The straight line 5x+4y=0 passes through the point of intersection of the straight lines x+2y-10=0 and 2x+y+5=

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5x+4y=0

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Concept understanding — Concurrent Lines Condition

What Does "Concurrent Lines" Mean?

Imagine three friends standing in a field. Each friend holds a long, straight rope stretched tight. If all three ropes pass through exactly the same point — say, a flagpole in the centre — then the ropes are concurrent. That common point is called the point of concurrency.

In geometry, when three or more lines all pass through a single point, we say they are concurrent lines. That point is their concurrency point.

Note

Two lines are always concurrent (unless they are parallel) — they meet at exactly one point. The interesting case is three or more lines. Do they all happen to pass through the same spot?

The Intuition Behind the Condition

Suppose you have three lines:

$L_1: a_1x + b_1y + c_1 = 0$
$L_2: a_2x + b_2y + c_2 = 0$
$L_3: a_3x + b_3y + c_3 = 0$

If they are concurrent, there exists some point $(x_0, y_0)$ that satisfies all three equations at once. That means $(x_0, y_0)$ is a common solution.

Now, think about it this way:

The first two lines $L_1$ and $L_2$ intersect at some point $P$ (unless they are parallel). For the three lines to be concurrent, $L_3$ must also pass through that same point $P$ . So the condition boils down to: the point of intersection of any two lines must lie on the third line.

That is the simplest way to check concurrency: solve two equations, get the intersection, and plug it into the third equation. If it satisfies, the lines are concurrent.

The Precise Algebraic Condition

There is a cleaner, more powerful condition using determinants — it avoids solving for the intersection explicitly.

a_1x + b_1y + c_1 = 0

This determinant being zero is the necessary and sufficient condition for concurrency of three lines.

Watch out

This condition assumes that no two of the lines are parallel. If $L_1$ and $L_2$ are parallel, they never meet, so the three lines cannot be concurrent (unless all three are the same line, which is a degenerate case). The determinant condition will still give zero in that parallel case, but the lines are not concurrent — they are parallel. So always check that the lines actually intersect pairwise first.

Why Does the Determinant Work?

Here is the reasoning in plain steps:

For concurrency, there must exist $(x_0, y_0)$ such that:

a_1x_0 + b_1y_0 + c_1 = 0

a_2x_0 + b_2y_0 + c_2 = 0

a_3x_0 + b_3y_0 + c_3 = 0

Think of these as three equations in three unknowns: $x_0$ , $y_0$ , and the constant $1$ . Yes, the constant 1 is treated as a variable here.
For a non-trivial solution to exist (i.e., a solution where the "variables" are not all zero), the determinant of the coefficient matrix must be zero. That is a standard result from linear algebra: a homogeneous system has a non-zero solution only when the determinant is zero.
The determinant being zero is exactly the condition that the three equations are linearly dependent — meaning one equation can be written as a combination of the other two. That is another way to say: the third line passes through the intersection of the first two.

Tip

For quick checks in exams, use the determinant. But if the numbers are simple, solving two equations and substituting into the third is often faster and less error-prone.

Example

Check if these lines are concurrent:

$L_1: 2x + 3y - 5 = 0$

$L_2: x - y + 2 = 0$

$L_3: 3x + 2y - 3 = 0$

Method 1 (substitution):

Solve $L_1$ and $L_2$ : …

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To determine if a line passes through the intersection of two other lines, first find the coordinates of the intersection point by solving the system of equations for the first two lines. Then, substitute these coordinates into the equation of the third line; if the equation holds true, the line passes through the intersection. In this case, the line $5x+4y=0$ does pass through the intersection point.

When we say a straight line "passes through the point of intersection" of two other lines, it means that the coordinates of the point where the first two lines meet must also satisfy the equation of the third line. Essentially, this point is common to all three lines.

To verify the given statement, we need to follow a two-step process:

Find the coordinates of the point where the lines $x+2y-10=0$ and $2x+y+5=0$ intersect.
Check if these coordinates satisfy the equation of the third line, $5x+4y=0$ .

Let's work through this step by step.

Find the point of intersection of the first two lines.

We are given two linear equations:

Equation (1): $x+2y-10=0$

Equation (2): $2x+y+5=0$

We can solve this system using either substitution or elimination. Let's use the elimination method.

From Equation (2), we can express $y$ in terms of $x$ :

$y = -2x-5$

Now, substitute this expression for $y$ into Equation (1):

$x + 2(-2x-5) - 10 = 0$

$x - 4x - 10 - 10 = 0$

$-3x - 20 = 0$

$-3x = 20$

$x = -\frac{20}{3}$

Now, substitute the value of $x$ back into the expression for $y$ :

$y = -2\left(-\frac{20}{3}\right) - 5$

$y = \frac{40}{3} - 5$

To combine these, find a common denominator:

$y = \frac{40}{3} - \frac{15}{3}$

$y = \frac{25}{3}$

So, the point of intersection of the lines $x+2y-10=0$ and $2x+y+5=0$ is $\left(-\frac{20}{3}, \frac{25}{3}\right)$ .
Check if the third line passes through this intersection point.

The equation of the third line is $5x+4y=0$ .

We need to substitute the coordinates of the intersection point $\left(-\frac{20}{3}, \frac{25}{3}\right)$ into this equation and see if it holds true.

Substitute $x = -\frac{20}{3}$ and $y = \frac{25}{3}$ :

$5\left(-\frac{20}{3}\right) + 4\left(\frac{25}{3}\right)$

$= -\frac{100}{3} + \frac{100}{3}$

$= 0$

Since the left-hand side equals the right-hand side ( $0=0$ ), the point $\left(-\frac{20}{3}, \frac{25}{3}\right)$ lies on the line $5x+4y=0$ . …

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Q.The straight line 5x+4y=05x+4y=05x+4y=0 passes through the point of intersection of the straight lines x+2y−10=0x+2y-10=0x+2y−10=0 and 2x+y+5=02x+y+5=02x+y+5=0.

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Q.The straight line $5x+4y=0$ passes through the point of intersection of the straight lines $x+2y-10=0$ and $2x+y+5=0$ .

Unlock everything free for 14 days