Finding the Equation of Line PQ

The problem asks us to find the equation of a straight line connecting two points, P and Q. Point P is the intersection of a line passing through A(3, 2) and the line $2y = x + 3$, where these two lines are perpendicular. Point Q is the intersection of another line passing through A(3, 2) and the x-axis, where these two lines are also perpendicular.

Finding the Coordinates of Point P

Point P lies on the line $2y = x + 3$. We can rewrite this as $x - 2y + 3 = 0$.

The slope of this line is $m_1 = \frac{\text{coefficient of } x}{\text{coefficient of } y} = -\frac{1}{-2} = \frac{1}{2}$.

The line passing through A(3, 2) and P is perpendicular to $x - 2y + 3 = 0$. If two lines are perpendicular, the product of their slopes is -1. Let the slope of line AP be $m_{AP}$.

So, $m_{AP} \times m_1 = -1$, which means $m_{AP} \times \frac{1}{2} = -1$. Therefore, $m_{AP} = -2$.

Now we can write the equation of the line AP using the point-slope form $y - y_1 = m(x - x_1)$ with A(3, 2) and $m_{AP} = -2$:

$y - 2 = -2(x - 3)$

$y - 2 = -2x + 6$

$2x + y - 8 = 0$

Point P is the intersection of the lines $x - 2y + 3 = 0$ and $2x + y - 8 = 0$. We can solve this system of equations:

Equation (1): $x - 2y = -3$

Equation (2): $2x + y = 8$

Multiply Equation (2) by 2:

$2 \times (2x + y) = 2 \times 8$

$4x + 2y = 16$ (Equation 3)

Add Equation (1) and Equation (3):

$(x - 2y) + (4x + 2y) = -3 + 16$

$5x = 13$

$x = \frac{13}{5}$

Substitute the value of $x$ into Equation (2):

$2\left(\frac{13}{5}\right) + y = 8$

$\frac{26}{5} + y = 8$

$y = 8 - \frac{26}{5}$

$y = \frac{40 - 26}{5}$

$y = \frac{14}{5}$

So, the coordinates of point P are $\left(\frac{13}{5}, \frac{14}{5}\right)$.

Finding the Coordinates of Point Q

Point Q lies on the x-axis, which has the equation $y = 0$.

The x-axis is a horizontal line, and its slope is $m_x = 0$.

The line passing through A(3, 2) and Q is perpendicular to the x-axis. A line perpendicular to a horizontal line is a vertical line. A vertical line has an undefined slope and its equation is of the form $x = k$, where $k$ is the x-coordinate of any point on the line.

Since this line passes through A(3, 2), its equation must be $x = 3$.

Point Q is the intersection of the line $x = 3$ and the x-axis ($y = 0$).

So, the coordinates of point Q are $(3, 0)$.

Finding the Equation of Line PQ

We have the coordinates of P $\left(\frac{13}{5}, \frac{14}{5}\right)$ and Q $(3, 0)$. We can find the equation of the line PQ using the two-point form or by finding the slope and using the point-slope form.

Let's find the slope of PQ, $m_{PQ}$:

$m_{PQ} = \frac{y_2 - y_1}{x_2 - x_1} = \frac{0 - \frac{14}{5}}{3 - \frac{13}{5}}$

$m_{PQ} = \frac{-\frac{14}{5}}{\frac{15}{5} - \frac{13}{5}} = \frac{-\frac{14}{5}}{\frac{2}{5}}$

$m_{PQ} = \frac{-14}{2} = -7$

Now, using the point-slope form with point Q(3, 0) and slope $m_{PQ} = -7$:

$y - y_1 = m(x - x_1)$

$y - 0 = -7(x - 3)$

$y = -7x + 21$

Rearranging the terms to the general form $Ax + By + C = 0$:

$7x + y - 21 = 0$

This is the equation of the line PQ.

Checking the Options

Comparing our derived equation $7x + y - 21 = 0$ with the given options, we find that it matches Option 1.

Revision Table: Key Concepts in Coordinate Geometry

Additional Information: Lines and Their Properties

Straight lines are fundamental objects in coordinate geometry. Their properties, such as slope and intercepts, help us understand their orientation and position in the Cartesian plane. The equation of a line is a mathematical way to represent all the points that lie on that line.

• The slope of a line tells us how steep it is and its direction. A positive slope indicates the line rises from left to right, while a negative slope indicates it falls. A slope of zero means the line is horizontal, and an undefined slope means it is vertical.
• Perpendicular lines intersect at a right angle ($90^\circ$). The relationship between their slopes ($m_1 m_2 = -1$) is a key property used to find the equation of a line perpendicular to a given line. This relationship does not apply if one line is horizontal and the other is vertical (slopes 0 and undefined), but they are still perpendicular.
• Finding the point of intersection of two lines involves solving the system of equations representing those lines. This point satisfies both equations simultaneously.
• The x-axis is the horizontal line where the y-coordinate is always 0. Its equation is $y=0$.
• The y-axis is the vertical line where the x-coordinate is always 0. Its equation is $x=0$.

Understanding these basic concepts is crucial for solving problems involving lines and their intersections.