Straight Lines : Notes and Study Materials -pdf
CBSE Class 11 Maths Notes Chapter 10 Straight Lines
Distance Formula
The distance between two points A(x1, y1) and B (x2, y2) is given by
The distance of a point A(x, y) from the origin 0 (0, 0) is given by OA = x2+y2−−−−−−√
Section Formula
The coordinates of the point which divides the joint of (x1, y1) and (x2, y2) in the ratio m : n internally, is
Mid-point of the joint of (x1, y1) and (x2, y2) is
X-axis divides the line segment joining (x1, y1) and (x2, y2) in the ratio -y1 : y2.
Y-axis divides the line segment joining (x1, y1) and (x2, y2) in the ratio -x1 : x2.
The coordinates of the centroid of the triangle whose vertices are (x1, y1), (x2, y2) and (x3, y3) is
Area of Triangle
The area of the triangle, the coordinates of whose vertices are (x1, y1), (x2, y2)and (x3, y3) is the absolute value of
If the points (x1, y1), (x2, y2) and (x3, y3) are collinear, then x1 (y2 – y3) + x2 (y3 – y1) + x3 (y1 – y2) = 0.
Shifting of Origin
Let the origin is shifted to a point O'(h, k). If P(x, y) are coordinates of a point referred to old axes and P'(X, Y) are the coordinates of the same points referred to new axes, then x = X + h, y = Y + k.
Straight Line
Any curve is said to be a straight line if two points are taken on the curve such that every point on the line segment joining any two points on it lies on the curve. General equation of a line is ax + by + c = 0.
Slope or Gradient of Line
The inclination of angle θ to a line with a positive direction of X-axis in the anti-clockwise direction, the tangent of angle θ is said to be slope or gradient of the line and is denoted by m.
i.e. m = tan θ
The slope of a line passing through points P(x1, y1) and Q(x2, y2) is given by
Note: Slope of a line parallel to X-axis is zero and slope of a line parallel to Y-axis is not defined.
Angle between Two Lines
The angle θ between two lines having slope m1 and m2 is
- If two lines are parallel, their slopes are equal i.e. m1 = m2.
- If two lines are perpendicular to each other, then their product of slopes is -1 i.e. m1m2 = -1.
Various Forms of the Equation of a Line
If a line is at a distance k and parallel to X-axis, then the equation of the line is y = ± k.
If a line is parallel to Y-axis at a distance c from Y-axis, then its equation is x = ± c.
Slope-intercept form: The equation of line with slope m and making an intercept c on the y-axis, is y = mx + c.
One point-slope form: The equation of a line which passes through the point (x1, y1) and has the slope of m is given by y – y1 = m (x – x1).
Two points form: The equation of a line passing through the points (x1, y1) and (x2, y2) is given by
The Intercept form: The equation of a line which cuts off intercepts a and b respectively on the x and y-axes is given by xa+yb=1
The normal form: The equation of a straight line upon which the length of the perpendicular from the origin is p and angle made by this perpendicular to the x-axis is α, is given by x cos α + y sin α = p.
General Equation of a Line
Any equation of the form Ax + By + C = 0, where A and B are simultaneously not zero is called the general equation of a line.
Different Forms of Ax + By + C = 0
Slope intercept form: If B ≠ 0, then Ax + By + C = 0 can be written as
If B = 0, then x = – C / A which is a vertical line, whose slope is not defined and x-intercept is – C/A.
Intercept form If C ≠ 0, then Ax + By + C = 0 can be written as
where a = – C / A and b = – C/B
If C = 0, then Ax + By + C = 0 can be written as Ax + By = 0 which is a line passing through origin and therefore has zero intercept on the axes.
Normal form: The normal form of equation Ax + By + C = 0 is x cos α + y sin α = p where
Note: Proper choice of signs to be made so that p should be always positive.
Position of Points is Relative to a Given Line
Let the equation of the given line be ax + by + c = 0 and let the coordinates of the two given points be P(x1, y1) and Q(x2, y2).
The two points are on the same side of the straight line ax + by + c = 0, If ax1 + by1 + c and ax2 + by2 + c have the same sign.
The two points are on the opposite sides of the straight line ax + by + c = 0, If ax1 + by1 + c and ax2 + by2 + c have opposite sign.
Condition of concurrency for three given lines
a1x + b1y + c1 = 0, a2x + b2y + c2 = 0 and a3x + b3y+ c3 = 0 is a3(b1c2 – b2c1) + b3(a2c1 – a1c2) + c3(a1b2 – a2b1) = 0
Point of intersection of two lines
Let equation of lines be ax1 + by1 + c1 = 0 and a2x + b2y + c2 = 0, then their point of intersection is
Distance of a Point from a Line
The perpendicular distanced of a point P(x1, y1)from the line Ax + By + C = 0 is given by
Distance Between Two Parallel Lines
The distance d between two parallel lines y = mx + c1 and y = mx + c2 is given by
Straight Lines Class 11 MCQs Questions with Answers
Question 1.
