Two adjacent sides of a parallelogram are given by 4 x + 5 y = 0 and 7 x + 2 y = 0 and one diagonal is 11 x + 7 y = 9 . Find the equations of the remaining sides and the other diagonal.

Let 4 x + 5 y = 0 ...(i) and 7 x + 2 y = 0 ...(ii) respectively denote the side O A and O B of the parallelogram O A B C . Equation of the diagonal A B is 11 x + 7 y - 9 = 0 ...(iii) Solving (i) and (ii), vertex O = ( 0 , 0 ) Solving (i) and (iii), vertex A = 5 3 , - 4 3 Solving (ii) and (iii), vertex B = - 2 3 , 7 3 Midpoint of A B is P 1 2 , 1 2 Slope of O P is 1 2 - 0 1 2 - 0 = 1 . Equation to O C is y = ( 1 ) x ⇒ x - y = 0 Equation of A C be 4 x - 5 3 + 3 y + 4 3 = 0 ⇒ 4 x + 5 y = 9 Equation of B C be 7 x + 2 3 + 2 y - 7 3 = 0 ⇒ 7 x + 2 y = 9

The Cartesian equations of the sides B C , C A and A B of a triangle are respectively u r ≡ a r x + b r y + c r = 0 , r = 1 , 2 , 3 . Show that the equation of the straight line passing through A and bisecting the side B C is u 3 a 3 b 1 - a 1 b 3 = u 2 a 1 b 2 - a 2 b 1 .

A is the point of intersection of the lines u 2 = 0 and u 3 = 0 . ∴ Equation to a line passing through A is u 2 + λ u 3 = 0 ⇒ a 2 x + b 2 y + c 2 + λ a 3 x + b 3 y + c 3 = 0 ...(i) ⇒ a 2 + λ a 3 x + b 2 + λ b 3 y + c 2 + λ c 3 = 0 If this is parallel to a 1 x + b 1 y + c 1 = 0 , ⇒ a 2 + λ a 3 a 1 = b 2 + λ b 3 b 1 [using, condition that slopes of parallel lines are equal] ⇒ a 2 + λ a 3 b 1 = b 1 + λ b 3 a 1 ⇒ a 2 b 1 + λ a 3 b 1 = a 1 b 2 + λ a 1 b 3 ⇒ λ a 3 b 1 - a 1 b 3 = - a 2 b 1 - a 1 b 2 ⇒ λ = a 2 b 1 - a 1 b 2 a 3 b 1 - a 1 b 3 Substituting this value of λ in (i), the required equation is a 2 x + b 2 y + c 2 - a 2 b 1 - a 1 b 2 a 3 b 1 - a 1 b 3 a 3 x + b 3 y + c 3 = 0 ⇒ a 3 b 1 - a 1 b 3 a 2 x + b 2 y + c 2 - a 2 b 1 - a 1 b 2 a 3 x + b 3 y + c 3 = 0 ⇒ a 3 b 1 - a 1 b 3 u 2 - a 2 b 1 - a 1 b 2 u 3 = 0 ⇒ a 3 b 1 - a 1 b 3 u 2 = a 2 b 1 - a 1 b 2 u 3 ⇒ u 3 a 3 b 1 - a 1 b 3 = u 2 a 2 b 1 - a 1 b 2 Hence, proved.

E M B I B E

Intermediate First Year Mathematics Paper 1B>Chapter 3 - The Straight Line>Exercise 3(e)

Andhra Pradesh Board Solutions for Chapter: The Straight Line, Exercise 5: Exercise 3(e)

Author:Andhra Pradesh Board

Andhra Pradesh Board Mathematics Solutions for Exercise - Andhra Pradesh Board Solutions for Chapter: The Straight Line, Exercise 5: Exercise 3(e)

Attempt the practice questions on Chapter 3: The Straight Line, Exercise 5: Exercise 3(e) with hints and solutions to strengthen your understanding. Intermediate First Year Mathematics Paper 1B solutions are prepared by Experienced Embibe Experts.

Questions from Andhra Pradesh Board Solutions for Chapter: The Straight Line, Exercise 5: Exercise 3(e) with Hints & Solutions

HARD

11th Andhra Pradesh Board

IMPORTANT

Intermediate First Year Mathematics Paper 1B>Chapter 3 - The Straight Line>Exercise 3(e)>Q III

Find the circumcenter of the triangle whose sides are given by $x + y + 2 = 0, 5 x - y - 2 = 0$ and $x - 2 y + 5 = 0$ .

MEDIUM

11th Andhra Pradesh Board

IMPORTANT

Intermediate First Year Mathematics Paper 1B>Chapter 3 - The Straight Line>Exercise 3(e)>Q III

Find the equations of the straight lines passing through $(1, 1)$ and which are at a distance of $3$ units from $(- 2, 3)$ .

HARD

11th Andhra Pradesh Board

IMPORTANT

Intermediate First Year Mathematics Paper 1B>Chapter 3 - The Straight Line>Exercise 3(e)>Q III

If $p$ and $q$ are the lengths of the perpendiculars from the origin to the straight lines $x \sec α + y cosec α = a$ and $x \cos α - y \sin α = a \cos 2 α,$ prove that $4 p^{2} + q^{2} = a^{2}$ .

MEDIUM

11th Andhra Pradesh Board

IMPORTANT

Intermediate First Year Mathematics Paper 1B>Chapter 3 - The Straight Line>Exercise 3(e)>Q III

Two adjacent sides of a parallelogram are given by $4 x + 5 y = 0$ and $7 x + 2 y = 0$ and one diagonal is $S A = S C$ . Find the equations of the remaining sides and the other diagonal.

HARD

11th Andhra Pradesh Board

IMPORTANT

Intermediate First Year Mathematics Paper 1B>Chapter 3 - The Straight Line>Exercise 3(e)>Q III

Find the incentre of the triangle formed by the straight lines $x + 1 = 0, 3 x - 4 y = 5$ and $5 x + 12 y = 27$ .

HARD

11th Andhra Pradesh Board

IMPORTANT

Intermediate First Year Mathematics Paper 1B>Chapter 3 - The Straight Line>Exercise 3(e)>Q III

Find the incentre of the triangle formed by the straight lines $x + y - 7 = 0, x - y + 1 = 0$ and $x - 3 y + 5 = 0$ .

HARD

11th Andhra Pradesh Board

IMPORTANT

Intermediate First Year Mathematics Paper 1B>Chapter 3 - The Straight Line>Exercise 3(e)>Q III

A triangle is formed by the lines $a x + b y + c = 0, l x + m y + n = 0$ and $p x + q y + r = 0$ . Given that the triangle is not right-angled, show that the straight line $\frac{a x + b y + c}{a p + b q} = \frac{l x + m y + n}{l p + m q}$ passes through the orthocenter of the triangle.

HARD

11th Andhra Pradesh Board

IMPORTANT

Intermediate First Year Mathematics Paper 1B>Chapter 3 - The Straight Line>Exercise 3(e)>Q III

The Cartesian equations of the sides $B C, C A$ and $A B$ of a triangle are respectively $u_{r} \equiv a_{r} x + b_{r} y + c_{r} = 0, r = 1, 2, 3$ . Show that the equation of the straight line passing through $A$ and bisecting the side $B C$ is $\frac{u_{3}}{a_{3} b_{1} - a_{1} b_{3}} = \frac{u_{2}}{a_{1} b_{2} - a_{2} b_{1}}$ .

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