Answer:
[tex]\tt y + 6 =-\dfrac{5}{8}(x - 10) [/tex]
Step-by-step explanation:
The point-slope form of a linear equation is given by:
[tex]\boxed{\tt y - y_1 = m(x - x_1)} [/tex]
where [tex]\tt (x_1, y_1)[/tex] is a point on the line and [tex]\tt m[/tex] is the slope.
In this case, the given point is [tex]\tt (10, -6)[/tex] and the slope is [tex]\tt m = -\dfrac{5}{8}[/tex].
Substitute these values into the point-slope formula:
[tex]\tt y - (-6) = -\dfrac{5}{8}(x - 10) [/tex]
Simplify the equation:
[tex]\tt y + 6 =-\dfrac{5}{8}(x - 10) [/tex]
So, the answer is:
[tex]\tt y + 6 =-\dfrac{5}{8}(x - 10) [/tex]
Answer:
[tex]\textsf{3)}\quad y+6=-\dfrac{5}{8}(x+10)[/tex]
Step-by-step explanation:
To write an equation in point-slope form for a line that passes through a given point with a given slope, we can use the point-slope formula:
[tex]\boxed{\begin{array}{l}\underline{\textsf{Point-slope form of a linear equation}}\\\\y-y_1=m(x-x_1)\\\\\textsf{where:}\\ \phantom{ww}\bullet\;\textsf{$m$ is the slope.}\\\phantom{ww}\bullet\;\textsf{ $(x_1,y_1)$ is a point on the line.}\end{array}}[/tex]
In this case:
Substitute the given slope and point into the point-slope formula:
[tex]y-(-6)=-\dfrac{5}{8}(x-(-10))[/tex]
Distribute the negative signs to the terms inside the parentheses:
[tex]y+6=-\dfrac{5}{8}(x+10)[/tex]
Therefore, the equation in point-slope form is:
[tex]\Large\boxed{\boxed{y+6=-\dfrac{5}{8}(x+10)}}[/tex]
Additional Notes
There appears to be a typing error in your original question. I believe the point should be (-10, -6).
