# How do you find the general form of the line that passes through point A(-6, 8); x-intercept 2?

Jul 21, 2016

The equation of the line is $y = - x + 2$

#### Explanation:

The coordinate of x-intercept is $\left(2 , 0\right)$The slope of the line is $\frac{8 - 0}{- 6 - 2} = - 1$.The equation of the line passing through pt $A \left(- 6 , 8\right)$ having a slope of $- 1$ is $y - 8 = - 1 \left(x + 6\right) \mathmr{and} y - 8 = - x - 6 \mathmr{and} y = - x + 2$ graph{-x+2 [-10, 10, -5, 5]}[Ans]

Jul 21, 2016

$y = - x + 2$

#### Explanation:

$\textcolor{red}{\text{Assumption: This question is based on a straight line}}$

The standard form of this equation type is:$\text{ } y = m x + c$

Where $m$ is the gradient and $c$ is the y intercept

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$\textcolor{b l u e}{\text{To determine the gradient}}$

The x-intercept is at $y = 0$

Let point ${P}_{1}$ be the x-intercept

Then ${P}_{1} \to \left({x}_{1} , {y}_{1}\right) = \left(2 , 0\right)$

We are given a point A
Let point ${P}_{A} \to \left({x}_{A} , {y}_{A}\right) = \left(- 6 , 8\right)$

$\textcolor{b r o w n}{\text{It is not good practice to change the lettering in a question.}}$ $\textcolor{b r o w n}{\text{It makes a lot of work for the person marking your solution}}$

The gradient is $\left(\text{change in y")/("change in x}\right) = m$

=>m->P_A-P_1-> ("change in y")/("change in x")= (y_A-y_1)/(x_A-x_1

$\implies m = \frac{8 - 0}{- 6 - 2} = \frac{8}{- 8} = - 1$

$\textcolor{g r e e n}{\text{So the equation is now } y = - x + c}$
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$\textcolor{w h i t e}{.}$

$\textcolor{b l u e}{\text{Determine the value of } c}$

The line passes through ${P}_{A} \to \left(x , y\right) = \left(- 6 , 8\right)$

So by substitution we have

$8 = - \left(- 6\right) + c$

$\implies 8 = + 6 + c$

Subtract 6 from both sides

$\implies c = 2$

$\textcolor{g r e e n}{\text{So the equation is now } y = - x + 2}$