Question 1 1 / -0
The points A(- 4, - 1), B(- 2, - 4), C(4, 0) and D(2, 3) are the vertices of a
Solution
AB =
=
=
BC =
=
=
CD =
=
=
DA =
=
=
Since AB = CD and BC = DA. But as the four sides are not equal, it is neither a rhombus nor a square.
Also, AC =
=
=
Since AB
2 + BC
2 = AC
2 , so AB and BC are perpendicular.
Hence, ABCD is a rectangle. Since, every rectangle is parallelogram. Hence, option (4) is correct.
Question 2 1 / -0
The points A(1, 3) and C(5, 1) are opposite vertices of a rectangle ABCD. If the slope of BD is 2, then its equation is:
Solution
BD will pass through the point of intersection of diagonals.
The point of intersection of diagonals is the midpoint of BD or AC.
Coordinates of midpoint (point of intersection of diagonals) =
= (3, 2)
Equation of BD would be:
y - 2 = 2(x - 3)
y = 2x - 4 or 2x - y = 4
Question 3 1 / -0
To which quadrant does the point (1, -2) belong?
Solution
As x is positive and y is negative, so (1, – 2) lies in the 4th quadrant.
Question 4 1 / -0
Find the value of k if the area of the triangle formed by (1, k), (4, - 3) and (- 9, 7) is 15 sq. units.
Solution
|1(- 3 - 7) + 4(7 - k) - 9(k + 3)| = 15
|- 10 + 28 - 4k - 9k - 27| = 30
|- 13k - 9| = 30
- 13k - 9 = 30 or 13k + 9 = 30
k = - 3 or k =
.
Option 1 is correct.
Question 5 1 / -0
What would be the equation of one of the sides of an isosceles right-angled triangle, whose hypotenuse is 3x + 4y = 4 and the opposite vertex of the hypotenuse is (2, 2)?
Question 6 1 / -0
If a, b and c are in A.P, through which fixed point will the straight line ax + 2by + c = 0 always pass?
Solution
As 2b = 2 + C equation of the give is
Question 7 1 / -0
What is the value of k, if (-3, 12), (7, 6) and (k, 9) are collinear?
Solution
If ABC are collinear, then area of
ABC = 0.
(x
1 , y
1 ) = (-3, 12)
(x
2 , y
2 ) = (7, 6)
(x
3 , y
3 ) = (k, 9)
Area of triangle ABC = 0.
[-3(6 - 9) + 7(9 - 12) + k(12 - 6)] = 0
9 - 21 + 6k = 0
k = 2
Option 3 is correct.
Question 8 1 / -0
In which quadrant does the point (1, 2) lie?
Solution
As both x and y are positive, therefore the point lies in the 1st quadrant.
Question 9 1 / -0
The line through the points (a, b) and (- a, - b) passes through the point
Solution
Equation of line passing through given points is
y - b =
(x - a) or
= 0
From the given option only (a
2 , ab) lies on this line.
Question 10 1 / -0
The points A(2a, 4a), B(2a, 6a) and C(2a +
, 5a), where a > 0, are the vertices of
Solution
Length of side AB =
units
= 2a units
Length of side BC =
units= 2a units and length of side CA =
units= 2a units
Since AB = BC = CA, the triangle is equilateral.
Question 11 1 / -0
The four points P (0, 5), Q (- 2, - 2), R (5, 0) and S (7, 7) form a
Solution
Using distance formula, prove
PQ = QR = RS = SP and PR
SQ
Hence, it is a rhombus.
Question 12 1 / -0
If the lines ax + 2y + 1 = 0, bx + 3y + 1 = 0, cx + 4y + 1 = 0 are concurrent, then a, b and c are in
Solution
If all three lines are concurrent, then
= 0
= 0
2b - 2a + a - c = 0
2b = a + c
Therefore, a, b and c are in AP.
Question 13 1 / -0
To which quadrant does the point (-1, -4) belong?
Solution
As both x and y are negative,
They lie in the 3
rd quadrant.
Option 3 is correct.
Question 14 1 / -0
The coordinates of the point C, which divides the line segment joining the points A(2, 6) and B(5, 1) in the ratio of 2 : 3, is
Question 15 1 / -0
Find the equation of the straight line passing through the point (-2, -3) and perpendicular to the line passing through points (-2, 3) and (-5, -6).
Solution
Let slope of the line passing through points (-2, 3) and (-5, -6) is m
1 .
m
1 =
= 3
Let the slope of the required equation be m
2 .
As the lines are perpendicular, therefore
∴ m
1 m
2 = -1
m
2 =
∴ Equation of the line:
(y - y
1 ) = m(x - x
1 )
y + 3 =
(x + 2)
3y + 9 = -x - 2
3y + x + 11 = 0
Option 2 is correct.
Question 16 1 / -0
Find the area of
ABC with A(3, 2), B(5, - 3) and C(6, - 4).
Solution
Area of
ABC =
[3(-3 + 4) + 5(-4 - 2) + 6(2 + 3)]
|3 - 30 + 30|
=
sq. units
Option 4 is correct.
Question 17 1 / -0
What is the ratio in which P (4, 6) divides the line joining A (-2, 3) and B (6, 7)?
Solution
Let the ratio be m
1 : m
2 .
Coordinates of P =
= (4, 6)
= 4
6m
1 - 2m
2 = 4m
1 + 4m
2 2m
1 = 6m
2 Ratio = 3 : 1
Option 1 is correct.
Question 18 1 / -0
On which axis does the point (-1, 0) lie?
Solution
Point (- 1, 0) has negative x coordinate and 0 as y coordinate.
This point lies on negative x-axis.
Option 3 is correct.
Question 19 1 / -0
The equation of the line which passes through the point (1, -2) and cuts-off equal intercepts on the axes is
Solution
Intercept form of line =
= 1
Since intercepts are equal, so a = b.
= 1
x + y = a
As it passes through (1, -2),
1 - 2 = a
a = -1
Therefore, equation of line is x + y + 1 = 0.
Question 20 1 / -0
The points A (7, 9), B (3, - 7) and C (- 3, 3) are the vertices of
Solution
Show that BC = AC and AB2 = AC2 + BC2 . Hence an isosceles right triangle.
×
Sign in
Profile
Signout
Get latest Exam Updates 
