Q 31 :

The line passing through the point of intersection of x+y=2, x-y=0 and is parallel to x+2y=5 is ax+by=c, then value of a+b+c is _________.



(6)

Required equation of line which is parallel to given line x+2y=5 is x+2y+k=0               …(i)

Given equations of lines are

          x+y=2                                                             …(ii)

          x-y=0                                                             …(iii)

Adding (ii) and (iii), we get 2x=2⇒x=1

Substituting x=1 in (iii), we get y=1

∴   Point of intersection is (1,1)

Substituting x=1, y=1 in (i), we get

1+2+k=0⇒k=-3

∴   Required equation of line is x+2y=3

Here, a=1, b=2 and c=3

∴  a+b+c=1+2+3=6



Q 32 :

A1,A2,…,An are points on the line y=x lying in the positive quadrant such that OAn=n·OAn-1, O being the origin. If OA1=1 and the coordinates of An are (25202, 25202), then the value of n is _______.



(7)

We have, OAn=n·OAn-1

=n(n-1)·OAn-2=…=n!·OA1=n!

∴  OAn=n!

⇒2×25202=n!

⇒n!=5040

⇒n=7



Q 33 :

The area of the rhombus ABCD is 24. The equation of the diagonal BD is 4x+3y+2=0 and A=(3, 2). The length of the side of the rhombus is ______.



(5)

Let AC and BD intersect at P.

Now, AP=12+6+216+9=4

∴  Area of ∆ABD=AP×BP=242=12⇒BP=3

∴  AB=AP2+BP2=5



Q 34 :

In ΔABC, the equation of altitudes AM and BN are x+5y-3=0, x+y-k=0. If the altitude CL is given by 3x-y-1=0, then k= ______.



(1)

Solving the altitudes AM and CL, we get

Orthocentre=(12,12)  lies on x+y-k=0⇒k=1



Q 35 :

The centroid of the triangle formed by (a,b), (b,c), (c,a) is the origin and a3+b3+c3=kabc, then k is _______.



(3)

Given,  G=(a+b+c3,a+b+c3)=(0,0)

⇒a+b+c=0

⇒a3+b3+c3=3abc

⇒k=3



Q 36 :

If P(6, 1) be the orthocentre of the triangle whose vertices are A(5,-2), B(8, 3) and C(h, k), then the point C lies on the circle:

  • x2+y2-61=0

     

  • x2+y2-52=0

     

  • x2+y2-65=0

     

  • x2+y2-74=0

     

(3)

Slope of AD=3,        Slope of BC=-13,    Equation of BC=3y+x-17=0

Slope of BE=1,    Slope of AC=-1,        Equation of AC is x+y-3=0,    point C is (-4,7)



Q 37 :

If a straight line having negative slope passing through the point P = (8, 2) meets the x-axis at A and y-axis at B respectively. Then the minimum value of OA + OB (O being origin) is _____

  • 20

     

  • 18

     

  • 10

     

  • 15

     

(2)

Equation of line is y-2=m(x-8),   OA=8+2-m and OB=8(-m)+2

OA+OB=10+8(-m)+2-m≥18 (AM≥GM)



Q 38 :

In a ∆ABC, AB=AC where A is (3, 1) and the equation of the base BC is 2x+y=4. Also, B lies on x+3y=7. Then sum of coordinates of vertex 'C' is ______

  • 135

     

  • 185

     

  • 165

     

  • 75

     

(4)

B=(1,2) & Let C≡(α,4-2α)

∴ AB=22+12=(2α-3)2+(3-α)2,  

5α2-18α+13=0



Q 39 :

Consider the following Statements

Statement I: General form ax+by+c=0 (here a,b and c are the parameters and a,b,c∈R) always represents a straight line.

Statement II: A line ax+by+1=0, is such that the algebraic sum of the perpendiculars on it, from a number of points (xi,yi), (i=1,2,…,n), is zero, then the line always passes through a fixed point (x¯,y¯) where x¯=∑xin and y¯=∑yin

  • Only Statement I is true

     

  • Only Statement II is true

     

  • Both Statement I and II are True

     

  • Both Statement I and II are false

     

(2)

S1: a=b=c=0  Then it is not linear

S2 is obviously True



Q 40 :

The equations of the sides AB, BC and CA of a triangle ABC are:

2x+y=0,  x+py=21a, (a≠0) and x-y=3 respectively. Let P(2,a) be the centroid of ∆ABC. Then (BC)2 is equal to

  • 25

     

  • 36

     

  • 121

     

  • 122

     

(4)

2x+y=0    ------(1)

  x-y=0    ------(2)

x+py=21a    ------(3)

Solving (1) & (2)⇒A(1,-2)

Centroid of triangle ABC is (4+s+t3,-2-2s+t3)=(2,a)

