What is the slope-intercept form for (1,6) and (2,5)

1 answer:

5 0

$\bf (\stackrel{x_1}{1}~,~\stackrel{y_1}{6})\qquad (\stackrel{x_2}{2}~,~\stackrel{y_2}{5}) \\\\\\ slope = m\implies \cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{5-6}{2-1}\implies \cfrac{-1}{1}\implies -1 \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-6=-1(x-1) \\\\\\ y-6=-x+1\implies y=-x+7$

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Angle (theta) is in standard position. If (8,-15) is on the terminal ray of angle (theta), find the values of the trigonometric

mariarad [96]

Answer:

$\sin \theta =\frac{-15}{17}\\cos \theta =\frac{8}{17}\\\tan \theta =\frac{-15}{8}\\\csc \theta =\frac{-17}{15}\\\sec \theta =\frac{17}{8}\\\cot \theta =\frac{-8}{15}$

Step-by-step explanation:

Given: (8,-15) is on the terminal ray of angle

To find: All the trigonometric ratios

Solution:

Trigonometry is a branch of mathematics that explain relationship between the sides and angles of triangles.

If (x, y) lies on the terminal side of θ then $r=\sqrt{x^2+y^2}$

$r=\sqrt{(8)^2+(-15)^2}=\sqrt{64+225}=\sqrt{289}=17$ units

$\sin \theta =\frac{y}{r}=\frac{-15}{17}\\cos \theta =\frac{x}{r}=\frac{8}{17}\\\tan \theta =\frac{y}{x}=\frac{-15}{8}\\\csc \theta =\frac{1}{\sin \theta }=\frac{-17}{15}\\\sec \theta =\frac{1}{cos \theta}=\frac{17}{8}\\\cot \theta =\frac{1}{\tan \theta}=\frac{-8}{15}$