The orthocentre of the triangle formed by the lines x+y=1, 2x + 3y = 6 and 4x - y + 4 = 0 lies in qu

Q: Wait, why do we need to find the vertices first? Can't we just use the orthocenter formula?

While there are general formulas for the orthocenter, they are often complex and prone to memory errors. In JEE Advanced, the most robust approach is to derive it from first principles: find the vertices, find the slopes of the sides, find the slopes of the altitudes, and then find the intersection. This method is foolproof and builds a deeper understanding of the geometry.

Q: Why is the orthocenter outside the triangle? Did I make a mistake?

You haven't made a mistake! The orthocenter of an obtuse triangle always lies outside the triangle. If you calculate the slopes of the sides and find that the triangle has an obtuse angle, it is perfectly natural for the orthocenter to fall outside the boundary of the triangle. Trust your calculations.

Q: How do I know which line is which side? The question just gives three lines.

The three lines intersect to form the triangle. Any two lines intersect at a vertex. By solving the system of equations for each pair of lines, you will naturally find the three vertices. For example, the intersection of $L_1$ and $L_3$ gives you one vertex, and so on. Just be systematic in your pairing.

Question

The orthocentre of the triangle formed by the lines $x+y=1, 2x + 3y = 6$ and $4x - y + 4 = 0$ lies in quadrant number .........

Sigma Percentile · Accepted Answer

### Analyzing the Setup

To find the orthocenter of the triangle formed by the lines $L_1: x+y=1$, $L_2: 2x+3y=6$, and $L_3: 4x-y+4=0$, we must first determine the coordinates of the vertices $A$, $B$, and $C$.

### Pinning Down the Vertices

Vertex $A$ is the intersection of $L_1$ and $L_3$. Solving the system:
$$x+y=1$$
$$4x-y=-4$$
Adding these equations yields $5x = -3$, so $x = -\frac{3}{5}$. Substituting back, we find $y = 1 - (-\frac{3}{5}) = \frac{8}{5}$. Thus, $A = \left(-\frac{3}{5}, \frac{8}{5}\right)$.

Vertex $B$ is the intersection of $L_1$ and $L_2$. Substituting $x = 1-y$ into $2x+3y=6$:
$$2(1-y) + 3y = 6 \Rightarrow 2 - 2y + 3y = 6 \Rightarrow y = 4$$
With $y=4$, we find $x = 1-4 = -3$. Thus, $B = (-3, 4)$.

Vertex $C$ is the intersection of $L_2$ and $L_3$. Solving $2x+3y=6$ and $y=4x+4$:
$$2x + 3(4x+4) = 6 \Rightarrow 2x + 12x + 12 = 6 \Rightarrow 14x = -6$$
This gives $x = -\frac{3}{7}$. Substituting into $y=4x+4$, we get $y = 4(-\frac{3}{7}) + 4 = \frac{16}{7}$. Thus, $C = \left(-\frac{3}{7}, \frac{16}{7}\right)$.

### The Art of Altitudes

The orthocenter is the intersection of the altitudes. An altitude is a line passing through a vertex perpendicular to the opposite side.

For the altitude from $A$ to side $BC$ (which lies on $L_2$):
The slope of $L_2$ is $-\frac{2}{3}$, so the altitude slope is $m_1 = \frac{3}{2}$. Using point $A\left(-\frac{3}{5}, \frac{8}{5}\right)$:
$$y - \frac{8}{5} = \frac{3}{2}\left(x + \frac{3}{5}\right)$$
Multiplying by $10$ gives $10y - 16 = 15x + 9$, which simplifies to:
$$3x - 2y + 5 = 0$$

For the altitude from $B$ to side $AC$ (which lies on $L_3$):
The slope of $L_3$ is $4$, so the altitude slope is $m_2 = -\frac{1}{4}$. Using point $B(-3, 4)$:
$$y - 4 = -\frac{1}{4}(x + 3)$$
Multiplying by $4$ gives $4y - 16 = -x - 3$, which simplifies to:
$$x + 4y - 13 = 0$$

### Final Calculation

The orthocenter $H$ is the intersection of $3x - 2y + 5 = 0$ and $x + 4y - 13 = 0$. From the second equation, $x = 13 - 4y$. Substituting this into the first:
$$3(13 - 4y) - 2y + 5 = 0$$
$$39 - 12y - 2y + 5 = 0 \Rightarrow 14y = 44 \Rightarrow y = \frac{22}{7}$$

Now, solve for $x$:
$$x = 13 - 4\left(\frac{22}{7}\right) = \frac{91 - 88}{7} = \frac{3}{7}$$

The orthocenter of the triangle is **$H\left(\frac{3}{7}, \frac{22}{7}\right)$**.

Animated Solution for Mathematics - Straight Lines: The orthocentre of the triangle formed by the lines $x + y = 1, 2 x + 3 y = 6$ and $4 x - y + 4 = 0$ lies in quadrant number .........

Visualized Solution

The Geometric Setup

Finding Vertex $A$

Coordinates of $A$

Finding Vertex $B$

Coordinates of $B$

Finding Vertex $C$

Coordinates of $C$

Altitude from $A$ to $B C$

Equation of Altitude from $A$

Altitude from $B$ to $A C$

Equation of Altitude from $B$

Finding the Orthocenter $H$

Coordinates of $H$

Determining the Quadrant

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

Analyzing the Setup

Pinning Down the Vertices

The Art of Altitudes

Final Calculation

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Animated Solution for Mathematics - Straight Lines: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The orthocentre of the triangle formed by the lines x+y=1,2x+3y=6 and 4x−y+4=0 lies in quadrant number .........

Visualized Solution

The Geometric Setup

Finding Vertex .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A

Coordinates of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A

Finding Vertex .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } B

Coordinates of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } B

Finding Vertex .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } C

Coordinates of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } C

Altitude from .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A to .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } BC

Equation of Altitude from .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A

Altitude from .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } B to .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } AC

Equation of Altitude from .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } B

Finding the Orthocenter .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } H

Coordinates of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } H