The curve passing through the points of intersection of S 1 ≡ x 2 a 2 + y 2 b 2 - 1 = 0 and S 2 ≡ x 2 + y 2 + 2 g x + 2 f y + c = 0 represents a pair of straight lines which are

equally inclined to the x -axis

not equally inclined to y -axis

Find the number of integral values of λ if ( λ , λ + 1 ) is an interior point of Δ A B C , where A ≡ ( 0 , 3 ) B ≡ ( - 2 , 0 ) and C = ( 6 , 1 ) .

Given, A 0 , 3 , B - 2 , 0 and C 6 , 1 be the vertices of triangle A B C . Now, for the point to be an interior point P x , y , the basic conditions can be applied are, - 2 < x < 6 and 0 < y < 3 . Since λ , λ + 1 be an interior point of △ A B C . So, - 2 < λ < 6 … 1 and 0 < λ + 1 < 3 ⇒ - 1 < λ < 2 … 2 Combining conditions 1 & 2 , we get - 1 < λ < 2 So, the possible integral values of λ = 0 , 1 . Number of integral values of λ = 2 .

Let A B C be a triangle such that the coordinates of the vertex A are ( - 3 , 1 ) . Equation of the median through B is 2 x + y - 3 = 0 and equation of the angular bisector of C is 7 x - 4 y - 1 = 0 . Find the slope of line B C .

Given equation of angular bisector of point C be 7 x - 4 y - 1 = 0 . So, let C D be angular bisector of point C . Consider the coordinates of C as h , k , and since the point C lies on the angular bisector of C , therefore the equation becomes, 7 h - 4 k = 1 … 1 Since the median through point B is B E which is 2 x + y - 3 = 0 The midpoint of A C will be E h - 3 2 , k + 1 2 Hence, 2 h - 3 2 + k + 1 2 - 3 = 0 ⇒ 2 h + k = 11 … 2 From equation 1 & 2 , h = 3 , k = 5 So, Slope of line A C is = 5 - 1 3 - - 3 = 2 3 The slope of bisector C D 7 x - 4 y - 1 = 0 = 7 4 So, from diagram we can see that the angle between A C and the bisector is equal to angle between B C and the bisector. Let the angle be θ and slope of B C be m . ⇒ tan θ = 2 3 - 7 4 1 + 2 3 7 4 = m - 7 4 1 + 7 4 m ⇒ 4 m - 7 4 + 7 m = 1 2 ⇒ 4 m - 7 4 + 7 m = ± 1 2 If 4 m - 7 4 + 7 m = 1 2 , then m = 18 If 4 m - 7 4 + 7 m = - 1 2 , then m = 2 3 If m = 2 3 , then slope of B C & A C will be equal and the points A , B , C will be collinear and it will form no triangle. Therefore, slope of B C is 18 .

A ( 3 , 4 ) , B ( 0 , 0 ) and C ( 3 , 0 ) are vertices of Δ A B C . If ' P ' is the point inside the Δ A B C , such that d ( P , B C ) ≤ min . { d ( P , A B ) , d ( P , A C ) } . Then the maximum of d ( P , B C ) is. (where d ( P , B C ) represent distance between P and B C ).

Given the vertices of △ A B C be A 3 , 4 , B 0 , 0 and C 3 , 0 and P is the point inside the △ A B C such that d P , B C ≤ m i n d P , A B , d P , A C , where d P , B C is the distance between the point P & B C . Equation of line A B y - 4 = 0 - 4 0 - 3 x - 3 ⇒ 3 y - 4 x = 0 Equation of line B C y - 0 = 0 - 0 3 - 0 x - 0 ⇒ y = 0 Equation of line A C y - 4 = 0 - 4 3 - 3 x - 3 ⇒ x = 3 Let the point P be h , k . d P , B C = 0 + k 0 + 1 = k d P , A B = - 4 h + 3 k - 4 2 + 3 2 = 1 5 4 h - 3 k and d P , A C = h + 0 - 3 1 + 0 = 3 - h So, we have k ≤ m i n 1 5 4 h - 3 k , 3 - h ⇒ k ≤ 4 h - 3 k 5 & k ≤ 3 - h ⇒ 5 k ≤ 4 h - 3 k & k + h ≤ 3 ⇒ 2 k ≤ h & k + h ≤ 3 ⇒ k + 2 k ≤ k + h ≤ 3 ⇒ 3 k ≤ 3 ⇒ k ≤ 1 Hence, maximum value of k = 1 So, maximum value of d P , B C = 1

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Find the equation of the two straight lines which together with those given by the equation $6 x^{2} - x y - y^{2} + x + 12 y - 35 = 0$ will make a parallelogram whose diagonals intersect in the origin.

