y = mx


Hint:


Equation of the line passing through (x₁, y₁) and slope m is


(y - y₁) = m(x - x₁)


Now, required line is


(y - 0 ) = m(x - 0)


⇒ y = mx

x – y = 0

Hint:

Let the coordinate of the variable point P is (x, y)

Now, the abscissa of this point = x

and its ordinate = y

Given, abscissa = ordinate

⇒ x = y

⇒ x – y = 0

So, the locus of the point is x – y = 0

y – 2 = 3 × (x – 1)

Hint:

Given straight line is: y = 3x + 1

Slope = 3

Now, required line is parallel to this line.

So, slope = 3

Hence, the line is

y – 2 = 3 × (x – 1)

θ is an obtuse angle

Hint:

Let θ be the angle of inclination of the given line with the positive direction of x-axis in the anticlockwise sense.

Then its slope is given by m = tan θ

Given, slope is positive

⇒ tan θ < 0

⇒ θ lies between 0 and 180 degree

⇒ θ is an obtuse angle

x + y = α + β

Hint:

Let the equation of the line be x/a + y/b = 1 which cuts off intercepts a and b with

the coordinate axes.

It is given that a = b, therefore the equation of the line is

x/a + y/a = 1

⇒ x + y = a …..1

But it is passes through (α, β)

So, α + β = a

Put this value in equation 1, we get

x + y = α + β

a₁/a₂ = b₁/b₂ = c₁/c₂


Hint:
Two lines a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0 are coincedent if
a₁/a₂ = b₁/b₂ = c₁/c₂

2x - y - 1 = 0

Hint:

Given, the point (2, 3) and slope of the line is 2

By, slope-intercept formula,

y - 3 = 2(x - 2)

⇒ y - 3 = 2x - 4

⇒ 2x - 4 - y + 3 = 0

⇒ 2x - y - 1 = 0

-a/b

Hint:

Give, equation of line is ax + by + c = 0

⇒ by = -ax - c

⇒ y = (-a/b)x - c/b

It is in the form of y = mx + c

Now, slope m = -a/b

tan^-1 (5/4)


Hint:


Given, lines are:


x - 2y = 5 .......... 1


and y - 2x = 5 .......... 2


From equation 1,


x - 5 = 2y


⇒ y = x/2 - 5/2


Here, m₁ = 1/2


From equation 2,


y = 2x + 5


Here. m₂ = 2


Now, tan θ = |(m₁ + m₂)/{1 + m₁ × m₂}|


= |(1/2 + 2)/{1 + (1/2) × 2}|


= |(5/2)/(1 + 1)|


= |(5/2)/2|


= 5/4


⇒ θ = tan^-1 (5/4)

a₁/a₂ = b₁/b₂ ≠ c₁/c₂


Hint:


Two lines a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0 are parallel if
a₁/a₂ = b₁/b₂ ≠ c₁/c₂

(7, - 2)


Hint:


The equation of median through B is x + y = 5


The point B lies on it.


Let the coordinates of B are (x₁, 5 - x₁)


Now CF is a median through C,


So co-ordiantes of F i.e. mid-point of AB are


((x₁+1)/2, (5 – x₁+ 2)/2)


Now since this lies on x = 4


⇒ (x₁ + 1)/2 = 4


⇒ x₁ + 1 = 8


⇒ x₁ = 7


Hence, the co-oridnates of B are (7, -2)

√3x + y = 14

Hint:

Given, The length of the perpendicular from the origin to a line is 7 and the line makes an angle of 150 degrees with the positive direction of the y-axis.

