Question

Find the x -intercept of a line with a slope of - 2 and a point ( 3 , 4 ) .

Answer 1

We are given information that we can use to create a point-slope equation for the linear equation. Remember, point-slope requires one ordered pair (x, y) and the slope of the line.

Point-slope: (y-y₁)=m(x-x₁), where y₁ = first y-coordinate, m=slope, and x₁=first x-coordinate.

Let’s plug the given information into point-slope form:

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

*( ) indicate a value has been substituted in for a variable*

Now, let’s transform the equation into standard form. We will do this because standard form is the easiest, equivalent form to calculate the x-intercept.

Standard Form: Ax+By=C

Translating to standard form:

y-4=2(x-3)

Distribute the 2:

y-4=2x-6

Combine like terms - add 4 to both sides:

y=2x-6+4

y=2x-2

Subtract 2x from both sides:

-2x+y=-2

We are now in standard form. To calculate the x-intercept, y must equal 0 because there will not be any vertical change when y=0, so we can determine what value lies on the x-axis with 0 vertical change.

When y=0, x=….

-2x+(0)=-2

-2x=-2

Divide both sides by -2:

x=-2/-2

x=1


Therefore, x=1 is the x-intercept or the point (1, 0).

Answer 2

let's firstly get its equation and then find the x-intercept, keeping in mind that the x-intercept occurs when y = 0.

`(\stackrel{x_1}{3}~,~\stackrel{y_1}{4})\hspace{10em} \stackrel{slope}{m} ~=~ - 2 \\\\\\ \begin{array} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{4}=\stackrel{m}{- 2}(x-\stackrel{x_1}{3}) \qquad \implies\quad \stackrel{\textit{setting y = 0}}{0-4=-2(x-3)} \\\\\\ -4=-2x+6\implies -10=-2x\implies \cfrac{-10}{-2}=x\implies 5=x~\hfill \boxed{\stackrel{x-intercept}{(5~~,~~0)}}`