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Important Questions on Point and Straight Line

MEDIUM

JEE Main/Advance

IMPORTANT

Alpha Question Bank for Engineering: Mathematics>Chapter 15 - Point and Straight Line>Exercise-2>Q 19

The curve passing through the points of intersection of

S_{1} \equiv \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} - 1 = 0

and

S_{2} \equiv x^{2} + y^{2} + 2 g x + 2 f y + c = 0

represents a pair of straight lines which are

HARD

JEE Main/Advance

IMPORTANT

Alpha Question Bank for Engineering: Mathematics>Chapter 15 - Point and Straight Line>Exercise-2>Q 20

If the points

(\frac{a^{3}}{a - 1}, \frac{a^{2} - 3}{a - 1}), (\frac{b^{3}}{b - 1}, \frac{b^{2} - 3}{b - 1})

and

(\frac{c^{3}}{c - 1}, \frac{c^{2} - 3}{c - 1})

are collinear for three distinct values

a, b, c

and

a \neq 1, b \neq 1

and

c \neq 1,

then find the value of

a b c - (a b + b c + a c) + 3 (a + b + c) .

HARD

JEE Main/Advance

IMPORTANT

Alpha Question Bank for Engineering: Mathematics>Chapter 15 - Point and Straight Line>Exercise-2>Q 21

Find the number of integral values of

λ

(λ, λ + 1)

is an interior point of

Δ A B C,

where

A \equiv (0, 3) B \equiv (- 2, 0)

and

C = (6, 1) .

HARD

JEE Main/Advance

IMPORTANT

Alpha Question Bank for Engineering: Mathematics>Chapter 15 - Point and Straight Line>Exercise-2>Q 22

Let

A B C

be a triangle such that the coordinates of the vertex

A

are

(- 3, 1)

. Equation of the median through

B

2 x + y - 3 = 0

and equation of the angular bisector of

C

7 x - 4 y - 1 = 0

. Find the slope of line

B C .

HARD

JEE Main/Advance

IMPORTANT

Alpha Question Bank for Engineering: Mathematics>Chapter 15 - Point and Straight Line>Exercise-2>Q 23

A (3, 4), B (0, 0)

and

C (3, 0)

are vertices of

Δ A B C

. If

' P'

is the point inside the

Δ A B C,

such that

d (P, B C) \leq \min . {d (P, A B), d (P, A C)} .

Then the maximum of

d (P, B C)

is. (where

d (P, B C)

represent distance between

P

and

B C

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IMPORTANT

Alpha Question Bank for Engineering: Mathematics>Chapter 15 - Point and Straight Line>Exercise-2>Q 24

Drawn from the origin are two mutually perpendicular straight lines forming an isosceles triangle together with the straight line

2 x + y = 5 .

Then, find the area of the triangle.

HARD

JEE Main/Advance

IMPORTANT

Alpha Question Bank for Engineering: Mathematics>Chapter 15 - Point and Straight Line>Exercise-2>Q 25

On the straight-line

y = x + 2,

a point

(a, b)

is such that the sum of the square of distances from the straight lines

3 x - 4 y + 8 = 0

and

3 x - y - 1 = 0

is least, then find value of

11 (a + b) .

HARD

JEE Main/Advance

IMPORTANT

Alpha Question Bank for Engineering: Mathematics>Chapter 15 - Point and Straight Line>Exercise-2>Q 26

A

is a variable point on

x

-axis and

B (0, b)

is a fixed point. An equilateral triangle

A B C

is completed with vertex

C

away from origin. If the locus of the point

C

α x + β y = b,

then

α^{2} + β^{2}

Find the equation of the two straight lines which together with those given by the equation 6x2-xy-y2+x+12y-35=0 will make a parallelogram whose diagonals intersect in the origin.

Important Questions on Point and Straight Line

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Find the equation of the two straight lines which together with those given by the equation $6 x^{2} - x y - y^{2} + x + 12 y - 35 = 0$ will make a parallelogram whose diagonals intersect in the origin.