12th Grade > Mathematics
STRAIGHT LINES MCQs
Straight Lines
Total Questions : 60
| Page 2 of 6 pages
Answer: Option C. -> (2, - 3)
:
C
Putting x = x' + h, y = y' + k, the given equation transforms to
x′2+y′2+x′(2h−4)+y′(2k+6)+h2+k2−7=0
To eliminate linear terms, we should have
2h - 4 = 0, 2k + 6 = 0 ⇒ h = 2, k = -3
i.e., (h, k) = (2, -3).
:
C
Putting x = x' + h, y = y' + k, the given equation transforms to
x′2+y′2+x′(2h−4)+y′(2k+6)+h2+k2−7=0
To eliminate linear terms, we should have
2h - 4 = 0, 2k + 6 = 0 ⇒ h = 2, k = -3
i.e., (h, k) = (2, -3).
Answer: Option C. -> 2x + y – 12 = 0
:
C
Let AD be the internal bisector of angle BAC cutting BC at D.
Now,AB=√(5−2)2+(2−3)2=√10andAC=√(5−6)2+(2−5)2=√10
since AD is the internal bisector of angle BAC,
∴BDDC=ABAC=√10√10=11
∴ Coordinates of D are (2+62,3+52) i.e. (4, 4)
So, the equation of AD is
y−2=2−45−4 (x – 5) or 2x + y – 12 = 0
:
C
Let AD be the internal bisector of angle BAC cutting BC at D.
Now,AB=√(5−2)2+(2−3)2=√10andAC=√(5−6)2+(2−5)2=√10
since AD is the internal bisector of angle BAC,
∴BDDC=ABAC=√10√10=11
∴ Coordinates of D are (2+62,3+52) i.e. (4, 4)
So, the equation of AD is
y−2=2−45−4 (x – 5) or 2x + y – 12 = 0
Answer: Option B. -> (2,4)
:
B
In any triangle centroid divides the line joining orthocenter and circumcentre internally in the ratio 2 : 1.
Applying section formula to find the point which divides the line joining (0,0) in the ratio 2:1 , we get the coordinated of centroid equal to (2,4).
:
B
In any triangle centroid divides the line joining orthocenter and circumcentre internally in the ratio 2 : 1.
Applying section formula to find the point which divides the line joining (0,0) in the ratio 2:1 , we get the coordinated of centroid equal to (2,4).
Answer: Option D. -> p and q may have any value
:
D
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. Herere a + b = 0 for every p and q.
:
D
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. Herere a + b = 0 for every p and q.
Answer: Option C. -> 2x + y – 12 = 0
:
C
Let AD be the internal bisector of angle BAC cutting BC at D.
Now,AB=√(5−2)2+(2−3)2=√10andAC=√(5−6)2+(2−5)2=√10
since AD is the internal bisector of angle BAC,
∴BDDC=ABAC=√10√10=11
∴ Coordinates of D are (2+62,3+52) i.e. (4, 4)
So, the equation of AD is
y−2=2−45−4 (x – 5) or 2x + y – 12 = 0
:
C
Let AD be the internal bisector of angle BAC cutting BC at D.
Now,AB=√(5−2)2+(2−3)2=√10andAC=√(5−6)2+(2−5)2=√10
since AD is the internal bisector of angle BAC,
∴BDDC=ABAC=√10√10=11
∴ Coordinates of D are (2+62,3+52) i.e. (4, 4)
So, the equation of AD is
y−2=2−45−4 (x – 5) or 2x + y – 12 = 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 B. -> 12x2−4y2=3
:
B
According to the given condition
√(x−1)2+y2−√(x+1)2+y2 = ±1
On squaring both sides, we get
2x2+2y2+1=2√(x−1)2+y2√(x+1)2+y2
Again on squaring, we get 12x2−4y2=3.
:
B
According to the given condition
√(x−1)2+y2−√(x+1)2+y2 = ±1
On squaring both sides, we get
2x2+2y2+1=2√(x−1)2+y2√(x+1)2+y2
Again on squaring, we get 12x2−4y2=3.
Answer: Option A. -> y2=2x
:
A
a = x - intercept, b =y - intercept
2h=a+b,k2=ab
xa+yb=1, substitute (1, 1)
1a+1b=1
a + b = ab
2h=k2⇒y2=2x
:
A
a = x - intercept, b =y - intercept
2h=a+b,k2=ab
xa+yb=1, substitute (1, 1)
1a+1b=1
a + b = ab
2h=k2⇒y2=2x
Answer: Option C. -> (3,1)
:
C
Let the centroid of the triangle be (x, y).
The centroid of a triangle is given by (x1+x2+x33,y1+y2+y33)
x=4+3+23=3
y=−3−2+83=1
:
C
Let the centroid of the triangle be (x, y).
The centroid of a triangle is given by (x1+x2+x33,y1+y2+y33)
x=4+3+23=3
y=−3−2+83=1
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