Coordinate GeometrySL

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1.A line has intercepts a and b on the coordinate axes. If keeping the origin fixed, the coordinate axes are rotated through 90(, the same line has intercepts p and q, then

(A) p =a, q = b

(B) p = b, q = a

(C) p = -b, q = -a

(D) p = b, q = -a

1.(D) Obviously, new x-axis is old y-axis and new y-axis is old (-x) axis. Thus p =b, q = -a.

2.The equations of the lines representing the sides of a triangle are 3x 4y = 0, x + y = 0 and 2x 3y = 7. The line 3x + 2y = 0 always passes through the

(A) incentre

(B) centroid

(C) circumcentre

(D) orthocentre

2.(D) 3x + 2y =0 passes through the point of intersection of two lines ( i.e. origin) and is perpendicular to the third.

3.If the mid-points P, Q and R of the sides of the ( ABC are (3, 3) , ( 3, 4) and (2, 4) respectively, then ( ABC is

(A) right angled

(B) acute angled

(C) obtuse angled

(D) isosceles

3.(A), (D)

( ABC is similar to ( PQR, which is isosceles right triangle.

4.If the line y =x cuts the curve x4 + ax2y + bxy + cx + dy + 6 = 0 at A, B, C and D, then OA.OB.OC.OD ( where O is the origin) is

(A) a 2b +c

(B) 2c2d

(C) 96

(D) 6

4.(C)

The line y = can be written as x = , y = . If this line cuts the given curve, then

.

Therefore OA. OB.OC.OD = |r1| |r2| |r3| |r4| = |r1 r2 r3 r4| = 96

5.A curve with equation of the form y = ax4 + bx3+cx +d has zero gradient at the point (0, 1) and also touches the x-axis at the point ( -1, 0). Then the values of x for which the curve has negative gradients are

(A) x > -1(B) x < 1

(C) x < -1(D) -1 ( x ( 1

5. (C)

y = ax4 + bx3 +cx +d

. . . . (1)

y touches x-axis at (-1, 0)

so, ( -1, 0) lies on it and dy/ dx =0

so, 0 = a b c +d

. . . . (2)

From (1) dy/ dx = 4ax3 + 3bx2 +c

. . . . (3)

Hence ( - 4a + 3b +c =0. . . . (4)

Also = c =0

(since curve touches (0, 1) )

( ( 0, 1) also lies on it

hence d = 1

Putting values of c and d in (2) and (4), solving for a and b we get a = 3, b = 4

Therefore (3) becomes = 12x3 +12x2

Now dy/dx < 0 ( 12 x3 + 12x2 < 0

( 12x2 (x+1) < 0

( x < -1

6.If 5a + 4b + 20c = t, then the value of t for which the line ax + by + c ( 1 = 0 always passes through a fixed point is

(A) 0 (B) 20

(C) 30 (D) none of these

6.B

Equation of line has two independent parameters. It can pass through a fixed point if it contains only one independent parameter. Now there must be one relation between and independent of a, b and c so that can be expressed in terms of and straight line contains only one independent parameter. Now that given relation can be expressed as .

Now RHS be independent of c if t = 20.

7.Let y = mx + 5 be a line. Then the range of m so that point (5, 6) and ((6, 6) always lie on the opposite side of origin is

(A) m > and m < (

(B) < m <

(C) m >

(D) none of these

7.B

For (0, 0)

y mx 5 = 5 < 0

so, for (5, 6) and (6, 6) to lie on opposite side we have

6 5m 5 > 0 and 6 + 6m 5 > 0

( < m < .

8.As system of line is given as y = mix + ci, where mi can take any value out of 0, 1, (1 and when mi is positive then ci can be 1 or (1 when mi equal 0, ci can be 0 or 1 and when mi equals (1, ci can take 0 or 2. Then the area enclosed by all these straight line is

(A)

(B)

(C)

(D) none of these

8.C

Lines are y = 1, y = 0

y = (x, y = (x + 2

y = x + 1, y = x ( 1

Area of OABCDE = area of OBGF

=

9.Two sides of a rhombus OABC ( lying entirely in first quadrant or fourth quadrant ) of area equal to 2 sq. units, are y = , y = x . Then possible coordinates of B is / are (O being the origin)

(A)

(B)

(C)

(D) none of these

9.(A), (B)

Let the length of the sides of the rhombus be l

( Area = l2 sin30( =2 ( l = 2 unit

OB2 = OA2 + AB2 2OA .AB cos150(

= 4 + 4 2 (4). = 4 (2 + )

OB = 2

B ( (

Hence coordinates of B can be or

10.If (OAB is an equilateral triangle (O is the origin and A is a point on the x-axis), then centroid of the triangle will be

(A) always rational (B) rational if B is rational

(C) rational if A is rational(D) never rational

(a point P(x, y) is said to be rational if both x and y are rational)

10.(D)

Centroid is . Now clearly and cannot be rational simultaneously for any real a.

