Question

9.What is the sum of −−→ −−→ −−→ −−→
AB,- CB,- CD,- ED


a.−−→
AC
b.−−→
AB
c.−−→
AE
d.−−→
BE

10.What is the gradient the line that passes through points A(-3,-2) and B(1,0)?
a.(1/3)
b.(1/2)
c.−1 ans
d.1

Answer 1

Answer to Question #87798 - Math - Analytic Geometry

Question:

9. What is the sum of $\overrightarrow{AB}, -\overrightarrow{CB}, \overrightarrow{CD}, -\overrightarrow{ED}$ ?

a. $\overrightarrow{AC}$

b. $\overrightarrow{AB}$

c. $\overrightarrow{AE}$

d. $\overrightarrow{BE}$

10. What is the gradient of the line that passes through points $A(-3, -2)$ and $B(1, 0)$ ?

a. $\frac{1}{3}$

b. $\frac{1}{2}$

c. $-1$

d. $1$

Solution:

9. What is the sum of $\overrightarrow{AB}, -\overrightarrow{CB}, \overrightarrow{CD}, -\overrightarrow{ED}$ ?

By Parallelogram law of vectors, $\overrightarrow{AB} + \overrightarrow{BC} = \overrightarrow{AC}$ .

Therefore, $\overrightarrow{AB} + -\overrightarrow{CB} + \overrightarrow{CD} + -\overrightarrow{ED} = \overrightarrow{AB} + \overrightarrow{BC} + \overrightarrow{CD} + \overrightarrow{DE} = \overrightarrow{AC} + \overrightarrow{CD} + \overrightarrow{DE} = \overrightarrow{AD} + \overrightarrow{DE} = \overrightarrow{AE}$ .

Answer: (c)

10. What is the gradient of the line that passes through points $A(-3, -2)$ and $B(1, 0)$ ?

Gradient of the line $AB$ , $m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{0 - (-2)}{1 - (-3)} = \frac{2}{4} = \frac{1}{2}$ .

Answer: (b)

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