A software designer is mapping the streets for a new racing game. All of the streets are depicted as either perpendicular or par
allel lines. The equation of the lane passing through A and B is -7x + 3y = -21.5. What is the equation of the central street PQ?
1 answer:
The equation of the line segment through A and B is given as
-7x + 3y = -21.5
In standard form,
3y = 7x - 21.5
y = (7/3)x - 7.1667
Line AB has a slope of 7/3.
Let the equation of line segment PQ be
y = mx + b
Because line segments AB and PQ are perpendicular, therefore
(7/3)*m = -1
m = -3/7
The equation of PQ is
y = -(3/7)x + b
To find b, note that the line passes through the point (7,6). Therefore
6 = -(3/7)*7 + b
6 = -3 + b
b = 9
The equation of PQ is
y = - (3/7)x + 9
or
7y = -3x + 63
3x + 7y = 63
Answer: 3x + 7y = 63
-7x + 3y = -21.5
In standard form,
3y = 7x - 21.5
y = (7/3)x - 7.1667
Line AB has a slope of 7/3.
Let the equation of line segment PQ be
y = mx + b
Because line segments AB and PQ are perpendicular, therefore
(7/3)*m = -1
m = -3/7
The equation of PQ is
y = -(3/7)x + b
To find b, note that the line passes through the point (7,6). Therefore
6 = -(3/7)*7 + b
6 = -3 + b
b = 9
The equation of PQ is
y = - (3/7)x + 9
or
7y = -3x + 63
3x + 7y = 63
Answer: 3x + 7y = 63
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Answer:
A
Step-by-step explanation:
The equation of a line passing through the origin is
y = mx ( m is the slope )
To find m use the slope formula
m = ( y₂ - y₁ ) / ( x₂ - x₁ )
with (x₁, y₁ ) = (1, 1) and (x₂, y₂ ) = (- 1, - 1) ← 2 points on the line
m = =
= 1
y = x ← is the equation of the line → A