Straight Lines Class 11 MCQ Online Test With Answers Questions

Answer: (b) 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

Answer: (c) 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)

Answer: (b) 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

Answer: (a) 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

Answer: (d) 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

Answer: (c) 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

Answer: (b) -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

Answer: (c) 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

Answer: (c) 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)

Answer: (a) 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

Answer: (b) 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

Answer: (b) (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)

Answer: (c) √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

Answer: (d) (-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)

Answer: (c) 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

Answer: (c) 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

Answer: (c) 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.

Answer: (c) -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

Answer: (d) 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

Answer: (b) 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