Prove that the diagonals of the parallelogram formed by the four straight lines 3 x + y = 0 , 3 y + x = 0 , 3 x + y = 1 and 3 y + x = 1 are at right angles to one another.

Given: Sides of a parallelogram are 3 x + y = 0 , 3 y + x = 0 , 3 x + y = 1 & 3 y + x = 1 Now distance between sides O A & B C is d 1 = 0 - - 1 3 2 + 1 2 = 1 2 and distance between sides A B & O C is d 2 = 0 - - 1 1 2 + 3 2 = 1 2 ( ∵ Distance between two parallel lines a x + b y + c 1 = 0 & a x + b y + c 2 is given as d = c 1 - c 2 a 2 + b 2 ) Parallelogram O A B C is rhombus and diagonals of a rhombus are at right angles. Hence proved.

Prove that the diagonals of the parallelogram whose sides are x a + y b = 1 , x b + y a = 1 , x a + y b = 2 and x b + y a = 2 are at right angles to one another.

Given sides of a parallelogram are x a + y b = 1 ; x b + y a = 1 ; x a + y b = 2 and x b + y a = 2 Now distance between sides O A & B C is, d 1 = 2 - 1 1 a 2 + 1 b 2 = a b a 2 + b 2 and distance between sides A B & O C is, d 2 = 2 - 1 1 b 2 + 1 a 2 = a b a 2 + b 2 ( ∵ Distance between two parallel lines a x + b y + c 1 = 0 and a x + b y + c 2 = 0 given as d = c 1 - c 2 a 2 + b 2 ) So, parallelogram O A B C is a rhombus and diagonals of a rhombus are at right angles. Hence proved.

Find the equations to the straight lines, bisecting the angles between the pairs of straight lines x + 3 y = 6 + 2 3 and x - 3 y = 6 - 2 3 , placing first the bisector of the angle in which the origin lies.

The equations of the lines are given as, x + 3 y - 6 + 2 3 = 0 x - 3 y - 6 - 2 3 = 0 Now, making the first constant term positive, we get, - x - 3 y + 6 + 2 3 = 0 - x + 3 y + 6 - 2 3 = 0 The general equation of the angle bisectors to the lines a 1 x + b 1 y + c 1 = 0 and a 2 x + b 2 y + c 2 = 0 is given as, a 1 x + b 1 y + c 1 a 1 2 + b 1 2 = ± a 2 x + b 2 y + c 2 a 2 2 + b 2 2 Hence, the equation of the given angle bisectors can be written as, - x - 3 y + 6 + 2 3 - 1 2 + - 3 2 = ± - x + 3 y + 6 - 2 3 - 1 2 + - 3 2 ⇒ - x - 3 y + 6 + 2 3 2 = ± - x + 3 y + 6 - 2 3 2 Now, the origin lies in the bisector, when we consider the positive sign in the above equation. This can be written as, - x - 3 y + 6 + 2 3 = - x + 3 y + 6 - 2 3 ⇒ y - 2 = 0 By considering the negative sign, we get, - x - 3 y + 6 + 2 3 = x - 3 y - 6 + 2 3 ⇒ x - 6 = 0 Hence, the equations of the bisectors are given as, ⇒ y - 2 = 0 ⇒ x - 6 = 0

MEDIUM

JEE Advanced

IMPORTANT

Earn 100

Find the equation of the straight line passing through the intersection of the lines $2 x - 3 y = 10$ and $x + 2 y = 6$ and the intersection of the lines $16 x - 10 y = 33$ and $12 x + 14 y + 29 = 0$ .

Important Questions on The Straight Line, Rectangular Coordinates

HARD

JEE Advanced

IMPORTANT

The Elements of COORDINATE GEOMETRY Part 1 Cartesian Coordinates>Chapter 3 - The Straight Line, Rectangular Coordinates>EXAMPLES VIII>Q 32

If through the angular points of a triangle straight lines be drawn parallel to the sides, and if the intersections of these lines be joined to the opposite angular points of the triangle, show that the joining lines so obtained will meet in a point.