11.If a b, b c, c a are in A.P., then the straight line (a b)x + ( b c)y +(c a) = 0 will pass through

(A) (1, 2)(B) (2, 1)

(C) (2, 3)(D) (3, 1)

11.(A)

a b +c a = 2(b c)

c b = 2( b c) ( c= b

a b = -( c- a)

So, equation of straight line is x 1 = 0.

12.If the quadratic equation ax2 + bx + c = 0 has 2 as one of its roots then ax + by + c = 0 represents

(A)A family of concurrent lines(B)A family of parallel lines

(C)A single line(D)A line perpendicular to x-axis

12.(A)

Since 2 is a root of ax2 + bx + c = 0 ( 4a 2b + c = 0

So, ax + by + c = 0 always passes through (4, - 2)

13.Two particles start from the same point (2, -1), one moving 2 units along the line x + y = 1 and the other 5 units along x 2y = 4. If the particles move towards increasing y, then their new position will be

(A)(2 - (2, (2 1)(B)(2(5 + 2, (5 1)

(C) (2 + (2, (2 + 1)(D)(2(5-2, (5 1)

13.(A), (B)

Let P (2, -1) goes 2 units along x + y = 1 upto A and 5 units along x 2y = 4 up to B

Slope of PA = -1= tan135 , slope of PB = = tan(

(sin( = , cos( =

( A ( (x1 + rcos135, y1 = rsin135) = (2 + 2x , -1 + 2 ( )

= (2 - (2, (2 1)

B ( (x1 + rcos(, y1 + rsin() = (2 + 5 ( , -1 + ) = (2(5 + 2, (5 1)

14.In an isosceles right angled triangle, a straight line drawn from the mid-point of one of equal sides to the opposite angle . It divides the angle into two parts, ( and ((/4 - (). Then tan( and tan[((/4) -(] are equal to

(A)

(B)

(C)

(D) none of these

14.(A)

Let ABC be the isosceles triangle right angled at A. E is the mid-point of AC and (ABE = (. B =C = 45(,

AB = AC,AE = EC = AC and

(CBE = -( . From the right angled triangle ABE,

we get tan( =

tan( -() = .

15.Consider a line passing through (, 1) and (1, ) then number of rational points lying in the line is

(A) 1(B) at most 2

(C) ((D) none of these

15.(D) slope m =

Let there be a rational point P (x, y)

Now slope of line joining (, 1) and (x, y) is

It is obvious that any rational choice for x, y it is not equal to .16.A ray of light travelling along the line x + y = 1 is inclined on the x-axis and after refraction it enters the other side of the x-axis by turning 15( away from the x-axis. The equation of the line along which the refraction ray travels is

(A) y - x +1 = 0(B) y + x +1 = 0

(C) y + x -1 = 0(D) none of these .

16.(D)

The line of the refracted ray passes through the point (1, 0) and its slope is tan120(

( The equation of the line of the refracted ray is

y - 0 = tan120( ( x 1)

y = -( x 1) ( y + x = 0.

17.If P ( ; Q ( ; R ( where xk ( 0, k = p,q, r ( N, denotes the kth term of a Harmonic progression, then

(A) Area((PQR) =

(B) (PQR is a right angled triangle

(C) the points P, Q and R are collinear.

(D) none of these .

17.(C)

consider

= = 0

( Points P, Q and R are collinear.

18.If 3l + 5m + 7n = 0 then the family of straight line lx + my + n = 0 passes through the fixed point

(A) (2, 5)(B)

(C)

(D)

18.(D) 3l + 5m + 7n = 0

(

(1)

also lx + my + n = 0(2)

( l

This shows that the family of straight lines passes through the fixed point .

19.The number of points on the line 3 x + 4y = 5, which are at a distance of sec2( +2 cosec2(, ( ( R, from the point (1, 3), is

(A) 1(B) 2

(C) 3(D) infinite

19.(B)

The perpendicular distance of (1, 3) from the line 3x + 4y = 5 is 2 units while sec2( +2cosec2( ( 3 (as sec2( , cosec2( ( 1)

Evidently, there will be two such points on the line.

20.If the median AD of a (ABC is bisected at E and BE intersects AC in F, then is equal to

(A)

(B)

(C)

(D)

20.(B)

Let A be the origin and position vectors of B and C be respectively. Now , , and

Equation of

(1)

Equation of

(2)

For the point F, , for some particular ( and ( .

Since are non-collinear,

( ( = 4/3, ( = 1/3

( Position vector of F = ( = .

Thus .

21.The base BC of (ABC is bisected at (p, q) and equation of sides AB and AC are px + qy = 1 and qx + py = 1. Then equation of median through A is

(A)(2q 1) (px + qy) = pq

(B)(2pq 1) (px + qy 1) = (p2 + q2 1) (qx + py 1)

(C)(px + qy