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All questions of Coordinate Geometry- Straight Lines for A Level Exam
The equation of a line whose perpendicular distance from the origin is 3 units and the angle made by perpendicular with positive x-axis is 30° is:- a) x + √3y = 16
- b) x + √3y = 6
- c) √3x + y + 6 = 0
- d) √3x + y - 2 = 0
Correct answer is option 'D'. Can you explain this answer?
The equation of a line whose perpendicular distance from the origin is 3 units and the angle made by perpendicular with positive x-axis is 30° is:
a)
x + √3y = 16
b)
x + √3y = 6
c)
√3x + y + 6 = 0
d)
√3x + y - 2 = 0
|
Krishna Iyer answered |
xcos 30° + y sin 30° = 3
⇒ √3x + y - 6 = 0
⇒ √3x + y - 6 = 0
For a line whose equation is √3x + y = 8, the length of the perpendicular from the origin is- a)4
- b)8
- c)16
- d)2
Correct answer is option 'A'. Can you explain this answer?
For a line whose equation is √3x + y = 8, the length of the perpendicular from the origin is
a)
4
b)
8
c)
16
d)
2
|
Rajesh Gupta answered |
√3x + y - 8
√3x + y = 8
Dividing by √[(√3)2 + (1)2]
= √[3+1]
= √4
= 2
√3x/2 + y/2 = 8/2
√3x/2 + y/2 = 4
x(√3/2) + y(1/2) = 4.....(1)
Normal form of any line : xcos w + ysin w = p....(2)
Comparing (1) and (2)
p = 4
√3x + y = 8
Dividing by √[(√3)2 + (1)2]
= √[3+1]
= √4
= 2
√3x/2 + y/2 = 8/2
√3x/2 + y/2 = 4
x(√3/2) + y(1/2) = 4.....(1)
Normal form of any line : xcos w + ysin w = p....(2)
Comparing (1) and (2)
p = 4
If A(–2, – 4), B(2, 3) and C(2, – 2) are three points, then the angle between the lines AB and AC is:- a)3π/4
- b)π/4
- c)
- d)
Correct answer is option 'C'. Can you explain this answer?
If A(–2, – 4), B(2, 3) and C(2, – 2) are three points, then the angle between the lines AB and AC is:
a)
3π/4
b)
π/4
c)
d)
|
Lavanya Menon answered |
Let the slope of the line AB and AC are m1 and m2 respectively.
Then,
Let θ be the angle between AB and BC. Then,
If the slope of the line passing through the points (2, 5) and (x, 1) is 2, then x = …a)-2b)0c)8d)6Correct answer is option 'B''Can you explain this answer?
|
|
Krishna Iyer answered |
Slope = Change in y coordinates÷change in x coodinates
= (5-1)÷(2-x) =2
5-1 = 4-2x
0 = 2x
x=0
= (5-1)÷(2-x) =2
5-1 = 4-2x
0 = 2x
x=0
If area of ΔABC = 0 ,three points A,B,C are- a)Coincide
- b)Collinear
- c)Non collinear
- d)None of these
Correct answer is option 'B'. Can you explain this answer?
If area of ΔABC = 0 ,three points A,B,C are
a)
Coincide
b)
Collinear
c)
Non collinear
d)
None of these
|
|
Riya Banerjee answered |
The three points A, B and C are collinear if and only if area of ΔABC = 0.
If the slope of line m = tan 0°. Therefore, the line is …… to the X-axis.- a)Perpendicular
- b)Parallel
- c)Con current
- d)Co-incident
Correct answer is option 'B'. Can you explain this answer?
If the slope of line m = tan 0°. Therefore, the line is …… to the X-axis.
a)
Perpendicular
b)
Parallel
c)
Con current
d)
Co-incident
|
Knowledge Hub answered |
Slope of x-axis is m = tan 0° = 0.
Since the inclination of every line parallel to x-axis is 0°, so its slope (m) = tan 0° = 0. Therefore, the slope of every horizontal line is 0.
Since the inclination of every line parallel to x-axis is 0°, so its slope (m) = tan 0° = 0. Therefore, the slope of every horizontal line is 0.
Let A(2, 12) and B(6,4) be two points. The slope of a line perpendicular to the line AB is:- a)2
- b)-2
- c)1/2
- d)-1/2
Correct answer is option 'C'. Can you explain this answer?
Let A(2, 12) and B(6,4) be two points. The slope of a line perpendicular to the line AB is:
a)
2
b)
-2
c)
1/2
d)
-1/2
|
Suresh Reddy answered |
Slope of the points A and B is -2 and the lines are perpendicular then
m1×m2 = -1.
∴ the slope is 1/2
What is the distance of the point (3,3) from the line 2(x-3) = 3(y+5)?- a)5/3
- b)6
- c)
- d)
Correct answer is option 'C'. Can you explain this answer?
What is the distance of the point (3,3) from the line 2(x-3) = 3(y+5)?
a)
5/3
b)
6
c)
d)
|
Naina Sharma answered |
2(x-3) = 3(y+5)
2x - 6 = 3y + 15
2x - 3y = 21
2x - 3y - 21 = 0
Using distance formula = |ax1 + by1 + c|/(a2 + b2)1/2
= (6-9-21)/((2)2 + (3)2)1/2
= 24/(13)1/2
2x - 6 = 3y + 15
2x - 3y = 21
2x - 3y - 21 = 0
Using distance formula = |ax1 + by1 + c|/(a2 + b2)1/2
= (6-9-21)/((2)2 + (3)2)1/2
= 24/(13)1/2
Find the equation of the line parallel to the line 5x – 4y + 3 = 0 and passing through the point (2, 5) is:- a)5x – 4y + 20 = 0.
