Straight Lines and Pair of Straight Lines Questions — JEE Mathematics

Q1

If in a parallelogram ABDC, the coordinates of A, B and C are respectively (1, 2), (3, 4) and (2, 5), then the equation of the diagonal AD is :

A 5x + 3y – 11 = 0

B 5x – 3y + 1 = 0

C 3x – 5y + 7 = 0

D 3x + 5y – 13 = 0

Correct Answer

Option B

Solution

co-ordinates of point D are (4, 7) $\Rightarrow$ line AD is 5x $-$ 3y + 1 = 0

Q2

If non zero numbers

a, b, c

are in

H.P.,

then the straight line

{x \over a} + {y \over b} + {1 \over c} = 0

always passes through a fixed point. That point is :

A

(-1,2)

B

(-1, -2)

C

(1, -2)

D

\left( {1, - {1 \over 2}} \right)

Correct Answer

Option C

Solution

a,b,c

are in

H.P. \Rightarrow {1 \over a}.{1 \over b},{1 \over c}

are in

A.P.

\Rightarrow {2 \over b} = {1 \over a} + {1 \over c}

\Rightarrow {1 \over a} - {2 \over b} + {1 \over c} = 0

$\therefore$

{x \over a} + {y \over a} + {1 \over c} = 0

passes through

\left( {1, - 2} \right)

Q3

If a

\Delta

ABC has vertices A(–1, 7), B(–7, 1) and C(5, –5), then its orthocentre has coordinates :

A (–3, 3)

B (3, –3)

C

\left( {{3 \over 5}, - {3 \over 5}} \right)

D

\left( { - {3 \over 5},{3 \over 5}} \right)

Correct Answer

Option A

Solution

equation of CD y + 5 = -1 (x - 5) $\Rightarrow$ x + y = 0 .....(1) equation of AE y - 7 = 2 (x + 1) $\Rightarrow$ 2x - y = -9 ......(2) from (1) & (2) x = -3, y = 3 Othocentre = (-3, 3)

Q4

The image of the point (3, 5) in the line x

-

y + 1 = 0, lies on :

A (x

-

4)2 + (y

-

4)2 = 8

B (x

-

4)2 + (y

+

2)2 = 16

C (x

-

2)2 + (y

-

2)2 = 12

D (x

-

2)2 + (y

-

4)2 = 4

Correct Answer

Option D

Solution

So, let the image is (x, y) So, we have

{{x - 3} \over 1} = {{y - 5} \over { - 1}} = - {{2(3 - 5 + 1)} \over {1 + 1}}

$\Rightarrow$ x = 4, y = 4 $\Rightarrow$ Point (4, 4) Which will satisfy the curve (x $-$ 2)2 + (y $-$ 4)2 = 4 as (4 $-$ 2)2 + (4 $-$ 4)2 = 4 + 0 = 4

Q5

Let the three sides of a triangle are on the lines

4 x-7 y+10=0, x+y=5

and

7 x+4 y=15

. Then the distance of its orthocentre from the orthocentre of the tringle formed by the lines

x=0, y=0

and

x+y=1

is

A

\sqrt{20}

B

20

C

\sqrt{5}

D

5

Correct Answer

Option C

Solution

A is orthocentre of above $\Delta$ .

\begin{aligned} &O \text{ is orthocentre of above } \Delta \text{. }\\ &O A=\sqrt{5} \end{aligned}

Q6

Two sides of a parallelogram are along the lines, x + y = 3 & x – y + 3 = 0. If its diagonals intersect at (2, 4), then one of its vertex is :

A (2, 1)

B (2, 6)

C (3, 5)

D (3, 6)

