Slope\, of \ AB}\) 

The slope of PQ, m = \(\frac{1}{2-4}\) (-1 - 3) = \(\frac{4}{-2}\) 

So, The slope of PQ, m = – 2 

The required equation of PQ is (y – y₁) = m(x – x₁)

 y – 3 = – 2(x – 1) 

y – 3 = – 2x + 2 

y + 2x = 5 

Hence, The equation of line is y + 2x = 5

AB be the given line which makes an angle of 150° with the positive direction of y–axis and OQ be the perpendicular drawn from the origin on the line. 

Given:

p = 7 and α = 30° 

Explanation: 

So, the equation of the line AB is 

Formula Used: x cos α + y sin α = p 

⇒ x cos 30° + y sin 30° = 7

⇒ \(\frac{\sqrt{3x}}{2}+\frac{y}{2}\)

⇒ √3 x + y = 14 

Hence, the equation of line in normal form is √3 x + y = 14

Given: the normal form of a line is x cos θ + y sin θ = p …..… (1) 

To find: P and θ. 

Explanation: 

Let us try to write down the equation √3 + y + 2 = 0 in its normal form. 

Now √3 + y + 2 = 0 

⇒ √3 + y = – 2 

Dividing both sides by 2, 

⇒ – √3/2 – y/2 = 1

⇒   \(\Big(-\frac{\sqrt{3}}{2}\Big)x+\Big(-\frac{1}{2}\Big)y\) = 1 ......(2)

Comparing equations (1) and (2) we get, 

cosθ = \(-\frac{\sqrt{3}}{2}\) and p = 1 

⇒ θ = 210° = 7π/6 and p = 1 

Hence, θ = 210° = 7π/6 and p = 1

Given line is x + 4y – 5 = 0

Here, A = 1, B = 4, C = – 5 and the given point is (x₁, y₁) = (2, 3)

∴ Distance, d = \(\bigg|\frac{A_1 x + B_1 y + C_1}{\sqrt{A^2 + B^2}}\bigg|\)

= \(\bigg|\frac{1.2 + 4.3 + 5}{\sqrt{1^2 + 4^2}}\bigg|\)

= \(\frac{9}{\sqrt{17}}\)

= \(\frac{9\sqrt{17}}{17}\) units