Question

Can someone answer #4 and 5 please

Answer 1

Answer:  $\bold{4)\quad y=-\dfrac{1}{2}x+8}$

                $\bold{5)\quad y=\dfrac{1}{2}x+4}$

Step-by-step explanation:

Use the Point-Slope formula: y - y₁ = m(x - x₁)   where

• (x₁ , y₁) is a point
• m is the slope

$y-5=-\dfrac{1}{2}(x-6)\\\\\\y-5=-\dfrac{1}{2}x+3\\\\\\.\quad \large\boxed{y=-\dfrac{1}{2}x+8}$

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Find the slope using the formula: $m=\dfrac{y_2-y_1}{x_2-x_1}$

Then use the Point-Slope formula using ONE of the points as (x₁ , y₁)

$m=\dfrac{5-3}{2-(-2)}=\dfrac{2}{4}=\boxed{\dfrac{1}{2}}$

$y-5=\dfrac{1}{2}(x-2)\\\\\\y-5=\dfrac{1}{2}x-1\\\\\\.\quad \large\boxed{y=\dfrac{1}{2}x+4}$

Answer 2

4)

If you know the slope $m$ and a point $(x_0,y_0)$ belonging to a line, then the equation of the line is

$y-y_0=m(x-x_0)$

In your case, you have $m=-\frac{1}{2}$ and $(x_0,y_0)=(6,5)$. So, the equation of the line is

$y-5=-\dfrac{1}{2}(x-6) \iff y = -\dfrac{1}{2}x+8$

5)

If you know two points $(x_1,y_1),\ (x_2,y_2)$ belonging to a line, then the equation of the line is

$\dfrac{y-y_2}{y_1-y_2}=\dfrac{x-x_2}{x_1-x_2}$

In your case, you have $(x_1,y_1)=(-2,3)$ and $(x_2,y_2)=(2,5)$. So, the equation of the line is

$\dfrac{y-5}{3-5}=\dfrac{x-2}{-2-2} \iff \dfrac{y-5}{-8}=\dfrac{x-2}{-4} \iff y-5 = 2x-4 \iff y = 2x+1$