Correct Answer

Option D

Solution

Solving

\begin{array}{ll}{x + y = 3} \\ {x - y = - 3} \end{array} \,\, > \,\,A\left( {0,3} \right)

and

{{{x_1} + 0} \over 2} = 2;\,\,{x_i} = 4

similarly y1 = 5 C $\Rightarrow$ (4, 5) Now equation of BC is x $-$ y = $-$ 1 and equation of CD is x + y = 9 Solving x + y = 9 and x $-$ y = $-$ 3 Point D is (3, 6)

Q7

If the line 3x + 4y – 24 = 0 intersects the x-axis at the point A and the y-axis at the point B, then the incentre of the triangle OAB, where O is the origin, is :

A (3, 4)

B (2, 2)

C (4, 4)

D (4, 3)

Correct Answer

Option B

Solution

\left| {{{3r + 4r - 24} \over 5}} \right| = r

7r - 24 = \pm 5r

2r = 24

or

12r + 24

r = 14,\,\,\,r = 2

then incentre is

(2,2)

Q8

A straight line through origin O meets the lines 3y = 10 − 4x and 8x + 6y + 5 = 0 at points A and B respectively. Then O divides the segment AB in the ratio :

A 2 : 3

B 1 : 2

C 4 : 1

D 3 : 4

Correct Answer

Option C

Solution

The lines 4x + 3y $-$ 10 = 0 and 8x + 6y + 5 = 0 , are parallel as

{4 \over 8}

=

{3 \over 6}

Now length of perpendicular from (0, 0, 0) to 4x + 3y $-$ 10 = 0 is, P1 =

\left| {{{4\left( 0 \right) + 3\left( 0 \right) - 10} \over {\sqrt {{4^2} + {3^2}} }}} \right|

=

{{10} \over 5}

= 2 Length of perpendicular from 0 (0, 0) to 8x + 6y + 5 = 0 is P2 =

\left| {{{8\left( 0 \right) + 6\left( 0 \right) + 5} \over {\sqrt {{6^2} + {8^2}} }}} \right|

=

{5 \over {10}}

=

{1 \over 2}

\therefore\,\,\,

P1 : P2 = 2 :

{1 \over 2}

= 4 : 1

Q9

A point P moves on the line 2x – 3y + 4 = 0. If Q(1, 4) and R (3, – 2) are fixed points, then the locus of the centroid of

\Delta

PQR is a line :

A parallel to y-axis

B with slope

{2 \over 3}

C parallel to x-axis

D with slope

{3 \over 2}

Correct Answer

Option B

Solution

Let the centroid of

\Delta

PQR is (h, k) & P is ( $\alpha$ , $\beta$ ), then

{{\alpha + 1 + 3} \over 3} = h\,

and

{{\beta + 4 - 2} \over 3} = k

\alpha = \left( {3h - 4} \right)

\beta = \left( {3k - 4} \right)

Point P( $\alpha$ , $\beta$ ) lies on the line 2x $-$ 3y + 4 = 0 $\therefore$ 2(3h $-$ 4) $-$ 3 (3k $-$ 2) + 4 = 0 $\Rightarrow$ locus is 6x $-$ 9y + 2 = 0

Q10

In a triangle PQR, the co-ordinates of the points P and Q are (

-

2, 4) and (4,

-

2) respectively. If the equation of the perpendicular bisector of PR is 2x

-

y + 2 = 0, then the centre of the circumcircle of the

\Delta

PQR is :

A (

-

1, 0)

B (1, 4)

C (0, 2)

D (

-

2,

-

2)

Correct Answer

Option D

Solution

Mid point of

PQ \equiv \left( {{{ - 2 + 4} \over 2},{{4 - 2} \over 2}} \right) \equiv (1,1)

Slope of

PQ = {{4 + 2} \over { - 2 - 4}} = - 1

Slope of perpendicular bisector of PQ = 1 Equation of perpendicular bisector of PQ

y - 1 = 1(x - 1)

\Rightarrow y = x

Solving with perpendicular bisector of PR, 2x $-$ y + 2 = 0 Circumcentre is ( $-$ 2, $-$ 2)