Free Objective Test 01 Practice Test - 11th and 12th
Question 1
If A(4, -3), B(3, -2) and C(2, 8) are the vertices of a triangle, then its centroid will be
(-3,3)
(3,3)
(3,1)
(1,3)
SOLUTION
Solution : 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=3y=−3−2+83=1
Question 2
If two vertices of a triangle are (6,4), (2,6) and its centroid is (4, 6), then the third vertex is
(4,8)
(8,4)
(6,4)
(0,0)
SOLUTION
Solution : 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=6Centroid (4,6)=(x1+x2+x33,y1+y2+y33)
⇒4=6+2+x33 and 6=4+6+y33
⇒ x3=4 and y3=8∴ Third vertex is (4,8).
Question 3
The equation of the internal bisector of ∠BAC of ΔABC with vertices A(5, 2), B(2, 3) and C(6, 5), is
SOLUTION
Solution : C
Let AD be the internal bisector of angle BAC cutting BC at D.Now, AB=√(5−2)2+(2−3)2=√10and AC=√(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
Question 4
A line through the point A(2, 0), which makes an angle of 30∘ with the positive direction of x-axis is rotated about A in clockwise direction through an angle 15∘. The equation of the straight line in the new position is
(2−√3)x−y−4+2√3=0
(2−√3)x+y−4+2√3=0
(2−√3)x−y+4+2√3=0
(2−√3)x−9y+4+2√3=0
SOLUTION
Solution : A
Let AB be the initial position of the line and AC be its new position.
Slope of the line
AC=tan 15∘=(2−√3)
∴ Equation of the line AC is
(y−0)=(2−√3)(x−2)or y=(2−√3)x−4+2√3or (2−√3)x−y−4+2√3=0
Question 5
If the coordinates of the points A, B, C, be (4,4), (3,-2) and (3,-16) respectively, then the area of the triangle ABC is
27
15
18
7
SOLUTION
Solution : D
△ = 12[4(- 2 + 16) + 3(-16 - 4) + 3(4 + 2)]
= 12 [56 - 60 + 18] = 7.
Question 6
If PM is the perpendicular from P (2,3) on to the line x+y=3 then the co-ordinates of M are
(2,1)
(−1,4)
(1,2)
(4,−1)
SOLUTION
Solution : 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)
Question 7
If 4a+5b+6c=0 then the set of lines ax+by+c=0 are concurrent at the point
(23,56)
(13,12)
(12,43)
(13,73)
SOLUTION
Solution : 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)
Question 8
If the line
(3x+14y+7)+k(5x+7y+6)=0
is parallel to the y-axis, then the value of k is
SOLUTION
Solution : C
Given line is (3x+14y+7)+k(5x+7y+6)=0
⇒(3+5k)x+(14+7k)y+(7+6k)=0
If it is parallel to y- axis, then coefficient of y =0
⇒14+7k=0
⇒k=−2
Question 9
The angle between the pair of straight lines x2−y2−2y−1=0, is
90∘
60∘
75∘
36∘
SOLUTION
Solution : A
Pair of straight lines represented by a second degree equation with coefficient of x2 as a and coefficient of y2 as b are perpendicular if a+b = 0
Here a + b = 0, so they are perpendicular to each other
Question 10
If the lines (p−q)x2+2(p+q)xy+(q−p)y2=0 are mutually perpendicular, then
p = q
q = 0
p = 0
p and q may have any value
SOLUTION
Solution : D
Pair of straight lines represented by a second degree equation with coefficient of x2 as a and coefficient of y2 as b are perpendicular if a+b = 0. Herere a + b = 0 for every p and q.
Question 11
The lines joining the points of intersection of line x + y = 1 and curve x2+y2−2y+λ=0 to the origin are perpendicular, then the value of λ will be
12
−12
1√2
0
SOLUTION
Solution : 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
Question 12
Two points A and B have coordinates (1, 0) and (-1, 0) respectively and Q is a point which satisfies the relation
AQ - BQ = ± 1. The locus of Q is
12x2+4y2=3
12x2−4y2=3
12x2−4y2+3 = 0
12x2+4y2+3 = 0
SOLUTION
Solution : 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.
Question 13
The orthocenter of the triangle formed by the lines x + y = 1, 2x + 3y = 6 and 4x - y + 4 = 0 lies in
SOLUTION
Solution : 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.
Question 14
If h denote the A.M, k denote G.M of the intercepts made on axes by the lines passing through (1, 1) then (h, k) lies on
SOLUTION
Solution : 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
Question 15
If the orthocenter and circumcentre of a triangle are (0,0) and (3,6) respectively then the centroid of the triangle is
SOLUTION
Solution : 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).