Answer:
(i) gradient: 2/3
(ii) equation: y -1 = 2/3(x +1)
(iii) area: 10 square units
(iv) length: 4.12 units
Explanation:
The gradient of a line is found by using the slope formula with two points on the line. The length between points is found using the distance formula, which is based on the Pythagorean theorem. The equation of a line can be written in several forms. One is the point-slope form:
y -k = m(x -h) . . . . . . line with slope m through point (h, k)
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(i)
The points A and B can be used in the slope formula:
m = (y2 -y1)/(x2 -x1)
m = (5 -1)/(5 -(-1)) = 4/6 = 2/3
The gradient of the line is 2/3.
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(ii)
Using the found slope in the point-slope equation with point A, we find the equation of the line to be ...
y -1 = 2/3(x +1)
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(iii)
The area of a triangle is found from the formula ...
A = 1/2bh . . . . . triangle with base b and height h
The base length will be the difference in x-coordinates of the horizontal segment AC: 4 -(-1) = 5.
The height will be the difference in y-coordinates from that horizontal segment and point B: 5 -1 = 4
Then the area of ABC is ...
A = 1/2(5)(4) = 10 . . . . square units
The area of ΔABC is 10 square units.
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(iv)
The length of BC is found from the distance formula:
d = √((x2 -x1)² +(y2 -y1)²)
d = √((5 -4)² +(5 -1)²) = √(1 +16) = √17
d ≈ 4.12 . . . . units
The length of BC is about 4.12 units.