⇒s+t=2    ------(4)

⇒s=-a, t=2+a  

-2s+t=3a+2    ------(5)

Solving (4) & (5) we get

B(-a,2a), C(a+5,a+2)

∴  Distance (BC)2=122



Q 41 :

Let the circumcenter of a triangle with vertices A(a,3), B(b,5) and C(a,b), ab>0 be P(1, 1). If the line AP intersects the line BC at the point Q(k1,k2), then k1+k2 is equal to

  • 2

     

  • 47

     

  • 27

     

  • 4

     

(2)

slope of AC=∞

Slope of PD=0

D(a+a2,b+32)=D(a,b+32)

b+32-1=0

b+3-2=0

⇒b=-1

b=-1

E(b+a2,5+b2)=(a-12,2)

slope of BC×slope of EP=-1

(5-bb-a)×(2-1a-12-1)=-1

⇒(6-1-a)×(2a-3)=-1

⇒12=(1+a)(a-3)

⇒12=a2-3a+a-3⇒a2-2a-15=0

a=-3 accept

Equation of AP,  A(-3,3), P(1,1)

y-1=(3-1-3-1)(x-1)⇒x+2y=3

Equation of BC,  B(-1,5), C(-3,-1)

y+1=(5+1-1+3)(x+3)=3x+9

3x-y+8=0

Q≡(-137,177)



Q 42 :

Consider the following statements:

S1: For the line 2x+3y-7=0, the points origin and (1, 1) are on the same side.

S2: Two points (x1,y1) and (x2,y2) lie on the same side of the line ax+by+c=0 (a,b,c∈R-{0}) if (ax1+by1+c)(ax2+by2+c)>0

  • Both statements are true

     

  • Both statements are false

     

  • S1 is true and S2 is false

     

  • S1 is false and S2 is true

     

(1)

Substituting the points in the given line gives negative values for both points



Q 43 :

If A(-2, -1), B(-2, 1), C(2, 1) are the vertices of a triangle ABC, then the perpendicular distance from its circumcenter to the side BC is

  • 1

     

  • 2

     

  • 3

     

  • 4

     

(1)

△ABC is a right angled triangle at B and midpoint of hypotenuse AC is the circumcentre

Circumcentre of △ABC is S(0,0)



Q 44 :

Let the line L1:x+3=0 intersect the lines L2:x–y=0 and L3:3x+y=0 at the points A and B, respectively. Let the bisector of the obtuse angle between the lines L2 and L3 intersect the line L1 at the point C. Then BC2:AC2 is equal to:          [2026]

  • 5 : 1

     

  • 1 : 5

     

  • 2 : 3

     

  • 3 : 2

     

(1)

We have, A : (–3, –3) and B : (–3, 9)

The equations of the angle bisectors given by:

          x–y12+(–1)2=±3x+y32+12

⇒ x–y2=±3x+y10

⇒ 5(x–y)=±(3x+y)

To find the obtuse angle bisector, we check the sign of a1a2+b1b2 for the lines L2(x–y=0) and L3(3x+y=0), where a1=1, b1=–1, a2=3, b2=1 : a1a2+b1b2=(1)(3)+(–1)(1)=3–1=2

Since, the result is positive, the obtuse angle bisector is given by the equation with the minus sign (opposite the sign of a1a2+b1b2):

         5(x–y)=–(3x+y)

⇒ 5x–5y=–3x–y

⇒ y(1–5)=–x(3x+5)

⇒ y=x3+55–1

Point C lies on L1=(x=–3), so substituting x = –3 into the bisector equations:

          yC=(–3)3+55–1=–3(3+5)(5+1)(5–1)(5+1)

⇒ yC=–335+3+3+55–1

                =–38+454=–3(2+5)

                C=(–3,–6–35)

∴   BC2AC2=90(3+5)18(3+5)=5:1



Q 45 :

Let the vertex A of a triangle ABC be (1, 2) and the midpoint of the side AB be(5, –1). If the centroid of this triangle is (3, 4) and its circumcenter is (α,β), then 21(α+β) is equal to:          [2026]

  • 309

     

  • 403

     

  • 497

     

  • 524

     

(3)

From the given information:

FIGURE

Coordinates of B:x2+12=5; y2+22=–1

⇒ x2=9, y2=–4

∴  B≡(9,–4)

Coordinates of C:1+9+x33=3

⇒ x3=–1 and 2–4+y33=4

⇒ y3=14

∴  C≡(–1,14)

Slope of ⊥r bisector of AB, m1=43

Equation of ⊥r bisector of AB

y+1=43(x–5)

⇒ 4x–3y+23=0          ... (i)

Midpoint of BC is N (4, 5)

