12th Grade > Mathematics
STRAIGHT LINES MCQs
Straight Lines
Total Questions : 60
| Page 1 of 6 pages
Answer: Option A. -> 90∘
:
A
Pair of straight lines represented by a second degree equation with coefficient of x2 as a and coefficient of y2 as bare perpendicular if a+b = 0
Here a + b = 0,so they are perpendicular to each other
:
A
Pair of straight lines represented by a second degree equation with coefficient of x2 as a and coefficient of y2 as bare perpendicular if a+b = 0
Here a + b = 0,so they are perpendicular to each other
Answer: Option C. -> (1,2)
:
C
If(x,y) is the foot of the perpendicular from (x1,y1) to the line ax + by + c = 0 then
x−x1a=y−y1b=−(ax1+by1+c)a2+b2
Here (x1,y1)=(2,3);ax+by+c=x+y−3
∴ x−21=y−31=−((1×2)+(1×3)+(−3))12+12
⇒ x−21=y−31=−((2)+(3)+(−3))2
⇒ x−21=y−31=−22
⇒ x−21=y−31=−1
∴ x - 2 = -1 ; y-3 = -1
⇒ x = -1 + 2 ; y = -1 + 3
⇒ x = 1 ; y = 2
∴ (x,y) = (1,2)
:
C
If(x,y) is the foot of the perpendicular from (x1,y1) to the line ax + by + c = 0 then
x−x1a=y−y1b=−(ax1+by1+c)a2+b2
Here (x1,y1)=(2,3);ax+by+c=x+y−3
∴ x−21=y−31=−((1×2)+(1×3)+(−3))12+12
⇒ x−21=y−31=−((2)+(3)+(−3))2
⇒ x−21=y−31=−22
⇒ x−21=y−31=−1
∴ x - 2 = -1 ; y-3 = -1
⇒ x = -1 + 2 ; y = -1 + 3
⇒ x = 1 ; y = 2
∴ (x,y) = (1,2)
Answer: Option A. -> (4,8)
:
A
Given:
Centroid =(4,6)
Vertices(6,4)&(2,6)
Let the Co-ordinates of C be (x3,y3)
x1=6,x2=32,y1=4&y2=6
Centroid (4,6)=(x1+x2+x33,y1+y2+y33)
⇒4=6+2+x33 and6=4+6+y33
⇒ x3=4 andy3=8
∴ Third vertex is (4,8).
:
A
Given:
Centroid =(4,6)
Vertices(6,4)&(2,6)
Let the Co-ordinates of C be (x3,y3)
x1=6,x2=32,y1=4&y2=6
Centroid (4,6)=(x1+x2+x33,y1+y2+y33)
⇒4=6+2+x33 and6=4+6+y33
⇒ x3=4 andy3=8
∴ Third vertex is (4,8).
Answer: Option A. -> (23,56)
:
A
4a+5b+6c=0
(6x)a+(6y)b+6c=0 →[Multiply given set of lines ax+by+c=0 with '6']
Now on comparing 6x=4 and 6y =5
⇒ (x,y) = (23,56)
∴ ax+by+c=0 must passes through(23,56)
:
A
4a+5b+6c=0
(6x)a+(6y)b+6c=0 →[Multiply given set of lines ax+by+c=0 with '6']
Now on comparing 6x=4 and 6y =5
⇒ (x,y) = (23,56)
∴ ax+by+c=0 must passes through(23,56)
Answer: Option D. -> 7
:
D
△ = 12[4(- 2 + 16) + 3(-16 - 4) + 3(4 + 2)]
= 12 [56 - 60 + 18] = 7.
:
D
△ = 12[4(- 2 + 16) + 3(-16 - 4) + 3(4 + 2)]
= 12 [56 - 60 + 18] = 7.
Answer: Option B. -> On x-axis
:
B
x = a+b+c3, y = b−c+c−a+a−b3 = 0
Hence, centroid lies on x - axis.
:
B
x = a+b+c3, y = b−c+c−a+a−b3 = 0
Hence, centroid lies on x - axis.
Answer: Option A. -> 90∘
:
A
Pair of straight lines represented by a second degree equation with coefficient of x2 as a and coefficient of y2 as bare perpendicular if a+b = 0
Here a + b = 0,so they are perpendicular to each other
:
A
Pair of straight lines represented by a second degree equation with coefficient of x2 as a and coefficient of y2 as bare perpendicular if a+b = 0
Here a + b = 0,so they are perpendicular to each other
Answer: Option D. -> 0
:
D
Making the equation of curve homogeneous with the help of line x + y =1,we get
x2+y2−2y(x+y)+λ(x+y)2=0
⇒x2(1+λ)+y2(−1+λ)−2yx=0
Therefore the lines be perpendicular, if A +B = 0.
⇒1+λ−1+λ=0⇒λ=0
:
D
Making the equation of curve homogeneous with the help of line x + y =1,we get
x2+y2−2y(x+y)+λ(x+y)2=0
⇒x2(1+λ)+y2(−1+λ)−2yx=0
Therefore the lines be perpendicular, if A +B = 0.
⇒1+λ−1+λ=0⇒λ=0
Answer: Option A. -> I quadrant
:
A
Coordinates of A and B are (-3, 4) and (−35,85) if orthocenter P(h, k)
Then, (slope of PA)× (slope of BC) = - 1
k−4h+3×4=−1
⇒ 4k - 16 = -h - 3
⇒ h + 4k = 13....(i)
and slope of PB× slope of AC = - 1
⇒k−85h+35×−23=−1
⇒5k−85h+3×23=1
⇒ 10k - 16 = 15th + 9
15th - 10k + 25 = 10
3h - 2k + 5 = 0 ...(ii)
Solving Eqs. (i) and (ii), we get h=37,k=227
Hence, orthocentre lies in I quadrant.
:
A
Coordinates of A and B are (-3, 4) and (−35,85) if orthocenter P(h, k)
Then, (slope of PA)× (slope of BC) = - 1
k−4h+3×4=−1
⇒ 4k - 16 = -h - 3
⇒ h + 4k = 13....(i)
and slope of PB× slope of AC = - 1
⇒k−85h+35×−23=−1
⇒5k−85h+3×23=1
⇒ 10k - 16 = 15th + 9
15th - 10k + 25 = 10
3h - 2k + 5 = 0 ...(ii)
Solving Eqs. (i) and (ii), we get h=37,k=227
Hence, orthocentre lies in I quadrant.
Answer: Option C. -> 3¯x =2α−u and 3¯y =2β+v
:
C
We know that, the centroid of a triangle divides the segment joining the orthocentre and circumcentre internally in the ratio 2 : 1. Therefore,
¯x=2α+u2+1and¯y=2β+v2+1
⇒3¯x=2α+uand¯y=2β+v
:
C
We know that, the centroid of a triangle divides the segment joining the orthocentre and circumcentre internally in the ratio 2 : 1. Therefore,
¯x=2α+u2+1and¯y=2β+v2+1
⇒3¯x=2α+uand¯y=2β+v
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