JEE Advanced 1979

LEVELJEE Advanced

Animated Solution for Mathematics - Straight Lines: (a) Two vertices of a triangle are $(5, - 1)$ and $(- 2, 3)$ . If the orthocentre of the triangle is the origin, find the coordinates of the third point. (b) Find the equation of the line which bisects the obtuse angle between the lines $x - 2 y + 4 = 0$ and $4 x - 3 y + 2 = 0$ .

Visualized Solution

Visualizing the Triangle and Orthocenter

Given vertices: $A (5, - 1)$ and $B (- 2, 3)$
Orthocenter: $H (0, 0)$ (the origin)
Let the third vertex be $C (h, k)$

Property of Orthocenter: $A H ⊥ B C$

Property: The altitude from $A$ passes through the orthocenter $H$ and is perpendicular to the opposite side $B C$ .
Therefore, $Slope of A H \times Slope of B C = - 1$

Setting up the First Equation

Slope of $A H = \frac{- 1 - 0}{5 - 0} = - \frac{1}{5}$
Slope of $B C = \frac{k - 3}{h - ( - 2 )} = \frac{k - 3}{h + 2}$
Substituting into the perpendicularity condition:
$(- \frac{1}{5}) \times (\frac{k - 3}{h + 2}) = - 1$

Simplifying to Linear Form

Multiply both sides by $- 5 (h + 2)$ :
$k - 3 = 5 (h + 2)$
$k - 3 = 5 h + 10$
Equation 1: $5 h - k + 13 = 0$

Property of Orthocenter: $B H ⊥ A C$

Property: Similarly, the altitude from $B$ passes through $H$ and is perpendicular to $A C$ .
Therefore, $Slope of B H \times Slope of A C = - 1$

Setting up the Second Equation

Slope of $B H = \frac{3 - 0}{- 2 - 0} = - \frac{3}{2}$
Slope of $A C = \frac{k - ( - 1 )}{h - 5} = \frac{k + 1}{h - 5}$
Substituting into the perpendicularity condition:
$(- \frac{3}{2}) \times (\frac{k + 1}{h - 5}) = - 1$

Simplifying the Second Equation

Multiply both sides by $- \frac{2}{3} (h - 5)$ :
$3 (k + 1) = 2 (h - 5)$
$3 k + 3 = 2 h - 10$
Equation 2: $2 h - 3 k - 13 = 0$

Solving for Vertex $C (h, k)$

From Equation 1: $k = 5 h + 13$
Substitute into Equation 2:
$2 h - 3 (5 h + 13) - 13 = 0$
$- 13 h - 39 - 13 = 0 \Rightarrow - 13 h = 52 \Rightarrow h = - 4$
Substitute $h = - 4$ back: $k = 5 (- 4) + 13 = - 7$
The third vertex is $C (- 4, - 7)$

Part (b): Lines and Angle Bisectors

Given lines:
$L_{1} : x - 2 y + 4 = 0$
$L_{2} : 4 x - 3 y + 2 = 0$
Goal: Find the equation of the obtuse angle bisector.

The $a_{1} a_{2} + b_{1} b_{2}$ Test

Step 1: Ensure constant terms are positive: $c_{1} = 4 > 0$ and $c_{2} = 2 > 0$ .
Step 2: Calculate $a_{1} a_{2} + b_{1} b_{2}$ :
$a_{1} a_{2} + b_{1} b_{2} = (1) (4) + (- 2) (- 3) = 4 + 6 = 10$
Since $a_{1} a_{2} + b_{1} b_{2} > 0$ , the positive sign in the formula gives the obtuse angle bisector.