- b)5x – 4y + 10 = 0.
- c)4x – 5y + 10 = 0.
- d)4x – 5y + 20 = 0.
Correct answer is option 'B'. Can you explain this answer?
Find the equation of the line parallel to the line 5x – 4y + 3 = 0 and passing through the point (2, 5) is:
a)
5x – 4y + 20 = 0.
b)
5x – 4y + 10 = 0.
c)
4x – 5y + 10 = 0.
d)
4x – 5y + 20 = 0.
|
|
Neha Joshi answered |
5x – 4y + 3 = 0
5x – 4y = -3
(5/4)x + (-4/5)y = -3
Line is parallel m = 5/4
Points (2,5)
= 5 = (5/4)(2) + b
5 = 5/2 + b
b = 5/2
y = 5x/4 - 5/2
4y = 5x - 10
5x - 4y + 10 = 0
5x – 4y = -3
(5/4)x + (-4/5)y = -3
Line is parallel m = 5/4
Points (2,5)
= 5 = (5/4)(2) + b
5 = 5/2 + b
b = 5/2
y = 5x/4 - 5/2
4y = 5x - 10
5x - 4y + 10 = 0
For the line x+y= 1, what is the angle made with the positive direction of the x axis?- a)60°
- b)30°
- c)135°
- d)45°
Correct answer is option 'C'. Can you explain this answer?
For the line x+y= 1, what is the angle made with the positive direction of the x axis?
a)
60°
b)
30°
c)
135°
d)
45°
|
|
Krishna Iyer answered |
We have in Equation = mx+c the slope of line as m and m=tanθ, θ is the angle made by (+ve) in X-axis
Hence, y = 1 - x
Here m = -1 = tanθ
θ = 135o
Hence, y = 1 - x
Here m = -1 = tanθ
θ = 135o
The distance between the parallel lines 4x-3y+5 = 0 and 4x-3y+15 = 0 is :- a)4 units
- b)2 units
- c)3.5 units
- d)5 units
Correct answer is option 'B'. Can you explain this answer?
The distance between the parallel lines 4x-3y+5 = 0 and 4x-3y+15 = 0 is :
a)
4 units
b)
2 units
c)
3.5 units
d)
5 units
|
|
Raghav Bansal answered |
4x - 3y + 5 = 0, 4x - 3y + 15 = 0
A = 4, B = -3 c1 = 5, c2 = 15
|c1 - c2|/[A2 + B2]1/2
= |5 - 15|/[(4)2 + (-3)2]½
= 10/5
= 2
A = 4, B = -3 c1 = 5, c2 = 15
|c1 - c2|/[A2 + B2]1/2
= |5 - 15|/[(4)2 + (-3)2]½
= 10/5
= 2
A point P is in the interior of angle BAC, such that P lies on the bisector of angle BAC. What can be said about the distance of PM if PN = 2cm where PM and PN are perpendiculars from P on the lines BA and AC?- a)PM = 2 PN
- b)PM = 3 PN
- c)PM =2 cm
- d)PM =3.5 cm
Correct answer is option 'C'. Can you explain this answer?
A point P is in the interior of angle BAC, such that P lies on the bisector of angle BAC. What can be said about the distance of PM if PN = 2cm where PM and PN are perpendiculars from P on the lines BA and AC?
a)
PM = 2 PN
b)
PM = 3 PN
c)
PM =2 cm
d)
PM =3.5 cm
|
|
Krishna Iyer answered |
In Δ PAM & Δ PAN, angle (PAM) = angle (PAN) (Since AP bisects the angle (BAC)) angle (AMP) = angle (ANP) = 900
PA = PA (common)
By AAS congruence, Δ PAM and Δ PAN are congruent.
MP = NP = 2 cm
PA = PA (common)
By AAS congruence, Δ PAM and Δ PAN are congruent.
MP = NP = 2 cm
The tangent of the angle which the part of the line above the X-axis makes with the positive direction of the X-axis is:- a)Perpendicular line
- b)Slope of a line
- c)Concurrent line
- d)Parallel line
Correct answer is option 'B'. Can you explain this answer?
The tangent of the angle which the part of the line above the X-axis makes with the positive direction of the X-axis is:
a)
Perpendicular line
b)
Slope of a line
c)
Concurrent line
d)
Parallel line
|
Preeti Iyer answered |
The gradient or slope of a line (not parallel to the axis of y) is the trigonometrical tangent of the angle which the line makes with the positive direction of the x-axis. Thus, if a line makes an angle θ with the positive direction of the x-axis, then its slope will be tan θ.
The equation of the line parallel to the line 2x – 3y = 1 and passing through the middle point of the line segment joining the points (1, 3) and (1, –7), is:- a)2x – 3y – 8 = 0
- b)2x + 3y – 5 = 0
- c)4x – 6y + 7 = 0
- d)3x – 2y + 8 = 0
Correct answer is option 'A'. Can you explain this answer?
