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# Which equation is the equation of a line that passes through (-10, 3) and is perpendicular to y = 5x - 7

y = 5x + 53

y = -1/5x - 7

y = -1/5x + 1

y = 1/5x + 5

**Solution:**

Given line equation y = 5x - 7 --- [a]

This is in the standard slope-intercept form y = mx + c

Where m is slope

The slope of equation [a] is 5.

We know that the product of slopes of two perpendicular lines is ‘-1’

Let m_{1 }= 5 and m_{2} be the required slope of the line

⇒ m_{1} × m_{2} = -1

⇒ 5(m_{2}) = -1

⇒ m_{2 }= -1/5

The slope of required line is -1/5

We have line equation passing through a point and slope

y - y_{1} = m(x - x_{1})

Here x_{1} = -10, y_{1} = 3

⇒ y - 3 = (-1/5)(x - (-10))

⇒ 5(y - 3) = -1(x + 10)

⇒ 5y - 15 = -x - 10

⇒ 5y = -x - 10 + 15

⇒ 5y = -x + 5

⇒ y = -x/5 + 1

hence, y = -x/5 + 1 is the required line equation.

## Which equation is the equation of a line that passes through (-10, 3) and is perpendicular to y = 5x - 7

**Summary:**

y = -x/5 + 1 is the equation of a line that passes through (-10, 3) and is perpendicular to y = 5x - 7.

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