The locus of a point, whose abscissa and ordinate are always equal is
(a) x + y + 1 = 0
(b) x – y = 0
(c) x + y = 1
(d) none of these.
Answer
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
Question 2.
The equation of straight line passing through the point (1, 2) and parallel to the line y = 3x + 1 is
(a) y + 2 = x + 1
(b) y + 2 = 3 × (x + 1)
(c) y – 2 = 3 × (x – 1)
(d) y – 2 = x – 1
Answer
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)
Question 3.
What can be said regarding if a line if its slope is negative
(a) θ is an acute angle
(b) θ is an obtuse angle
(c) Either the line is x-axis or it is parallel to the x-axis.
(d) None of these
Answer
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
Question 4:
The equation of the line which cuts off equal and positive intercepts from the axes and passes through the point (α, β) is
(a) x + y = α + β
(b) x + y = α
(c) x + y = β
(d) None of these
Answer
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
Question 5.
Two lines a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0 are coincedent if
(a) a1/a2 = b1/b2 ≠ c1/c2
(b) a1/a2 ≠ b1/b2 = c1/c2
(c) a1/a2 ≠ b1/b2 ≠ c1/c2
(d) a1/a2 = b1/b2 = c1/c2
Answer
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
Question 6:
The equation of the line passing through the point (2, 3) with slope 2 is
(a) 2x + y – 1 = 0
(b) 2x – y + 1 = 0
(c) 2x – y – 1 = 0
(d) 2x + y + 1 = 0
Answer
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
Question 7.
The slope of the line ax + by + c = 0 is
(a) a/b
(b) -a/b
(c) -c/b
(d) c/b
Answer
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
Question 8.
Equation of the line passing through (0, 0) and slope m is
(a) y = mx + c
(b) x = my + c
(c) y = mx
(d) x = my
Answer
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
Question 9.
The angle between the lines x – 2y = y and y – 2x = 5 is
(a) tan-1 (1/4)
(b) tan-1 (3/5)
(c) tan-1 (5/4)
(d) tan-1 (2/3)
Answer
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)
Question 10.
Two lines a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0 are parallel if
(a) a1/a2 = b1/b2 ≠ c1/c2
(b) a1/a2 ≠ b1/b2 = c1/c2
(c) a1/a2 ≠ b1/b2 ≠ c1/c2
(d) a1/a2 = b1/b2 = c1/c2
Answer
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
Question 11.
The locus of a point, whose abscissa and ordinate are always equal is
(a) x + y + 1 = 0
(b) x – y = 0
(c) x + y = 1
(d) none of these.
Answer
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
Question 12.
In a ΔABC, if A is the point (1, 2) and equations of the median through B and C are respectively x + y = 5 and x = 4, then B is
(a) (1, 4)
(b) (7, – 2)
(c) none of these
(d) (4, 1)
Answer
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)
Question 13.
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. Then the equation of line is
(a) x + y = 14
(b) √3y + x = 14
(c) √3x + y = 14
(d) None of these
Answer
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
Question 14.
If two vertices of a triangle are (3, -2) and (-2, 3) and its orthocenter is (-6, 1) then its third vertex is
(a) (5, 3)
(b) (-5, 3)
(c) (5, -3)
(d) (-5, -3)
Answer
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)
Question 15.
The sum of squares of the distances of a moving point from two fixed points (a, 0) and (-a, 0) is equal to 2c² then the equation of its locus is
(a) x² – y² = c² – a²
(b) x² – y² = c² + a²
(c) x² + y² = c² – a²
(d) x² + y² = c² + a²
Answer
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
Question 16.
The equation of the line through the points (1, 5) and (2, 3) is
(a) 2x – y – 7 = 0
(b) 2x + y + 7 = 0
(c) 2x + y – 7 = 0
(d) x + 2y – 7 = 0
Answer
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
Question 17.
What can be said regarding if a line if its slope is zero
(a) θ is an acute angle
(b) θ is an obtuse angle
(c) Either the line is x-axis or it is parallel to the x-axis.
(d) None of these
Answer
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.
Question 18.
Two lines are perpendicular if the product of their slopes is
(a) 0
(b) 1
(c) -1
(d) None of these
Answer
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
Question 19.
y-intercept of the line 4x – 3y + 15 = 0 is
(a) -15/4
(b) 15/4
(c) -5
(d) 5
Answer
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
Question 20.
The equation of the locus of a point equidistant from the point A(1, 3) and B(-2, 1) is
(a) 6x – 4y = 5
(b) 6x + 4y = 5
(c) 6x + 4y = 7
(d) 6x – 4y = 7
Answer
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