Which equation in point-slope form contains the point (–2, –6) and has slope –4?

[tex]\bf \begin{array}{lllll} &x_1&y_1\\ % (a,b) &({{ -2}}\quad ,&{{ -6}})\quad \end{array} \\\\\\ % slope = m slope = {{ m}}= \cfrac{rise}{run} \implies -4 \\\\\\ % point-slope intercept y-{{ y_1}}={{ m}}(x-{{ x_1}})\qquad \begin{array}{llll} \textit{plug in the values for } \begin{cases} y_1=-6\\ x_1=-2\\ m=-4 \end{cases}\\ \end{array}\\ \left. \qquad \right. \uparrow\\ \textit{point-slope form}[/tex]

now, on another posting, you were asking what's the slope of a line perpendicular to another through (-3,2)(0,-2)?

ok, lemme post that here then

[tex]\bf \begin{array}{lllll} &x_1&y_1&x_2&y_2\\ % (a,b) &({{ -3}}\quad ,&{{ 2}})\quad % (c,d) &({{ 0}}\quad ,&{{ -2}}) \end{array} \\\\\\ % slope = m slope = {{ m}}= \cfrac{rise}{run} \implies \cfrac{{{ y_2}}-{{ y_1}}}{{{ x_2}}-{{ x_1}}}\implies \cfrac{-2-2}{0-(-3)}\implies \cfrac{-4}{3} \\\\ -------------------------------\\\\[/tex]

[tex]\bf \textit{perpendicular, negative-reciprocal slope for slope}\quad \cfrac{a}{b}\\\\ slope=\cfrac{a}{{{ b}}}\qquad negative\implies -\cfrac{a}{{{ b}}}\qquad reciprocal\implies - \cfrac{{{ b}}}{a}\\\\ -------------------------------\\\\ \textit{so a perpendicular slope to a line with }-\cfrac{4}{3}\textit{ will be} \\\\\\ negative\implies \cfrac{4}{3}\qquad reciprocal\implies \boxed{\cfrac{3}{4}}[/tex]

OrethaWilkison OrethaWilkison · Answer 2 · 2019-05-27T19:23:12+02:00

Answer:

[tex]y+6 = -4(x+2)[/tex]

Step-by-step explanation:

Point slope form:

The equation of straight line passes through the given point [tex](x_1, y_1)[/tex] is given by:-

[tex]y-y_1=m(x-x_1)[/tex] .....[1]

where, m is the slope of the line.

As per the statement:

Given the point (-2, -6) and has slope -4.

Substitute the given values in [1] we have;

[tex]y-(-6)=-4(x-(-2))[/tex]

[tex]y+6 = -4(x+2)[/tex]

Therefore, the equation in point-slope form contains the point (–2, –6) and has slope –4 is, [tex]y+6 = -4(x+2)[/tex]