Let the three sides of a triangle are on the lines 4x - 7y + 10 = 0, x + y = 5 and 7x + 4y = 15. The

Q: Wait, why do we reject the need to find the third vertex of the triangle?

That is a brilliant observation! In a right-angled triangle, the orthocenter is always the vertex where the right angle is formed. Since we identified that the lines $4x - 7y + 10 = 0$ and $7x + 4y = 15$ are perpendicular, their intersection point is the orthocenter. The third line simply acts as the hypotenuse, which does not affect the location of the orthocenter.

Q: How can I quickly check if two lines are perpendicular without calculating slopes?

You can use the coefficient test! For lines $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$, they are perpendicular if $a_1a_2 + b_1b_2 = 0$. Here, $(4)(7) + (-7)(4) = 28 - 28 = 0$. It is a massive time-saver in the exam hall!

Q: Why is the orthocenter of the triangle formed by $x=0, y=0, x+y=1$ at the origin?

The lines $x=0$ (y-axis) and $y=0$ (x-axis) are perpendicular and meet at $(0,0)$. This forms a right-angled triangle at the origin. Just like the first triangle, the orthocenter must be the vertex of the right angle.

Q: Is there any scenario where the orthocenter lies outside the triangle?

Yes, in an obtuse-angled triangle, the orthocenter lies outside. However, for right-angled triangles, it is always on the vertex, and for acute triangles, it is always inside. Always check the angles first!

Q: I calculated the distance as $\sqrt{5}$, but the options show 5. Did I miss a square?

Double-check your arithmetic! The distance is $\sqrt{(1-0)^2 + (2-0)^2} = \sqrt{1+4} = \sqrt{5}$. If you are seeing 5 as an option, ensure you haven't squared the distance by mistake. Trust your derivation!

Question

Let the three sides of a triangle are on the lines $4x - 7y + 10 = 0, x + y = 5$ and $7x + 4y = 15$. Then the distance of its orthocentre from the orthocentre of the triangle formed by the lines $x = 0, y = 0$ and $x + y = 1$ is

Options:

(A) 5

(B) $\sqrt{5}$

(C) $\sqrt{20}$

(D) 20

Sigma Percentile · Accepted Answer

### The Geometry of Hidden Perfection

Welcome, fellow traveler of the mathematical landscape. Today, we are not just solving a problem; we are uncovering a hidden symmetry.

When you look at a set of lines, do you see just equations, or do you see the skeleton of a geometric shape waiting to be revealed? Let us peel back the layers of this problem together.

### Phase 1: The Detective Work

We are given three lines: $L_1: 4x - 7y + 10 = 0$, $L_2: x + y = 5$, and $L_3: 7x + 4y = 15$. At first glance, they look like a standard system of linear equations.

As an elite student, you know that the secret often lies in the slopes. Let us calculate the slopes $m_1$ and $m_3$ for $L_1$ and $L_3$.

For $L_1$, we rearrange to $y = \frac{4}{7}x + \frac{10}{7}$, giving us $m_1 = \frac{4}{7}$. For $L_3$, we rearrange to $y = -\frac{7}{4}x + \frac{15}{4}$, giving us $m_3 = -\frac{7}{4}$.

Look at that! The product $m_1 \cdot m_3 = \frac{4}{7} \cdot (-\frac{7}{4}) = -1$. This is the 'Aha!' moment.

In the world of coordinate geometry, when the product of slopes is $-1$, the lines are perpendicular. We have discovered that our triangle is a right-angled triangle.

This changes everything. We no longer need to find the intersection of all three lines to locate the orthocenter. We know that in a right-angled triangle, the orthocenter is simply the vertex where the right angle resides.

### Phase 2: Pinpointing the Orthocenter

Since $L_1$ and $L_3$ are the perpendicular sides, their intersection point is our orthocenter $H_1$. We solve the system:

$$4x - 7y = -10$$
$$7x + 4y = 15$$

To eliminate $y$, we multiply the first by $4$ and the second by $7$:

$$16x - 28y = -40$$
$$49x + 28y = 105$$

Adding these equations yields $65x = 65$, so $x = 1$. Substituting $x=1$ back into $4x - 7y = -10$, we get $4 - 7y = -10$, which simplifies to $7y = 14$, or $y = 2$.

Thus, our first orthocenter is $H_1(1, 2)$.

### Phase 3: The Second Triangle and the Final Leap

Now, let us look at the second triangle formed by $x=0$, $y=0$, and $x+y=1$. These are the coordinate axes and a line connecting them.

This is the most fundamental right-angled triangle, with the right angle sitting comfortably at the origin $(0, 0)$. Therefore, the orthocenter $H_2$ is simply $(0, 0)$.

We have arrived at the final stage of our journey. We need the distance between $H_1(1, 2)$ and $H_2(0, 0)$. Using the distance formula:

$$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$$

$$d = \sqrt{(1 - 0)^2 + (2 - 0)^2} = \sqrt{1^2 + 2^2} = \sqrt{5}$$

There it is. The beauty of this problem lies not in the complexity of the algebra, but in the elegance of the geometric properties.

You didn't need to struggle with altitudes; you only needed to recognize the perpendicularity. Keep this mindset—always look for the simplest path through the forest of equations.

**The final answer is $\sqrt{5}$.**

Animated Solution for Mathematics - Straight Lines: Let the three sides of a triangle are on the lines $4 x - 7 y + 10 = 0, x + y = 5$ and $7 x + 4 y = 15$ . Then the distance of its orthocentre from the orthocentre of the triangle formed by the lines $x = 0, y = 0$ and $x + y = 1$ is

Select Answer:

Visualized Solution

Analyze Triangle 1 Lines

Check for Perpendicularity

Right-Angled Triangle Property

Find Orthocenter $H_{1}$ (Setup)

Find Orthocenter $H_{1}$ (Solve)

Find Orthocenter $H_{1}$ (Y-coordinate)

Analyze Triangle 2

Find Orthocenter $H_{2}$

Distance Between Orthocenters (Setup)

Final Calculation

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

The Geometry of Hidden Perfection

Phase 1

Phase 2

Phase 3

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Select Answer:

Visualized Solution

Analyze Triangle 1 Lines

Check for Perpendicularity

Right-Angled Triangle Property

Find Orthocenter .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } H1​ (Setup)

Find Orthocenter .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } H1​ (Solve)

Find Orthocenter .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } H1​ (Y-coordinate)