HARD

JEE Advanced

IMPORTANT

The Elements of COORDINATE GEOMETRY Part 1 Cartesian Coordinates>Chapter 3 - The Straight Line, Rectangular Coordinates>EXAMPLES VIII>Q 33

Find the equations to the straight lines passing through the point of intersection of the straight lines $A x + B y + C = 0$ and $A^{'} x + B^{'} y + C^{'} = 0$
and

$(i)$ Passing through the origin.

$(i i)$ Parallel to the axis of $y$ .

$(i i i)$ Cutting off a given distance $a$ from the axis of $y$ and

$(i v)$ Passing through the given point $(x^{'}, y^{'})$ .

EASY

JEE Advanced

IMPORTANT

The Elements of COORDINATE GEOMETRY Part 1 Cartesian Coordinates>Chapter 3 - The Straight Line, Rectangular Coordinates>EXAMPLES VIII>Q 34

Prove that the diagonals of the parallelogram formed by the four straight lines

\sqrt{3} x + y = 0, \sqrt{3} y + x = 0, \sqrt{3} x + y = 1

and

\sqrt{3} y + x = 1

are at right angles to one another.

HARD

JEE Advanced

IMPORTANT

The Elements of COORDINATE GEOMETRY Part 1 Cartesian Coordinates>Chapter 3 - The Straight Line, Rectangular Coordinates>EXAMPLES VIII>Q 35

Prove that the diagonals of the parallelogram whose sides are

\frac{x}{a} + \frac{y}{b} = 1

\frac{x}{b} + \frac{y}{a} = 1

\frac{x}{a} + \frac{y}{b} = 2

and

\frac{x}{b} + \frac{y}{a} = 2

are at right angles to one another.

HARD

JEE Advanced

IMPORTANT

The Elements of COORDINATE GEOMETRY Part 1 Cartesian Coordinates>Chapter 3 - The Straight Line, Rectangular Coordinates>EXAMPLES VIII>Q 36

One side of a square is inclined to the axis of

x

at an angle

α

and one of its extremities is at the origin; prove that the equations to its diagonals are

y (\cos α - \sin α) = x (\sin α + \cos α)

and

y (\sin α + \cos α) + x (\cos α - \sin α) = a

, where

a

is the length of the side of the square.

HARD

JEE Advanced

IMPORTANT

The Elements of COORDINATE GEOMETRY Part 1 Cartesian Coordinates>Chapter 3 - The Straight Line, Rectangular Coordinates>EXAMPLES VIII>Q 37

Find the equations to the straight lines, bisecting the angles between the pairs of straight lines

x + \sqrt{3} y = 6 + 2 \sqrt{3}

and

x - \sqrt{3} y = 6 - 2 \sqrt{3}

, placing first the bisector of the angle in which the origin lies.

MEDIUM

JEE Advanced

IMPORTANT

The Elements of COORDINATE GEOMETRY Part 1 Cartesian Coordinates>Chapter 3 - The Straight Line, Rectangular Coordinates>EXAMPLES VIII>Q 38

Find the equations to the straight lines, bisecting the angles between the pairs of straight lines

12 x + 5 y - 4 = 0

and

3 x + 4 y + 7 = 0

, placing first the bisector of the angle in which the origin lies.

EASY

JEE Advanced

IMPORTANT

The Elements of COORDINATE GEOMETRY Part 1 Cartesian Coordinates>Chapter 3 - The Straight Line, Rectangular Coordinates>EXAMPLES VIII>Q 39

Find the equations of the straight lines, bisecting the angles between the pair of straight lines

4 x + 3 y - 7 = 0

and

24 x + 7 y - 31 = 0

, placing first the bisector of the angle in which the origin lies.

Find the equation of the straight line passing through the intersection of the lines 2x-3y=10 and x+2y=6 and the intersection of the lines 16x-10y=33 and 12x+14y+29=0.

Important Questions on The Straight Line, Rectangular Coordinates

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Find the equation of the straight line passing through the intersection of the lines $2 x - 3 y = 10$ and $x + 2 y = 6$ and the intersection of the lines $16 x - 10 y = 33$ and $12 x + 14 y + 29 = 0$ .