The equation of the line parallel to the line 2x – 3y = 1 and passing through the middle point of the line segment joining the points (1, 3) and (1, –7), is:
a)
2x – 3y – 8 = 0
b)
2x + 3y – 5 = 0
c)
4x – 6y + 7 = 0
d)
3x – 2y + 8 = 0
|
|
Neha Joshi answered |
The midpoint of the line segment is (1+1/2, 3-7/2)
= (1,-2)
the equation of the line parallel to the line 2x-3y = 1 is of the form 2x-3y = k
since it passes through (1,-2)
2(1) - 3(-2) = k
k = 8
hence the required equation is 2x-3y=8
= (1,-2)
the equation of the line parallel to the line 2x-3y = 1 is of the form 2x-3y = k
since it passes through (1,-2)
2(1) - 3(-2) = k
k = 8
hence the required equation is 2x-3y=8
Can you explain the answer of this question below:The points A and B have coordinates (3, 2) and (1, 4) respectively. So, the slope of any line perpendicular to AB is- A:2
- B:1
- C:-1
- D:-2
The answer is b.
The points A and B have coordinates (3, 2) and (1, 4) respectively. So, the slope of any line perpendicular to AB is
A:
2
B:
1
C:
-1
D:
-2
|
Geetika Shah answered |
If the lines are perpendicular to each other then their slopes are in the form m1.m2 = -1.(since product of slopes of two perpendicular lines is -1) Therefore , m = 1.
The value of y, for the line passing through (3, y) and (2, 7) is parallel to the line passing through (-1 , 4) and (0, 6) is:- a)9/7
- b)3/7
- c)9
- d)3
Correct answer is option 'C'. Can you explain this answer?
The value of y, for the line passing through (3, y) and (2, 7) is parallel to the line passing through (-1 , 4) and (0, 6) is:
a)
9/7
b)
3/7
c)
9
d)
3
|
|
Suresh Reddy answered |
As A(3,y) and B(2,7) is parallel to C(-1,4) and D(0,6)
∴ Their slopes are equal
so, (y-7)/(3-1) = (4-6)/(-1-0)
y-7 = 2
y = 9
∴ Their slopes are equal
so, (y-7)/(3-1) = (4-6)/(-1-0)
y-7 = 2
y = 9
What will be the new equation of the straight line 3x + 4y = 15, if the origin gets shifted to (1,-3) ?
- a)3x+4y=4
- b)3x-4y=4
- c)4x+3y=6
- d)3x+4y=6
Correct answer is option 'D'. Can you explain this answer?
What will be the new equation of the straight line 3x + 4y = 15, if the origin gets shifted to (1,-3) ?
a)
3x+4y=4
b)
3x-4y=4
c)
4x+3y=6
d)
3x+4y=6
|
|
Neha Sharma answered |
Equation : 3x + 4y = 15
Points : (1,-3)
3(x-1) + 4(y-(-3)) = 15
3(x-1) + 4(y+3) = 15
3x - 3 + 4y + 12 = 15
3x + 4y = 6
Points : (1,-3)
3(x-1) + 4(y-(-3)) = 15
3(x-1) + 4(y+3) = 15
3x - 3 + 4y + 12 = 15
3x + 4y = 6
The ratio in which the point R (1, 2) divides the line segment joining points P (2, 3) and Q (3, 5) is:- a)1 : 2, externally
- b)2 : 1, externally
- c)1 : 2, internally
- d)2 : 1, internally
Correct answer is option 'A'. Can you explain this answer?
The ratio in which the point R (1, 2) divides the line segment joining points P (2, 3) and Q (3, 5) is:
a)
1 : 2, externally
b)
2 : 1, externally
c)
1 : 2, internally
d)
2 : 1, internally
|
|
Gaurav Kumar answered |
P(2,3) Q(3,5) R(1,2)
R is at centre between P and Q, using section formula for internal division
Therefore, (1,2) = ((3λ+2)/(λ+1), (5λ+3)/(λ+1))
1 = (3λ+2)/(λ+1)
(λ+1) = (3λ+2)
λ = -1/2
- sign indicates the external division
R is at centre between P and Q, using section formula for internal division
Therefore, (1,2) = ((3λ+2)/(λ+1), (5λ+3)/(λ+1))
1 = (3λ+2)/(λ+1)
(λ+1) = (3λ+2)
λ = -1/2
- sign indicates the external division
Find the perpendicular distance from the origin of the line x + y – 2 = 0 is:- a)√2
- b)√3
- c)5√2
- d)3√3
Correct answer is option 'A'. Can you explain this answer?
Find the perpendicular distance from the origin of the line x + y – 2 = 0 is:
a)
√2
b)
√3
c)
5√2
d)
3√3
|
|
Raghav Bansal answered |
The given point is P(0,0) and the given line is x + y - 2 = 0
Let d be the length of the perpendicular from P(0,0) to the line x + y - 2 = 0
Then,
d = |(1 × 0) + (3 × 0) − 2|/(√12 + 12)
= 2/√2
= (2/√2) * (√2/√2)
= √2
Let d be the length of the perpendicular from P(0,0) to the line x + y - 2 = 0
Then,
d = |(1 × 0) + (3 × 0) − 2|/(√12 + 12)
= 2/√2
= (2/√2) * (√2/√2)
= √2
The line through the points (a , b) and (- a , - b) passes through the point- a)(1 , 1)
- b)(a2,ab)
- c)(3a , - 2b)
- d)none of these
Correct answer is option 'B'. Can you explain this answer?
