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## Sunday, 21 February 2021

### CBSE Class 11 Maths - MCQ and Online Tests - Unit 10 - Straight Lines

#### CBSE Class 11 Maths  – MCQ and Online Tests – Unit 10 – Straight Lines

Every year CBSE schools conducts Annual Assessment exams for 6,7,8,9,11th standards. These exams are very competitive to all the students. So our website provides online tests for all the 6,7,8,9,11th standard’s subjects. These tests are also very effective and useful for those who preparing for any competitive exams like Olympiad etc. It can boost their preparation level and confidence level by attempting these chapter wise online tests.

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#### CBSE Class 11 Maths  – MCQ and Online Tests – Unit 10 – Straight Lines

Question 1.
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: (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 2.
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: (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 3.
What can be said regarding if a line if its slope is negative
(a) ? is an acute angle
(b) ? is an obtuse angle
(c) Either the line is x-axis or it is parallel to the x-axis.
(d) None of these

Answer: (b) ? is an obtuse angle
Hint:
Let ? 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 ?
Given, slope is positive
? tan ? < 0
? ? lies between 0 and 180 degree
? ? 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 (a, ß) is
(a) x + y = a + ß
(b) x + y = a
(c) x + y = ß
(d) None of these

Answer: (a) x + y = a + ß
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 (a, ß)
So, a + ß = a
Put this value in equation 1, we get
x + y = a + ß

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: (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: (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

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

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)

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 ? = |(m1 + m2)/{1 + m1 × m2}|
= |(1/2 + 2)/{1 + (1/2) × 2}|
= |(5/2)/(1 + 1)|
= |(5/2)/2|
= 5/4
? ? = tan-1 (5/4)

Question 10.
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: (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 11.
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: (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 12.
In a ?ABC, 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)

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) v3y + x = 14
(c) v3x + y = 14
(d) None of these

Answer: (c) v3x + 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
? v3x/2 + y/2 = 7
? v3x + y = 7×2
? v3x + y = 14

Question 14.
Two lines are perpendicular if the product of their slopes is
(a) 0
(b) 1
(c) -1
(d) None of these

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 15.
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)

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 16.
y-intercept of the line 4x – 3y + 15 = 0 is
(a) -15/4
(b) 15/4
(c) -5
(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 17.
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: (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

Question 18.
What can be said regarding if a line if its slope is zero
(a) ? is an acute angle
(b) ? is an obtuse angle
(c) Either the line is x-axis or it is parallel to the x-axis.
(d) None of these

Answer: (c) Either the line is x-axis or it is parallel to the x-axis.
Hint:
Let ? 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 ?
Given, slope is zero
? tan ? = 0
? ? = 0°
? Either the line is x-axis or it is parallel to the x-axis.

Question 19.
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: (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 20.
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: (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

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