Coordinate Geometry - The x-intercept-advanced

Find the $x$-intercept of the linear equation $3x - 4y = 12$.

To find the $x$-intercept, we set $y = 0$ in the given equation: $3x - 4(0) = 12$ $3x = 12$ $x = \frac{12}{3}$ $x = 4$ The $x$-intercept is $4$, and the coordinates are $(4, 0)$.

By setting $y=0$, we isolate the point where the line must cross the horizontal axis.

Determine the $x$-intercept of a line that passes through the points $(2, 3)$ and $(4, 7)$.

First, find the equation of the line. The slope $m$ is: $m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{7 - 3}{4 - 2} = \frac{4}{2} = 2$ Using point-slope form $y - y_1 = m(x - x_1)$: $y - 3 = 2(x - 2)$ $y - 3 = 2x - 4$ $y = 2x - 1$ To find the $x$-intercept, set $y = 0$: $0 = 2x - 1$ $2x = 1$ $x = \frac{1}{2}$ The $x$-intercept is $\frac{1}{2}$ or $(0.5, 0)$.

We first derive the linear equation using the two-point formula and then solve for $x$ by setting $y=0$.

Find the area of the triangle formed by the line $2x + 5y = 20$ and the coordinate axes.

Find the $x$-intercept (set $y=0$): $2x + 5(0) = 20 \implies 2x = 20 \implies x = 10$ Find the $y$-intercept (set $x=0$): $2(0) + 5y = 20 \implies 5y = 20 \implies y = 4$ The vertices of the triangle are $(0, 0)$, $(10, 0)$, and $(0, 4)$. $\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}$ $\text{Area} = \frac{1}{2} \times 10 \times 4$ $\text{Area} = 20 \text{ sq. units}$

The $x$-intercept and $y$-intercept provide the lengths of the base and height of a right-angled triangle formed with the origin.

Substitute the coordinates $(2, 3)$ into the equation: $\frac{2}{k} + \frac{3}{5} = 1$ $\frac{2}{k} = 1 - \frac{3}{5}$ $\frac{2}{k} = \frac{2}{5}$ $k = 5$ Since the equation is in the form $\frac{x}{a} + \frac{y}{b} = 1$, the $x$-intercept is $a = k = 5$.

By substituting a known point into the intercept form equation, we can solve for the missing intercept parameter $k$. In this form, $k$ directly represents the $x$-intercept.

1. Let the $y$-intercept be $b$. According to the problem, the $x$-intercept $a$ is $2b$.
2. Use the intercept form of a line: $\frac{x}{a} + \frac{y}{b} = 1$
3. Substitute $a = 2b$: $\frac{x}{2b} + \frac{y}{b} = 1$
4. Multiply the entire equation by $2b$ to simplify: $x + 2y = 2b$
5. Since the line passes through $(3, 4)$, substitute $x = 3$ and $y = 4$ into the equation: $3 + 2(4) = 2b \implies 3 + 8 = 2b \implies 11 = 2b \implies b = 5.5$
6. Calculate the $x$-intercept: $a = 2b = 2(5.5) = 11$
7. The equation of the line is $x + 2y = 11$.

In this problem, we use the intercept form of the linear equation $\frac{x}{a} + \frac{y}{b} = 1$. By relating the two intercepts and using a given point on the line, we can solve for the specific values of the intercepts and the equation itself.