Applying the Bisector Formula

Formula: $\frac{a _{1} x + b _{1} y + c _{1}}{a _{1}^{2} + b _{1}^{2}} = + \frac{a _{2} x + b _{2} y + c _{2}}{a _{2}^{2} + b _{2}^{2}}$
Substitute the values:
$\frac{x - 2 y + 4}{1 ^{2} + ( - 2 ) ^{2}} = \frac{4 x - 3 y + 2}{4 ^{2} + ( - 3 ) ^{2}}$
$\frac{x - 2 y + 4}{5} = \frac{4 x - 3 y + 2}{5}$

Final Equation of the Bisector

Multiply both sides by $5$ :
$5 (x - 2 y + 4) = 4 x - 3 y + 2$
Rearrange terms to standard form $A x + B y + C = 0$ :
$(4 - 5) x + (25 - 3) y - (45 - 2) = 0$

Summary and Key Takeaways

Part (a) Answer: The coordinates of the third vertex $C$ are $(- 4, - 7)$ .
Part (b) Answer: The obtuse angle bisector is $(4 - 5) x + (25 - 3) y - (45 - 2) = 0$ .
Key Concept: Orthocenter implies perpendicular altitudes. For angle bisectors, use the $a_{1} a_{2} + b_{1} b_{2}$ sign test.

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The Sigma Insight: Centroid, Incenter, Orthocenter, and Circumcenter

Analyzing the Orthocenter (Part a)

The orthocenter

H (0, 0)

is the intersection of the altitudes of

△ A B C

. Given vertices

A (5, - 1)

and

B (- 2, 3)

, we seek the third vertex

C (h, k)

Since

A H

is an altitude, it is perpendicular to side

B C

. The slope of

A H

is calculated as:

m_{A H} = \frac{- 1 - 0}{5 - 0} = - \frac{1}{5}

The slope of side

B C

using points

C (h, k)

and

B (- 2, 3)

is:

m_{B C} = \frac{k - 3}{h - ( - 2 )} = \frac{k - 3}{h + 2}

Since

m_{A H} \cdot m_{B C} = - 1

, we have:

(- \frac{1}{5}) \cdot (\frac{k - 3}{h + 2}) = - 1 \Rightarrow k - 3 = 5 (h + 2)

This simplifies to the linear equation:

5 h - k + 13 = 0

Finding the Third Vertex

We repeat this logic for the altitude

B H

, which is perpendicular to side

A C

. The slope of

B H

is:

m_{B H} = \frac{3 - 0}{- 2 - 0} = - \frac{3}{2}

The slope of side

A C

is:

m_{A C} = \frac{k - ( - 1 )}{h - 5} = \frac{k + 1}{h - 5}

Setting the product of the slopes to

- 1

(- \frac{3}{2}) \cdot (\frac{k + 1}{h - 5}) = - 1 \Rightarrow 3 (k + 1) = 2 (h - 5)

This simplifies to:

2 h - 3 k - 13 = 0

Solving the system of equations

k = 5 h + 13

and

2 h - 3 k - 13 = 0

by substitution:

2 h - 3 (5 h + 13) - 13 = 0 \Rightarrow - 13 h = 52 \Rightarrow h = - 4

Substituting

h = - 4

back into the first equation gives

k = 5 (- 4) + 13 = - 7

. Thus, the third vertex is $C (- 4, - 7)$ .

The Obtuse Angle Bisector (Part b)

Given lines

L_{1} : x - 2 y + 4 = 0

and

L_{2} : 4 x - 3 y + 2 = 0

, we identify the obtuse angle bisector. First, we ensure the constant terms are positive, which they are.

We compute the value of

a_{1} a_{2} + b_{1} b_{2}

(1) (4) + (- 2) (- 3) = 4 + 6 = 10

Since

10 > 0

, the positive sign in the bisector formula corresponds to the obtuse angle bisector.

The equation is given by:

\frac{x - 2 y + 4}{1 ^{2} + ( - 2 ) ^{2}} = \frac{4 x - 3 y + 2}{4 ^{2} + ( - 3 ) ^{2}}

\frac{x - 2 y + 4}{5} = \frac{4 x - 3 y + 2}{5}

Multiplying both sides by

5

5 (x - 2 y + 4) = 4 x - 3 y + 2

Rearranging into standard form, we obtain the final equation:

(4 - 5) x + (25 - 3) y - (45 - 2) = 0

Common Traps & FAQs

Wait, why can't we just use the centroid formula to find the third vertex?