The line through the points (a , b) and (- a , - b) passes through the point
a)
(1 , 1)
b)
(a2,ab)
c)
(3a , - 2b)
d)
none of these
|
|
Naina Sharma answered |
Slope of line passing through (a,b) and (−a,−b) is given by (b+b)/(a+a) = b/a
So equation of line passing is given by (using slope point form)
y−b = b/a(x−a)
⇒ ay − ab = bx − ab
⇒ ay = bx
Clearly the point (a2,ab) lie on the above line
So equation of line passing is given by (using slope point form)
y−b = b/a(x−a)
⇒ ay − ab = bx − ab
⇒ ay = bx
Clearly the point (a2,ab) lie on the above line
m1 and m2 are the slope of two perpendicular lines, if- a)m1.m2 = 1
- b)m1= m2
- c)m1 + m2 = 0
- d)1 + m1.m2 = 0
Correct answer is option 'D'. Can you explain this answer?
m1 and m2 are the slope of two perpendicular lines, if
a)
m1.m2 = 1
b)
m1= m2
c)
m1 + m2 = 0
d)
1 + m1.m2 = 0
|
|
Lakshmi Roy answered |
Explanation:
Two lines are perpendicular if the product of their slopes is -1. So, if m1 and m2 are the slopes of two perpendicular lines, then:
m1.m2 = -1
Rewriting this equation, we get:
1/m1 . 1/m2 = -1
Multiplying both sides by m1.m2, we get:
m2/m1 + m1/m2 = 0
This can be simplified as:
m1.m2 = 0
Therefore, the correct answer is option D.
Two lines are perpendicular if the product of their slopes is -1. So, if m1 and m2 are the slopes of two perpendicular lines, then:
m1.m2 = -1
Rewriting this equation, we get:
1/m1 . 1/m2 = -1
Multiplying both sides by m1.m2, we get:
m2/m1 + m1/m2 = 0
This can be simplified as:
m1.m2 = 0
Therefore, the correct answer is option D.
Two lines 3x+4y=8 and lx+my=n are perpendicular. Which of the following is true?- a)3m+4l=0
- b)3m-4l=0
- c)3l-4m=0
- d)3l+4m=0
Correct answer is option 'D'. Can you explain this answer?
Two lines 3x+4y=8 and lx+my=n are perpendicular. Which of the following is true?
a)
3m+4l=0
b)
3m-4l=0
c)
3l-4m=0
d)
3l+4m=0
|
|
Preeti Iyer answered |
In perpendicular, the slope of product = -1
Slope of L1 * slope of L2 = -1
-3/4 * (-l/m) = -1
=> 3l/4m = -1
=> 3l = -4m
= 3l + 4m = 0
Slope of L1 * slope of L2 = -1
-3/4 * (-l/m) = -1
=> 3l/4m = -1
=> 3l = -4m
= 3l + 4m = 0
The distance of (2,3) from x+y=1 is- a)2 units
- b)3√2 units
- c)4√2 units
- d)2√2 units
Correct answer is option 'D'. Can you explain this answer?
The distance of (2,3) from x+y=1 is
a)
2 units
b)
3√2 units
c)
4√2 units
d)
2√2 units
|
|
Lavanya Menon answered |
Given circle = (2,3)
Given line (x+y-1) = 0
Distance between point to line is:
d = |ax1 + by1 + c|/√(a2 + b2)
where a = 1, b = 1, c = -1 and x1 = 2, y1 = 3
d = |1(2) + 1(3) - 1|/√(1+1)
d = 4/(√2)
d = 2√2
Given line (x+y-1) = 0
Distance between point to line is:
d = |ax1 + by1 + c|/√(a2 + b2)
where a = 1, b = 1, c = -1 and x1 = 2, y1 = 3
d = |1(2) + 1(3) - 1|/√(1+1)
d = 4/(√2)
d = 2√2
The equation of a line whose perpendicular distance from the origin is 8 units and the angle made by perpendicular with positive x-axis is 60 degree is:- a)5x - 3y + 6 = 0
- b)x + √3y = 16
- c)√3x + y - 2 = 0
- d)√3x + √2y - 5 = 0
Correct answer is option 'B'. Can you explain this answer?
The equation of a line whose perpendicular distance from the origin is 8 units and the angle made by perpendicular with positive x-axis is 60 degree is:
a)
5x - 3y + 6 = 0
b)
x + √3y = 16
c)
√3x + y - 2 = 0
d)
√3x + √2y - 5 = 0
|
|
Gaurav Kumar answered |
If p isthe length of the normal from the origin to a line and ωis the angle made by the normal with the positive direction of thex-axis,then the equation of the line is given by xcosω +ysinω= p.
Here, p = 8 units and ω= 60°
Thus, therequired equation of the given line is
xcos 60° + y sin 60° = 8
x(1/2) + y(√3/2) = 8
x/2 + √3y/2 = 8
x + √3y = 16
Here, p = 8 units and ω= 60°
Thus, therequired equation of the given line is
xcos 60° + y sin 60° = 8
x(1/2) + y(√3/2) = 8
x/2 + √3y/2 = 8
x + √3y = 16
The acute angle between the lines ax + by + c = 0 and (a + b)x = (a – b)y , a ≠ b , is- a)450
- b)300
- c)600
- d)150
Correct answer is option 'A'. Can you explain this answer?
