Triangle ABC has vertices A(0, 4), B(2, 1), and C(4, 3). Find the
1 answer:
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<u>ΔACB</u> <u>ΔCDA</u>
AC² + BC² = AB² AD² + CD² = AC²
BC² = AB² - AC² BC² + CD² = AC² (AD=BC is given)
BC² = AC² - CD²
AB² - AC² = AC² - CD² (both sides were = to BC²)
AB² + CD² = 2AC²
(3)² + (√2)² = 2AC² (AB=3 and CD=√2 were given)
9 + 2 = 2AC²
11 = 2AC²
= AC²
= AC
= AC
Answer:
Step-by-step explanation:
1). Equation of a line which has slope 'm' and y-intercept as 'b' is,
y = mx + b
If slope 'm' = 1 and y-intercept 'b' = -3
Equation of the line will be,
y = x - 3
x - y = 3
2). Equation of a line having slope 'm' and passing through a point (x', y') is,
y - y' = m(x - x')
If the slope 'm' = 1 and point is (-1, 2),
The the equation of the line will be,
y - 2 = 1(x + 1)
y = x + 1 + 2
y = x + 3
x - y = -3
3). Equation of a line passing through two points and
will be,
If this line passes through (-2, 3) and (-3, 4),
y - 3 = -1(x + 2)
y = -x - 2 + 3
y = -x + 1
x + y = 1