Sovi.AI - AI Math Tutor

Scan to solve math questions

QUESTION IMAGE

15) what is the slope of the line that is perpendicular to a line that …

Question

  1. what is the slope of the line that is perpendicular to a line that passes through the points a (-2,-3) and b (5, 3)? (g2c) pick the right answer from the possible selections on your screen in eduphoria. 16) find the mid - point of the points a (-2,-3) and b (5, 3). (g2b) midpoint formula: (\\(\frac{x_1 + x_2}{2},\frac{y_1 + y_2}{2}\\)) pick the right answer from the possible selections on your screen in eduphoria. 17) what is the distance between points a (-2,-3) and b (5, 3)? (g2b) distance formula: \\(a^{2}+b^{2}=c^{2}\\) or \\(\sqrt{(x_2 - x_1)^{2}+(y_2 - y_1)^{2}}\\) pick the right answer from the possible selections on your screen in eduphoria. 18) which is the inverse of the conditional statement below? if two angles are a linear pair, then they must be supplementary. a. if two angles must be supplementary, then they are a linear pair. b. if two angles are not a linear pair, then they must not be supplementary. c. if two are angles must not be supplementary, then they are not a linear pair. d. two angles are a linear pair, they are supplementary. pick the right answer from the possible selections on your screen in eduphoria. 19) bonus question what is the slope of a line parallel to the graph with equation 2x + 5y = 6? pick the right answer from the possible selections on your screen in eduphoria.

Explanation:

Response
15

Step1: Find slope of line through A and B

The slope formula is $m=\frac{y_2 - y_1}{x_2 - x_1}$. For points A(-2,-3) and B(5,3), $m_{AB}=\frac{3-(-3)}{5 - (-2)}=\frac{3 + 3}{5+2}=\frac{6}{7}$.

Step2: Find slope of perpendicular line

If two lines are perpendicular, the product of their slopes is - 1. Let the slope of the perpendicular line be $m_p$. Then $m_{AB}\times m_p=-1$. So $m_p=-\frac{7}{6}$.

Use the mid - point formula $(\frac{x_1 + x_2}{2},\frac{y_1 + y_2}{2})$. For points A(-2,-3) and B(5,3), $x_1=-2,x_2 = 5,y_1=-3,y_2 = 3$. Then the mid - point is $(\frac{-2 + 5}{2},\frac{-3+3}{2})=(\frac{3}{2},0)$.

Use the distance formula $d=\sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}$. For points A(-2,-3) and B(5,3), $x_1=-2,x_2 = 5,y_1=-3,y_2 = 3$. Then $d=\sqrt{(5-(-2))^2+(3 - (-3))^2}=\sqrt{(5 + 2)^2+(3 + 3)^2}=\sqrt{49 + 36}=\sqrt{85}$.

Answer:

$-\frac{7}{6}$

16