The acute angle between the lines ax + by + c = 0 and (a + b)x = (a – b)y , a ≠ b , is
a)
450
b)
300
c)
600
d)
150
|
Aryan Khanna answered |
ax + by + c = 0 and (a + b)x = (a – b)y
m1 = -a/b, m2 = (a+b)/(a-b)
tanx = [(m1-m2)/(1+m1×m2)]
=> {(-a/b)- (a+b)/(a-b)}/{1+(-a/b)[(a+b)/(a-b)]}
=> {-a2+ab-ab-b2}/{b(a-b)} * {ba-b2-a2-ab}/{b(a-b)}
=> (-a2-b2)/{1/(-a2-b2)
tanx = 1
x = tan-1(1)
Angle = 45o
m1 = -a/b, m2 = (a+b)/(a-b)
tanx = [(m1-m2)/(1+m1×m2)]
=> {(-a/b)- (a+b)/(a-b)}/{1+(-a/b)[(a+b)/(a-b)]}
=> {-a2+ab-ab-b2}/{b(a-b)} * {ba-b2-a2-ab}/{b(a-b)}
=> (-a2-b2)/{1/(-a2-b2)
tanx = 1
x = tan-1(1)
Angle = 45o
What will be the value of ‘p” if the equation of the straight line 3x + 5y = 10 gets changed to 3x + 5y = p after shifting the origin at (2,2) ?- a)6
- b)-6
- c)26
- d)none of these
Correct answer is option 'C'. Can you explain this answer?
What will be the value of ‘p” if the equation of the straight line 3x + 5y = 10 gets changed to 3x + 5y = p after shifting the origin at (2,2) ?
a)
6
b)
-6
c)
26
d)
none of these
|
|
Krishna Iyer answered |
3x + 5y = 10, at origin
But now, it’s (2,2),
3(x-2) + 5(y-2) = 10
Hence, 3x - 6 + 5y - 10 = 10
3x + 5y = 26
So, 3x + 5y = p
=> p = 26.
But now, it’s (2,2),
3(x-2) + 5(y-2) = 10
Hence, 3x - 6 + 5y - 10 = 10
3x + 5y = 26
So, 3x + 5y = p
=> p = 26.
Find the equation of the line whose intercepts on X and Y-axes are a2 and b2 respectively- a)bx + ay = ab
- b)b2x + a2b2y = a2
- c)b2a2x + a2y = b2
- d)b2x + a2y = a2b2
Correct answer is option 'D'. Can you explain this answer?
Find the equation of the line whose intercepts on X and Y-axes are a2 and b2 respectively
a)
bx + ay = ab
b)
b2x + a2b2y = a2
c)
b2a2x + a2y = b2
d)
b2x + a2y = a2b2
|
Defence Exams answered |
Consider the given points. (a2,0) and (0,b2)
We know that the equation of the line which is passing through the points
y−y1 =[ (y2−y1) / (x2−x1)] (x−x1)
So, y−0 = b2−0/0-a2(x-a2)
-ya2 - b2x = -b2a2
xb2 +ya2 = b2a2
We know that the equation of the line which is passing through the points
y−y1 =[ (y2−y1) / (x2−x1)] (x−x1)
So, y−0 = b2−0/0-a2(x-a2)
-ya2 - b2x = -b2a2
xb2 +ya2 = b2a2
The vertices of a triangle are (0 , 3) , (- 3 , 0) and (3 , 0). The orthocenter of the triangle is- a)(0 , 3)
- b)(- 3 , 0)
- c)(3 , 0)
- d)none of these.
Correct answer is option 'A'. Can you explain this answer?
The vertices of a triangle are (0 , 3) , (- 3 , 0) and (3 , 0). The orthocenter of the triangle is
a)
(0 , 3)
b)
(- 3 , 0)
c)
(3 , 0)
d)
none of these.
|
Dipika Choudhury answered |
Method to Solve :Triangle ABC, vertices are A(3,4), B(0,0), C(4,0)O is the Orthocentre of the triangleBy considering the coordinates of B, C, A ,we can conclude that:Equation of BC is y=0………..(1)Equati
Find the equation of line passing through the mid-point of line joining the points (3, 4) and (5, 6) and perpendicular to the equation of line 2x + 3y = 5.- a)2x – 3y – 2 = 0
- b)5x – 7y + 4 = 0
- c)3x – 2y – 2 = 0
- d)4x – 5y + 13 = 0
Correct answer is option 'C'. Can you explain this answer?
Find the equation of line passing through the mid-point of line joining the points (3, 4) and (5, 6) and perpendicular to the equation of line 2x + 3y = 5.
a)
2x – 3y – 2 = 0
b)
5x – 7y + 4 = 0
c)
3x – 2y – 2 = 0
d)
4x – 5y + 13 = 0
|
|
Suresh Iyer answered |
Mid Point [(3+5)/2, (4+6)/2]
= [8/2, 10/2]
= (4,5)
Perpendicular to the equation : 2x + 3y - 5 = 10
3x - 2y + k = 0
3(4) - 2(5) + k = 0
12 - 10 + k 0
k = -2
Therefore the equation is : 3x - 2y - 2 = 0
= [8/2, 10/2]
= (4,5)
Perpendicular to the equation : 2x + 3y - 5 = 10
3x - 2y + k = 0
3(4) - 2(5) + k = 0
12 - 10 + k 0
k = -2
Therefore the equation is : 3x - 2y - 2 = 0
The equation of the line having normal distance 2a from the origin and angle 60° which the normal makes with the positive direction of X-axis is:
- a)x + √2y = 5a
- b)x + √3y = 4a
- c)x + 4y +5 = 0
- d)x + 2y = 3a
Correct answer is option 'B'. Can you explain this answer?