Now, equation of line is

x × cos 30 + y × sin 30 = 7

⇒ √3x/2 + y/2 = 7

⇒ √3x + y = 7×2

⇒ √3x + y = 14

(-5, -3)

Hint:

Let the third vertex of the triangle is C(x, y)

Given, two vertices of a triangle are A(3,-2) and B(-2,3)

Now given orthocentre of the circle = H(-6, 1)

So, AH ⊥ BC and BH ⊥ AC

Since the product of the slope of perpendicular lines equal to -1

Now, AH ⊥ BC

⇒ {(-2 - 1)/(3 + 6)} × {(y + 2)/(x - 3)} = -1

⇒ (-3/9) × {(y + 2)/(x - 3)} = -1

⇒ (-1/3)×{(y - 3)/(x + 2)} = -1

⇒ (y - 3)/{3×(x + 2)} = 1

⇒ (y - 3) = 3×(x + 2)

⇒ y - 3 = 3x + 6

⇒ 3x + 6 - y = -3

⇒ 3x - y = -3 - 6

⇒ 3x - 2y = -9 ............ 1

Again, BH ⊥ AC

⇒ {(3 - 1)/(-2 + 6)} × {(y - 3)/(x + 2)} = -1

⇒ (2/4) × {(y - 3)/(x + 2)} = -1

⇒ (1/2)×{(y - 3)/(x + 2)} = -1

⇒ (y - 3)/{2×(x + 2)} = 1

⇒ (y - 3) = 2×(x + 2)

⇒ y - 3 = 2x + 4

⇒ 2x + 4 - y = -3

⇒ 2x - y = -3 - 4

⇒ 2x - y = -7 ............ 2

Multiply equation 2 by 2, we get

4x - 2y = -14 ......... 3

Subtract equation 1 and we get

-x = 5

⇒ x = -5

From equation 2, we get

2×(-5) - y = -7

⇒ -10 - y = -7

⇒ y = -10 + 7

⇒ y = -3

So, the third vertex of the triangle is (-5, -3)

x² + y² = c² - a²

Hint:

Let P(h, k) be any position of the moving point and let A(a, 0) and B(-a, 0) be the given points. Then

PA² + PB² = 2c²

⇒ (h - a)² + (k - 0)² + (h + a)² + (k - 0)² = 2c²

⇒ h² - 2ah + a² + k² + h² + 2ah + a² + k² = 2c²

⇒ 2h² + 2k² + 2a² = 2c²

⇒ h² + k² + a² = c²

⇒ h² + k² = c² - a²

Hence, the locus of (h, k) is x² + y² = c² - a²

2x + y - 7 = 0

Hint:

Given, points are: (1, 5) and (2, 3)

Now, equation of line is

y - y₁ = {(y₂ - y₁)/(x₂ - x₁)} × (x - x₁)

⇒ y - 5 = {(3 - 5)/(2 - 1)} × (x - 1)

⇒ y - 5 = (-2) × (x - 1)

⇒ y - 5 = -2x + 2

⇒ 2x + y - 5 - 2 = 0

⇒ 2x + y - 7 = 0

Either the line is x-axis or it is parallel to the x-axis.

Hint:

Let θ be the angle of inclination of the given line with the positive direction of x- axis in the anticlockwise sense.

Then its slope is given by m = tan θ

Given, slope is zero

⇒ tan θ = 0

⇒ θ = 0°

⇒ Either the line is x-axis or it is parallel to the x-axis.

-1


Hint:


Let m₁ is the slope of first line and m₂ is the slope of second line.


Now, two lines are perpendicular if m₁ × m₂ = -1


i.e. the product of their slopes is equals to -1

5

Hint:

Given, equation of line is 4x - 3y + 15 = 0

⇒ 4x - 3y = -15

⇒ 4x/(-15) + (-3)y/(-15) = 1

⇒ x/(-15/4) + 3y/15 = 1

⇒ x/(-15/4) + y/(15/3) = 1

⇒ x/(-15/4) + y/5 = 1

Now, compare with x/a + y/b = 1, we get

y-intercept b = 5

6x + 4y = 5

Hint:

Let P(h, k) be any point on the locus. Then

Given, PA = PB

⇒ PA² = PB²

⇒ (h - 1)² + (k - 3)² = (h + 2)² + (k - 1)²

⇒ h² - 2h + 1 + k² - 6k + 9 = h² + 4h + 4 + k² - 2k + 1

⇒ -2h - 6k + 10 = 4h - 2k + 5

⇒ 6h + 4k = 5

Hence, the locus of (h, k) is 6x + 4y = 5