Question

answer the question fast friends this question is 10std,
state board ​

Answer 1

Answer:

Step-by-step explanation:

1). $m_{BC}$ = $\frac{1-5}{-1-3}$ = 1 ; $m_{AC}$ = $\frac{2-5}{6-3}$ = - 1

Thus, AC ⊥ BC and ΔABC is right triangle.

2). $m_{AB}$ = 2 , $m_{CD}$ = 2 ===> AB║CD

$m_{BC}$ = 1/2 , $m_{AD}$ = 1/2 ===> BC║AD

$d_{AB}$ = $\sqrt{5}$ , $d_{AD}$ = √5

Thus, ABCD is rhombus.

3). d² = (-7 - 5)² + (x - 2)²

13² = 12² + (x - 2)²

x² - 4x - 21 = 0

D = 4² + 84 = 100

$x_{12}$ = 2 ± 5

There are two points (-3, -7) and (7, -7) are equidistant from point (2, 5)

4). (a - 2)² + (7 + 1)² = (- 3 - 2)² + (a + 1)²

a² - 4a + 4 + 64 = 25 + a² + 2a + 1

- 4a + 68 = 2a + 26

6a = 42

a = 7