Question

Help please!What is the x-intercept of the line?X=33 Y=-22X=52 Y=-33X=71 Y=-44

Answer 1

x intercept is -5

Explanation

Step 1

find the slope of the line:

when you know 2 points of the line, you can find the slope, by using:

$\begin{gathered} \text{slope}=\frac{change\text{ in y}}{\text{change in x}}=\frac{y_2-y_1}{x_2-x_1} \\ \text{where} \\ P1(x_1,y_1) \\ P2(x_2,y_2) \end{gathered}$

then,Let

P1(33,-22)

P2(52,-33)

replace,

$\begin{gathered} \text{slope}=\frac{change\text{ in y}}{\text{change in x}}=\frac{y_2-y_1}{x_2-x_1} \\ \text{slope}=\frac{-33-(-22)}{52-33}=\frac{-33+22}{19}=\frac{-11}{19} \\ \text{slope}=-\frac{11}{19} \end{gathered}$

Step 2

find the equation of the line

$y-y_1=slope(x-x_1)\rightarrow equation$

let

$\begin{gathered} \text{slope}=-\frac{11}{19} \\ P1(33,-22) \end{gathered}$

replace,

$\begin{gathered} y-y_1=slope(x-x_1)\rightarrow equation \\ y-(-22)=-\frac{11}{19}(x-33) \\ y+22=-\frac{11}{19}x+\frac{363}{19} \\ to\text{ isolate y, subtract 22 in both sides} \\ y+22-22=-\frac{11}{19}x+\frac{363}{19}-22 \\ y=-\frac{11}{19}x-\frac{55}{19}\rightarrow equation\text{ of the line} \end{gathered}$

now, we have the equation of the line, to get the x intercetp ( it is when y=0)

replace

$\begin{gathered} y=-\frac{11}{19}x-\frac{55}{19}\rightarrow equation\text{ of the line} \\ 0=-\frac{11}{19}x-\frac{55}{19} \\ \text{isolate x} \\ \frac{11}{19}x=-\frac{55}{19} \\ 11x=-\frac{55\cdot19}{19} \\ 11x=-55 \\ \text{divide both sides by 11} \\ \frac{11x}{11}=\frac{-55}{11} \\ x=-5 \end{gathered}$

so, the x intercetp is -5.

I hope this helps you