The equation of the line having normal distance 2a from the origin and angle 60° which the normal makes with the positive direction of X-axis is:
a)
x + √2y = 5a
b)
x + √3y = 4a
c)
x + 4y +5 = 0
d)
x + 2y = 3a
|
|
Knowledge Hub answered |
We know that, If p is the length of the normal from origin to a line and w is the angle made by the normal with the positive direction of x-axis then the equation of the line is given by xcosw + ysinw = p.
Here, p=2a units and w=60°
Thus, the required equation of the given line is
xcos60° + ysin60° = 2a
x(1/2) + y(√3/2) = 2a
x + √3y = 4a
Here, p=2a units and w=60°
Thus, the required equation of the given line is
xcos60° + ysin60° = 2a
x(1/2) + y(√3/2) = 2a
x + √3y = 4a
What will be the new equation of the straight line 5x + 8y = 10, if the origin gets shifted to (2,-3) ?- a)5x + 8y = -4
- b)5x + 8y = – 14
- c)5x – 8y = 14
- d)8x + 5y = 24
Correct answer is option 'A'. Can you explain this answer?
What will be the new equation of the straight line 5x + 8y = 10, if the origin gets shifted to (2,-3) ?
a)
5x + 8y = -4
b)
5x + 8y = – 14
c)
5x – 8y = 14
d)
8x + 5y = 24
|
|
Rajesh Gupta answered |
Equation : 5x + 8y = 10
Points (2, -3)
(x-2, y+3)
⇒ 5(x-2) + 8(y+3) = 10
= 5x - 10 + 8y + 24 = 10
⇒ 5x + 8 = - 4
Points (2, -3)
(x-2, y+3)
⇒ 5(x-2) + 8(y+3) = 10
= 5x - 10 + 8y + 24 = 10
⇒ 5x + 8 = - 4
The coordinates of centroid of triangle whose vertices are A(-1,-3), B(5,-6) and C(2,3) and origin gets shifted to (1,2) - a)(4,-1)
- b)(2,-2)
- c)(0,0)
- d)(-1,4)
Correct answer is option 'D'. Can you explain this answer?
The coordinates of centroid of triangle whose vertices are A(-1,-3), B(5,-6) and C(2,3) and origin gets shifted to (1,2)
a)
(4,-1)
b)
(2,-2)
c)
(0,0)
d)
(-1,4)
|
Ishani Menon answered |
Finding the Centroid of a Triangle:
The centroid of a triangle is the point where the three medians of the triangle intersect. A median is a line segment that connects a vertex of the triangle to the midpoint of the opposite side. To find the centroid, we need to find the midpoint of each side and then find the intersection point of the three medians.
Finding the Midpoint of a Side:
The midpoint of a line segment is the point that is halfway between the two endpoints. To find the midpoint of a line segment, we use the midpoint formula:
Midpoint = ((x1 + x2) / 2 , (y1 + y2) / 2)
where (x1, y1) and (x2, y2) are the coordinates of the endpoints.
Finding the Coordinates of the Centroid:
Once we have the midpoints of each side, we can find the equation of the three medians and then find their intersection point. The intersection point is the centroid of the triangle.
Let's find the midpoint of each side first:
Midpoint of AB = ((-1 + 5) / 2 , (-3 - 6) / 2) = (2, -4.5)
Midpoint of AC = ((-1 + 2) / 2 , (-3 + 3) / 2) = (0.5, 0)
Midpoint of BC = ((5 + 2) / 2 , (-6 + 3) / 2) = (3.5, -1.5)
Now we can find the equation of the medians. The equation of a line that passes through two points (x1, y1) and (x2, y2) is:
(y - y1) / (y2 - y1) = (x - x1) / (x2 - x1)
Let's find the equation of the median that passes through A:
Midpoint of BC = (3.5, -1.5)
Slope of BC = (-6 - 3) / (5 - 2) = -3
Equation of BC: y + 6 = -3(x - 5) => y = -3x + 21
Slope of median from A to BC = (3 - (-3)) / (2 - 5) = 2/3
Equation of median from A to BC:
y + 3 = (2/3)(x + 1) => y = (2/3)x + (5/3)
Similarly, we can find the equations of the medians that pass through B and C:
Equation of median from B to AC: y + 6 = (1/3)(x - 5) => y = (1/3)x + 11/3
Equation of median from C to AB: y - 3 = (-2/3)(x - 2) => y = (-2/3)x + 5/3
Now we need to find the intersection point of these three medians. We can solve the system of equations:
y = (2/3)x + (5/3)
y = (1/3)x + 11/3
y = (-2/3)x + 5/3
Solving these equations, we get x = -1 and y =
The centroid of a triangle is the point where the three medians of the triangle intersect. A median is a line segment that connects a vertex of the triangle to the midpoint of the opposite side. To find the centroid, we need to find the midpoint of each side and then find the intersection point of the three medians.
Finding the Midpoint of a Side:
The midpoint of a line segment is the point that is halfway between the two endpoints. To find the midpoint of a line segment, we use the midpoint formula:
Midpoint = ((x1 + x2) / 2 , (y1 + y2) / 2)
where (x1, y1) and (x2, y2) are the coordinates of the endpoints.
Finding the Coordinates of the Centroid:
Once we have the midpoints of each side, we can find the equation of the three medians and then find their intersection point. The intersection point is the centroid of the triangle.