Slope of ⊥r bisector of BC, m2=59

Equation of ⊥r bisector of BC,

y–5=59(x–4) ⇒ 9y–45=5x–20

⇒ 5x–9y+25=0          ... (ii)

Solving (i) & (ii), we get

x=947, y=4×9421–233=21521

∴  (α,β)=(947,21521)

Thus, 21(α+β)=497



Q 46 :

In an equilateral triangle PQR, let the vertex P be at (3, 5) and the side QR be along the line x + y = 4. If the orthocentre of the triangle PQR is (α,β), then 9(α+β) is equal to:          [2026]

  • 16

     

  • 27

     

  • 36

     

  • 48

     

(4)

AP is perpendicular to QR

FIGURE

Slope of QR = –1

Equation of AP is y – 5 = 1(x – 3)

⇒ y–5=x–3

⇒ x–y=–2

On solving, x + y = 4 and x – y = –2, we get

       x = 1 and y = 3

∴  A≡(1,3)

Now, G≡(2+33,6+53)=(5/3,11/3)=(α,β)

∴  9(α+β)=9(53+113)=9×163=48



Q 47 :

If a straight line drawn through the point of intersection of the lines 4x + 3y – 1 = 0 and 3x + 4y – 1 = 0, meets the coordinate axes at the points P and Q, then the locus of the midpoint of PQ is:          [2026]

  • x + y – 7 = 0

     

  • x + y – 14xy = 0

     

  • 2x + y + 14xy = 0

     

  • x + 2y – 14xy = 0

     

(2)

          L1:4x+3y–1=0

          L2:3x+4y–1=0

          L:L1+λL2=0

⇒ (4x+3y–1)+λ(3x+4y–1)=0

⇒ (4+3λ)x+(3+4λ)y–(1+λ)=0

Point of intersection with x-axis:

         P(λ+13λ+4,0)

Point of intersection with y-axis:

         Q(0,λ+13+4λ)

Let mid-point of PQ be (h, k)

          h=λ+13λ+4+02=λ+16λ+8          ... (i)

          k=0+λ+13+4λ2=λ+18λ+6          ... (ii)

From (i), λ=1–8h6h–1

Put 'λ' in eq (ii), we get

          k=1–8h6h–1+18(1–8h6h–1)+6

⇒ h+k–14hk=0

∴   Locus is x + y – 14xy = 0



Q 48 :

Let A, B be points on the two half-lines x–3|y|=α, α>0 at a distance of α from their point of intersection P. The line segment AB meets the angle bisector of the given half-lines at the point Q. If PQ=92 and R is the radius of the circumvircle of △PAB, then α2R is equal to ________.          [2026]



(9)

Two half-line are x–3|y|=α, α>0, meeting at point P.

Points A and B are on these half-lines and each is at distance α from P.

 ∴  Lines are x–3|y|=α, α>0, and x+3y=α(y<0)

Their slopes are 13,–13

So, angle between them is 60°.

Since PA = PB =α,

∴  △PAB is isosceles triangle with vertex angle 60°.

Hence, PAB is an equilateral triangle

FIGURE

        PQ=3α2=92          [∵  Given]

∴  α=93 ⇒ α=33

Now, α2R=cos30° ⇒ α32·2R

⇒ α=3R ⇒ 3R=33          [∵  α=33]

⇒ R=3

So, α2R=(33)23=273=9



Q 49 :

From the point (–1, –1), two rays are sent making angles of 45° with the line x + y = 0. These rays get reflected from the mirror x + 2y = 1. If the equations of the reflected rays are ax + by = 0 and cx + dy = 7, a, b, c, d∈Z, then the value of ad + bc is ________.          [2026]



(7)

We have, tan45°=|m+11–m|

⇒ ±1=m+11–m

⇒ m = 0 & one line is perpendicular to x-axis.

Rays are : L1:x=–1, L2:y=–1

Intersection points of L1 and L2 with x + 2y = 1 are (–1, 1) and (3, –1) respectively.

Now, taking mirror image of x + 1 = 0 in x + 2y = 1.

A'=x+11=y+12=–2(1–2–1)12+22=85

⇒ x=35, y=115 and A'≡(35,115)

Now, taking mirror image of y + 1 = 0 in x + 2y = 1.

A''=x+11=y+12=–2(1–2–1)12+22=85

⇒ x=35, y=115

A''≡(35,115)

Equation of ray passes through (3, –1) is

(y+1)=(115+1)(35–3)(x–3)

⇒ 4x+3y=9

Comparing 4x + 3y = 9 with ax + by = 9, we get

a = 4, b = 3

Similarly, equation of ray passes through (–1, 1) is

–3x + 4y = 7

Comparing with cx + dy = 7, we get

c = –3, d = 4

Thus ad + bc = 4×4+3×(–3)=16–9=7