Coordinate Geometry - Intercept Form-advanced

Find the intercepts made by the line $4x - 3y = 12$ on the coordinate axes. Also, find the area of the triangle formed by this line with the axes.

Given equation: $4x - 3y = 12$ Divide both sides by $12$ to make the right side equal to $1$: $\frac{4x}{12} - \frac{3y}{12} = \frac{12}{12}$ $\frac{x}{3} - \frac{y}{4} = 1$ This can be written as: $\frac{x}{3} + \frac{y}{-4} = 1$ Comparing with $\frac{x}{a} + \frac{y}{b} = 1$, we get: $x$-intercept $a = 3$ $y$-intercept $b = -4$ Area of triangle = $\frac{1}{2} |a \times b|$ $\text{Area} = \frac{1}{2} |3 \times (-4)|$ $\text{Area} = \frac{1}{2} \times 12 = 6 \text{ square units}$

To find the intercepts, we transform the equation into the form $\frac{x}{a} + \frac{y}{b} = 1$. The denominators $a$ and $b$ represent the intercepts. The area is calculated using the product of the magnitudes of these intercepts.

Find the equation of a line that passes through the point $(2, 3)$ such that the $x$-intercept is twice the $y$-intercept.

Let the $y$-intercept be $b$. Given that the $x$-intercept $a = 2b$. The equation of the line in intercept form is: $\frac{x}{a} + \frac{y}{b} = 1$ Substitute $a = 2b$: $\frac{x}{2b} + \frac{y}{b} = 1$ Multiply the entire equation by $2b$: $x + 2y = 2b$ Since the line passes through $(2, 3)$, substitute $x = 2$ and $y = 3$: $2 + 2(3) = 2b$ $2 + 6 = 2b$ $8 = 2b \Rightarrow b = 4$ Now, find $a$: $a = 2b = 2(4) = 8$ Substitute $a$ and $b$ back into the intercept form: $\frac{x}{8} + \frac{y}{4} = 1$ Alternatively, in general form: $x + 2y - 8 = 0$

We use the relationship between $a$ and $b$ to reduce the equation to one variable, then use the given point to solve for that variable.

1. Let the equation of the line be $\frac{x}{a} + \frac{y}{b} = 1$.
2. The intercepts are $(a, 0)$ and $(0, b)$.
3. The midpoint of the segment is $(\frac{a+0}{2}, \frac{0+b}{2}) = (\frac{a}{2}, \frac{b}{2})$.
4. Given the midpoint is $(1, 4)$, we have: $\frac{a}{2} = 1 \implies a = 2$ $\frac{b}{2} = 4 \implies b = 8$
5. The equation is $\frac{x}{2} + \frac{y}{8} = 1$, which simplifies to $4x + y = 8$.
6. Length $L = \sqrt{a^2 + b^2} = \sqrt{2^2 + 8^2} = \sqrt{4 + 64} = \sqrt{68} = 2\sqrt{17}$ units.

We use the midpoint formula because the problem states the segment between the axes is bisected at the given point. This directly gives us the values of the intercepts.

1. Let the intercepts be $a$ and $b$. We are given $a + b = 14$, so $b = 14 - a$.
2. The equation is $\frac{x}{a} + \frac{y}{14-a} = 1$.
3. Since it passes through $(3, 4)$: $\frac{3}{a} + \frac{4}{14-a} = 1$
4. Multiplying by $a(14-a)$: $3(14-a) + 4a = a(14-a)$ $42 - 3a + 4a = 14a - a^2$ $a^2 - 13a + 42 = 0$
5. Factorizing the quadratic: $(a-6)(a-7) = 0 \implies a = 6 \text{ or } a = 7$
6. Case 1: $a=6, b=8 \implies \frac{x}{6} + \frac{y}{8} = 1 \implies 4x + 3y = 24$
7. Case 2: $a=7, b=7 \implies \frac{x}{7} + \frac{y}{7} = 1 \implies x + y = 7$

We express one intercept in terms of the other using the sum condition and then substitute the coordinates of the given point to solve for $a$.