Let's find the midpoint of each side first:
Midpoint of AB = ((-1 + 5) / 2 , (-3 - 6) / 2) = (2, -4.5)
Midpoint of AC = ((-1 + 2) / 2 , (-3 + 3) / 2) = (0.5, 0)
Midpoint of BC = ((5 + 2) / 2 , (-6 + 3) / 2) = (3.5, -1.5)
Now we can find the equation of the medians. The equation of a line that passes through two points (x1, y1) and (x2, y2) is:
(y - y1) / (y2 - y1) = (x - x1) / (x2 - x1)
Let's find the equation of the median that passes through A:
Midpoint of BC = (3.5, -1.5)
Slope of BC = (-6 - 3) / (5 - 2) = -3
Equation of BC: y + 6 = -3(x - 5) => y = -3x + 21
Slope of median from A to BC = (3 - (-3)) / (2 - 5) = 2/3
Equation of median from A to BC:
y + 3 = (2/3)(x + 1) => y = (2/3)x + (5/3)
Similarly, we can find the equations of the medians that pass through B and C:
Equation of median from B to AC: y + 6 = (1/3)(x - 5) => y = (1/3)x + 11/3
Equation of median from C to AB: y - 3 = (-2/3)(x - 2) => y = (-2/3)x + 5/3
Now we need to find the intersection point of these three medians. We can solve the system of equations:
y = (2/3)x + (5/3)
y = (1/3)x + 11/3
y = (-2/3)x + 5/3
Solving these equations, we get x = -1 and y =
The slope of line when coordinates of any two point A (x1, y1) and B (x2, y2) on the line are given is:- a)
- b)
- c)
- d)
Correct answer is option 'D'. Can you explain this answer?
The slope of line when coordinates of any two point A (x1, y1) and B (x2, y2) on the line are given is:
a)
b)
c)
d)
|
Hansa Sharma answered |
Let the line make an angle P with the positive direction of the x-axis (see Fig.). Draw AL and BM perpendicular to x-axis and AN perpendicular to BM. Clearly, ∠NAB = θ
tanθ = BN/AN = BN/LM
= BM-MN/OM-OL
= BM-AL/OM-OL = y2-y1/x2-x1
Which of the following lines is parallel to the line with equation 2x+y=3?- a)x+2y=1
- b)x-y=2
- c)3x+y=3
- d)4x+2y=6
Correct answer is option 'D'. Can you explain this answer?
Which of the following lines is parallel to the line with equation 2x+y=3?
a)
x+2y=1
b)
x-y=2
c)
3x+y=3
d)
4x+2y=6
|
|
Raghav Bansal answered |
This question can be done by picking the options.
a1/a2 = b1/b2 = c1/c2
Equation : 2x + y = 3
(d) 4x + 2y = 6
a1 = 2⇒, b1 = 1, c1 = 3
a2 = 4, b2 = 2, c2 = 6
⇒ 2/4 = 1/2 = 3/6
=> 1/2 = 1/2 = 1/2
a1/a2 = b1/b2 = c1/c2
Equation : 2x + y = 3
(d) 4x + 2y = 6
a1 = 2⇒, b1 = 1, c1 = 3
a2 = 4, b2 = 2, c2 = 6
⇒ 2/4 = 1/2 = 3/6
=> 1/2 = 1/2 = 1/2
Two lines are said to be parallel when the difference of their slopes is- a)-1
- b)0
- c)1
- d)2
Correct answer is option 'B'. Can you explain this answer?
Two lines are said to be parallel when the difference of their slopes is
a)
-1
b)
0
c)
1
d)
2
|
|
Aryan Khanna answered |
Parallel lines and their slopes are easy. Since slope is a measure of the angle of a line from the horizontal, and since parallel lines must have the same angle, then parallel lines have the same slope and lines with the same slope are parallel.
Considering a fixed distance from the origin, how many points can be plotted on the y axis?- a)2
- b)3
- c)4
- d)no points
Correct answer is option 'A'. Can you explain this answer?
Considering a fixed distance from the origin, how many points can be plotted on the y axis?
a)
2
b)
3
c)
4
d)
no points
|
|
Neha Joshi answered |
Let the fixed distance be d. Since the point has to lie on Y axis, therefore abscissa of such points is 0. Hence the points can be (0, d) and (0, -d). So, Option A is the correct answer.PS: This is the link to my YouTube channel Mechnovashia:https://www.youtube.com/channel/UCI89XUuzHmY8xLV4ovBWWUwSubscribe to this channel for more such simplified explanations and concepts. ☺️
General equation of a straight line is:- a)y = mx + c
- b)Ax + By + C = 0
- c)Ax2 + By2 + C = 0
- d)-A/B + By + C/B = 0
Correct answer is option 'B'. Can you explain this answer?
General equation of a straight line is:
a)
y = mx + c
b)
Ax + By + C = 0
c)
Ax2 + By2 + C = 0
d)
-A/B + By + C/B = 0
|
|
Raghav Bansal answered |
The equation Ax + By + c = 0 is the most general equation for a straight line, and can be used where other forms of equation are not suitable.
Find the value of x for which the points (1,3) , (-2, 9) and (x, -1) are collinear.
- a) -3
- b) 3
- c) 1
- d) 1/2
Correct answer is option 'B'. Can you explain this answer?
