Coordinate Geometry - Straight Lines (Various Forms and General Equation)

Find the equation of the line passing through the point $(2, 3)$ and perpendicular to the line $3x - 4y + 5 = 0$.

Step 1: Find the slope of the given line. The equation is $3x - 4y + 5 = 0$. Rewriting in $y = mx + c$ form: $4y = 3x + 5 \implies y = \frac{3}{4}x + \frac{5}{4}$. Thus, $m_1 = \frac{3}{4}$. Step 2: Find the slope ($m_2$) of the perpendicular line. Since $m_1 \cdot m_2 = -1$, we have $\frac{3}{4} \cdot m_2 = -1 \implies m_2 = -\frac{4}{3}$. Step 3: Use the point-slope form with point $(2, 3)$ and $m = -\frac{4}{3}$. $y - 3 = -\frac{4}{3}(x - 2)$ $3(y - 3) = -4(x - 2)$ $3y - 9 = -4x + 8 \implies 4x + 3y - 17 = 0$.

To find the equation, we first determine the slope of the given line, then apply the perpendicularity condition ($m_1 m_2 = -1$) to find our required slope, and finally substitute the values into the point-slope formula.

Reduce the equation $\sqrt{3}x + y - 8 = 0$ into normal form and find the values of $p$ and $\alpha$.

Step 1: Write the equation as $\sqrt{3}x + y = 8$. Here, $A = \sqrt{3}$ and $B = 1$. Step 2: Calculate $\sqrt{A^2 + B^2} = \sqrt{(\sqrt{3})^2 + 1^2} = \sqrt{3 + 1} = 2$. Step 3: Divide the entire equation by 2. $\frac{\sqrt{3}}{2}x + \frac{1}{2}y = \frac{8}{2} \implies \frac{\sqrt{3}}{2}x + \frac{1}{2}y = 4$. Step 4: Compare with $x \cos \alpha + y \sin \alpha = p$. We get $\cos \alpha = \frac{\sqrt{3}}{2}$, $\sin \alpha = \frac{1}{2}$, and $p = 4$. Since both sine and cosine are positive, $\alpha$ is in the first quadrant: $\alpha = 30^{\circ}$ or $\frac{\pi}{6}$.

Normal form reduction requires dividing the general equation by $\sqrt{A^2 + B^2}$ to normalize the coefficients into trigonometric values (sine and cosine), while $p$ represents the perpendicular distance from the origin.

1. Find the slope of the given line $2x - 3y + 6 = 0$. By rearranging into $y = mx + c$: $3y = 2x + 6 \implies y = \frac{2}{3}x + 2$ So, $m = \frac{2}{3}$.
2. Parallel lines have equal slopes, so the required line also has $m = \frac{2}{3}$.
3. The x-intercept is given as $4$, which means the line passes through $(4, 0)$.
4. Using Point-Slope form: $y - y_1 = m(x - x_1)$ $y - 0 = \frac{2}{3}(x - 4)$ $3y = 2x - 8$ $2x - 3y - 8 = 0$

Two lines are parallel if their slopes are equal. We extracted the slope from the general form of the first line and applied it to the given point (the x-intercept) to find the new equation.

Let the intercepts on the $x$-axis and $y$-axis be $a$ and $b$ respectively. According to the problem, $a + b = 14 \Rightarrow b = 14 - a$. The equation of the line in intercept form is: $\frac{x}{a} + \frac{y}{b} = 1$ Substituting $b = 14 - a$: $\frac{x}{a} + \frac{y}{14 - a} = 1$ Since the line passes through $(3, 4)$: $\frac{3}{a} + \frac{4}{14 - a} = 1$ $3(14 - a) + 4a = a(14 - a)$ $42 - 3a + 4a = 14a - a^2$ $a^2 - 13a + 42 = 0$ $(a - 6)(a - 7) = 0$ Thus, $a = 6$ or $a = 7$. Case 1: If $a = 6$, then $b = 14 - 6 = 8$. The equation is $\frac{x}{6} + \frac{y}{8} = 1$, which simplifies to $4x + 3y = 24$. Case 2: If $a = 7$, then $b = 14 - 7 = 7$. The equation is $\frac{x}{7} + \frac{y}{7} = 1$, which simplifies to $x + y = 7$.

We use the intercept form of a straight line equation. By expressing one intercept in terms of the other using the given sum, we substitute the coordinates of the given point to form a quadratic equation in 'a'. Solving this gives two possible lines that satisfy the condition.