That is a very common trap! The centroid is the average of the vertices, but the orthocenter is a completely different beast. It is defined by the intersection of altitudes, not medians. You must use the perpendicularity property of the altitudes to find the third vertex. The centroid formula will lead you to the wrong point entirely.

How do I know if I am finding the acute or obtuse angle bisector?

This is the classic coordinate geometry challenge! You must first normalize the constant terms to be positive. Once they are positive, calculate the value of

a_{1} a_{2} + b_{1} b_{2}

. If this value is positive, the '+' sign in the bisector formula gives you the obtuse angle bisector. If it is negative, the '+' sign gives you the acute one. It is a simple test that saves you from guessing!

What if the constant terms in the line equations are negative? Does the formula still work?

Great question! The formula for the angle bisector relies on the constant terms being positive. If you have a negative constant, you must multiply the entire equation of the line by

- 1

to make the constant term positive before you start the calculation. If you skip this step, your sign test will fail, and you will end up with the wrong bisector.

Visualized Solution

Visualizing the Triangle and Orthocenter

Property of Orthocenter: $A H ⊥ B C$

Setting up the First Equation

Simplifying to Linear Form

Property of Orthocenter: $B H ⊥ A C$

Setting up the Second Equation

Simplifying the Second Equation

Solving for Vertex $C (h, k)$

Part (b): Lines and Angle Bisectors

The $a_{1} a_{2} + b_{1} b_{2}$ Test

Applying the Bisector Formula

Final Equation of the Bisector

Summary and Key Takeaways

The Sigma Insight: Centroid, Incenter, Orthocenter, and Circumcenter

Analyzing the Orthocenter (Part a)

Finding the Third Vertex

The Obtuse Angle Bisector (Part b)

Similar Questions

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Visualized Solution

Visualizing the Triangle and Orthocenter

Property of Orthocenter: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } AH⊥BC

Setting up the First Equation

Simplifying to Linear Form

Property of Orthocenter: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } BH⊥AC

Setting up the Second Equation

Simplifying the Second Equation

Solving for Vertex .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } C(h,k)

Part (b): Lines and Angle Bisectors

The .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a1​a2​+b1​b2​ Test

Applying the Bisector Formula

Final Equation of the Bisector

Summary and Key Takeaways

The Sigma Insight: Centroid, Incenter, Orthocenter, and Circumcenter

Analyzing the Orthocenter (Part a)

Finding the Third Vertex

The Obtuse Angle Bisector (Part b)

Similar Questions

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Property of Orthocenter: $A H ⊥ B C$

Property of Orthocenter: $B H ⊥ A C$

Solving for Vertex $C (h, k)$

The $a_{1} a_{2} + b_{1} b_{2}$ Test

Let the equations of two sides of a triangle be $3 x - 2 y + 6 = 0$ and $4 x + 5 y - 20 = 0$ . If the orthocentre of this triangle is at (1,1), then the equation of its third side is :

Two vertices of a triangle are $(0, 2)$ and $(4, 3)$ . If its orthocentre is at the origin, then its third vertex lies in which quadrant ?

The vertices of a triangle are $[a t_{1} t_{2}, a (t_{1} + t_{2})]$ , $[a t_{2} t_{3}, a (t_{2} + t_{3})]$ , $[a t_{3} t_{1}, a (t_{3} + t_{1})]$ . Find the orthocentre of the triangle.

Let k be an integer such that the triangle with vertices $(k, - 3 k)$ , $(5, k)$ and $(- k, 2)$ has area 28 sq. units. Then the orthocentre of this triangle is at the point:

The distance of the origin from the centroid of the triangle whose two sides have the equations $x - 2 y + 1 = 0$ and $2 x - y - 1 = 0$ and whose orthocenter is $(\frac{7}{3}, \frac{7}{3})$ is:

The orthocentre of the triangle formed by the lines $x y = 0$ and $x + y = 1$ is

If the orthocentre of the triangle formed by the lines $y = x + 1, y = 4 x - 8$ and $y = m x + c$ is at $(3, - 1)$ , then $m - c$ is :