Find the value of x for which the points (1,3) , (-2, 9) and (x, -1) are collinear.
a)
-3b)
3c)
1d)
1/2
|
Sushil Kumar answered |
Let A(1,3), B(-2,9), and C(x,-1)
For to be points collinear,
x1(y2-y3) + x2(y3-y1) + x3(y1-y2)=0
⇒ 1(9-(-1)) + (-2)(-1-3) + x(3-9)=0
⇒ 1(10)+(-2)(-4)+x(-6)=0
⇒ 10+8-6x=0
⇒ 18 = 6x
⇒ x = 3
For to be points collinear,
x1(y2-y3) + x2(y3-y1) + x3(y1-y2)=0
⇒ 1(9-(-1)) + (-2)(-1-3) + x(3-9)=0
⇒ 1(10)+(-2)(-4)+x(-6)=0
⇒ 10+8-6x=0
⇒ 18 = 6x
⇒ x = 3
The area of the triangle whose vertices are (2, 3), (3, 5) and (7, 5) is:- a)2 units
- b)7 units
- c)7/2 units
- d)4 units
Correct answer is option 'D'. Can you explain this answer?
The area of the triangle whose vertices are (2, 3), (3, 5) and (7, 5) is:
a)
2 units
b)
7 units
c)
7/2 units
d)
4 units
|
|
Preeti Iyer answered |
Area of triangle = 1/2 × base × height
= 1/2×(7−3)×(5−3)
= 1/2*(4×2)
= 4 sq units.
= 1/2×(7−3)×(5−3)
= 1/2*(4×2)
= 4 sq units.
The line through the point (a, b) and parallel to the line Ax + By + C = 0 is:- a)Ax + By + C (a + b) = 0
- b)Ax + By + k = 0
- c)Ax + Bb + (a + b) = 0
- d)A (x – a) + B (y – b) = 0
Correct answer is option 'D'. Can you explain this answer?
The line through the point (a, b) and parallel to the line Ax + By + C = 0 is:
a)
Ax + By + C (a + b) = 0
b)
Ax + By + k = 0
c)
Ax + Bb + (a + b) = 0
d)
A (x – a) + B (y – b) = 0
|
|
Geetika Shah answered |
Equation of line passing through point (a,b) and parallel to line Ax + By + C = 0
(y-y1) = -A/B(x-x1)
(y-b) = -A/B(x-a)
A (x – a) + B (y – b) = 0
(y-y1) = -A/B(x-x1)
(y-b) = -A/B(x-a)
A (x – a) + B (y – b) = 0
Projection (the foot of perpendicular) from (x , y) on the x – axis is- a)(0 , y)
- b)(0 , - y)
- c)(- x , 0)
- d)(x , 0)
Correct answer is option 'D'. Can you explain this answer?
Projection (the foot of perpendicular) from (x , y) on the x – axis is
a)
(0 , y)
b)
(0 , - y)
c)
(- x , 0)
d)
(x , 0)
|
|
Sarita Yadav answered |
Slope of a line is not defined if the line is parallel to x axis as all points on the line have same x coordinate.
The distance of the point (x , y) from Y axis is- a)x
- b)|x|
- c)|y|
- d)y
Correct answer is option 'B'. Can you explain this answer?
The distance of the point (x , y) from Y axis is
a)
x
b)
|x|
c)
|y|
d)
y
|
|
Geetika Shah answered |
Distance is a metric, a bifunction d, which is always non-negative, along with being symmetric, satisfies the triangle inequality, and the identity of indiscernibles (i.e., d(x,y)=0⟺x=y)
d(x,y)=0⟺x=y). The “nearest” distance from a point (x1,y1) to the y-axis in 2-space is along the line y=y1. (orthogonal to the y-axis). This line interests the y-axis exactly at the point (0,y1). Using the Euclidean distance metric on R2, one obtains:
d((x,y),(0,y))= √(x−0)2+(y−y)2
= |x|
d(x,y)=0⟺x=y). The “nearest” distance from a point (x1,y1) to the y-axis in 2-space is along the line y=y1. (orthogonal to the y-axis). This line interests the y-axis exactly at the point (0,y1). Using the Euclidean distance metric on R2, one obtains:
d((x,y),(0,y))= √(x−0)2+(y−y)2
= |x|
Slope of a line is not defined if the line is- a)parallel to the line x – y
- b)parallel to the line x+y
- c)parallel to Y axis
- d)parallel to X axis
Correct answer is option 'C'. Can you explain this answer?
Slope of a line is not defined if the line is
a)
parallel to the line x – y
b)
parallel to the line x+y
c)
parallel to Y axis
d)
parallel to X axis
|
|
Sanghamitra Maithani answered |
As the formula for slope is given by tan theta . where theta is angle from +ve X axis and if the line is parallel to y axis then the angle from X axis is 90 and tan 90 is not defined.
The line x + y – 6 = 0 is the right bisector of the segment [PQ]. If P is the point (4, 3) , then the point Q is- a)(4 , 4)
- b)(3, 4)
- c)(3 , 3)
- d)(3, 2)
Correct answer is option 'D'. Can you explain this answer?
The line x + y – 6 = 0 is the right bisector of the segment [PQ]. If P is the point (4, 3) , then the point Q is
a)
(4 , 4)
b)
(3, 4)
c)
(3 , 3)
d)
(3, 2)
|
|
Arjun Singhania answered |
X-4y-6 = 0
Express in the form of y=mx+c, where m is the slope
x-4y-6=0
-4y = -x+6
y=1/4 x - 3,2
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