Finding the Equation of a Straight Line

The problem asks us to find the equation of a straight line that satisfies two conditions:

1. It passes through the point where two given lines intersect.
2. It is perpendicular to a third given line.

Let's break this down into steps.

Step 1: Find the Point of Intersection of the Two Lines

We are given the equations of two lines:

• Equation 1: \(x + 2y = 5\)
• Equation 2: \(3x + 7y = 17\)

We can solve this system of linear equations to find the point \((x, y)\) where they intersect. Let's use the substitution method.

From Equation 1, we can express \(x\) in terms of \(y\):

\(x = 5 - 2y\)

Now, substitute this expression for \(x\) into Equation 2:

\(3(5 - 2y) + 7y = 17\)

Distribute the 3:

\(15 - 6y + 7y = 17\)

Combine the \(y\) terms:

\(15 + y = 17\)

Subtract 15 from both sides to find \(y\):

\(y = 17 - 15\)

\(y = 2\)

Now substitute the value of \(y\) back into the expression for \(x\):

\(x = 5 - 2(2)\)

\(x = 5 - 4\)

\(x = 1\)

So, the point of intersection of the two lines is \((1, 2)\). The required straight line passes through this point.

Step 2: Find the Slope of the Perpendicular Line

The required straight line is perpendicular to the line \(3x + 4y = 10\). Let's find the slope of this line.

We can rewrite the equation in the slope-intercept form \(y = mx + c\), where \(m\) is the slope.

\(3x + 4y = 10\)

Subtract \(3x\) from both sides:

\(4y = -3x + 10\)

Divide by 4:

\(y = -\frac{3}{4}x + \frac{10}{4}\)

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

The slope of this line is \(m_1 = -\frac{3}{4}\).

If two lines are perpendicular, the product of their slopes is -1. Let \(m_2\) be the slope of the required straight line. Then:

\(m_1 \cdot m_2 = -1\)

\(-\frac{3}{4} \cdot m_2 = -1\)

To find \(m_2\), multiply both sides by \(-\frac{4}{3}\):

\(m_2 = -1 \cdot \left(-\frac{4}{3}\right)\)

\(m_2 = \frac{4}{3}\)

The slope of the required straight line is \(\frac{4}{3}\).

Step 3: Find the Equation of the Required Straight Line

We have the point that the line passes through, \((x_1, y_1) = (1, 2)\), and the slope \(m = \frac{4}{3}\). We can use the point-slope form of the equation of a line, which is \(y - y_1 = m(x - x_1)\).

Substitute the values:

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

To eliminate the fraction, multiply both sides by 3:

\(3(y - 2) = 4(x - 1)\)

Distribute on both sides:

\(3y - 6 = 4x - 4\)

To write the equation in the standard form \(Ax + By + C = 0\), move all terms to one side, usually keeping the coefficient of \(x\) positive:

\(0 = 4x - 3y - 4 + 6\)

\(0 = 4x - 3y + 2\)

So, the equation of the required straight line is \(4x - 3y + 2 = 0\).

Step 4: Compare with Options

Let's compare our derived equation \(4x - 3y + 2 = 0\) with the given options:

• Option 1: \(4x + 3y + 2 = 0\) (Does not match)
• Option 2: \(4x – y + 2 = 0\) (Does not match)
• Option 3: \(4x – 3y – 2 = 0\) (Does not match)
• Option 4: \(4x – 3y + 2 = 0\) (Matches)

The equation of the straight line is \(4x - 3y + 2 = 0\).

Revision Table: Straight Line Concepts

Additional Information: Solving System of Linear Equations

Solving a system of two linear equations with two variables means finding the values of the variables that satisfy both equations at the same time. This point \((x, y)\) represents the point of intersection of the two lines represented by the equations.

Common methods to solve such systems include:

• Substitution Method: Solve one equation for one variable, and then substitute that expression into the other equation. This reduces the system to a single equation with one variable.
• Elimination Method: Multiply one or both equations by constants so that the coefficients of one variable are opposites. Then add the equations together to eliminate that variable, resulting in a single equation with one variable.
• Matrix Method: Using matrices (like Cramer's rule or Gaussian elimination) is another way, especially useful for larger systems.

In this problem, we used the substitution method, which is often straightforward when one of the variables in an equation has a coefficient of 1 or -1.