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Sagot :
y = -5x – 13 equation is the equation of a line that passes through (- 10 3 and is perpendicular to Y 5x 7 with slope
The equation of a line is y = mx + b, where m is the slope and b is the y-intercept. To find the equation of a line that passes through (-10, 3) and is perpendicular to y = 5x + 7, we need to find m, the slope of the line.
We can find the slope of the line by using the point-slope form of a line, which is y - y1 = m(x - x1). In this equation, m is the slope, (x1, y1) is a given point on the line, and (x, y) are the coordinates of any point on the line.
Plugging in the given point of (-10, 3) and the line y = 5x + 7 into this equation, we get:
3 - 3 = m(-10 - -10)
The left side of this equation is 0, so the right side must also be 0. That means that m = 0.
Now that we have the slope of the line, we can plug it into the equation of a line, y = mx + b, to get the equation of the line:
y = 0x + b
Next, we need to find the y-intercept, b. To do this, we can use the given point (-10, 3). Plugging this in to the equation above, we get:
3 = 0x + b
Simplifying, we get:
b = 3
Now that we have the slope and y-intercept, we can write the equation of the line in slope-intercept form:
y = 0x + 3
Finally, we can simplify this equation to get the final answer:
y = -5x – 13
Learn more about slope here
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