The equation of the line in point-slope form is: C. y − 5 = −(x − 2).
<h3>What is the Point-Slope Equation of a Line?</h3>
A line with a slope (m) and a point (a, b) is represented in point-slope form as, y - b = m(x - a).
Slopes of perpendicular lines are negative reciprocals of each other. Find the slope of the line in the graph given as follows:
Slope (m) = rise/run = 3 units/3 units = 1
Negative reciprocal of 1 is -1.
The slope of the line would be m = -1.
Substitute m = -1 and (a, b) = (2, 5) into y - b = m(x - a) to wrote the equation:
y - 5 = -1(x - 2)
y − 5 = −(x − 2)
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The points are; (7/2, -1/2).
<h3>What is the given point?</h3>
Now the tangent line is given as;20x 4y = 1. When we rewrite it in the slope intercept form, we have the equation as; y = 1 - 20x/4 or y = 1/4 - 5x.
Then to obtain the slope of the curve we have; y = 19 - 2x
dy/dx = 2
Using the relation;
m1m2 = -1
m2 = -1/2
Hence;
y = 1/4 - 5(-1/2)
y = 1/4 + 5/2
y = 7/2
Thus the points are; (7/2, -1/2)
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Five cubes.
Three with two adjoining it
Answer:
y=25-15x (i hope it is right)