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Please help!!

Write an equation in slope-intercept form for the line that passes through the given point and is perpendicular to the graph of the given equation.

Please help!! Write an equation in slope-intercept form for the line that passes through-example-1
User KoenV
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1 Answer

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Answer:

Explanation:

10) The equation of a straight line can be represented in the slope-intercept form, y = mx + c

Where c = intercept

For two lines to be perpendicular, the slope of one line is the negative reciprocal of the other line. The given equation is

4x + 7y = 6

7y = - 4x + 6

y = -4x/7 + 6/7

Comparing with the slope intercept form,

Slope, m = -4/7

This means that the slope of the line that is perpendicular to it is 7/4

The given points are (-4,1)

To determine c,

We will substitute m = 7/4, y = 1 and x = - 4 into the equation, y = mx + c. It becomes

1 = 7/4 × -4 + c

1 = - 7 + c

c = 8

The equation becomes

y = 7x/4 + 8

11) 5x + 4y = 8 (10,5)

For two lines to be perpendicular, the slope of one line is the negative reciprocal of the other line. The given equation is

5x + 4y = 8

4y = - 5x + 8

y = -5x/4 + 2

Comparing with the slope intercept form,

Slope, m = -5/4

This means that the slope of the line that is perpendicular to it is 4/5

The given points are (10,5)

To determine c,

We will substitute m = 4/5, y = 5 and x = 10 into the equation, y = mx + c. It becomes

5 = 4/5 × 10 + c

5 = 8 + c

c = 5 - 8 = - 3

The equation becomes

y = 4x/5 - 3

12) 2x - 5y = - 10 (4 ,-5)

For two lines to be perpendicular, the slope of one line is the negative reciprocal of the other line. The given equation is

2x - 5y = -10

5y = 2x + 10

y = 2x/5 + 2

Comparing with the slope intercept form,

Slope, m = 2/5

This means that the slope of the line that is perpendicular to it is - 5/2

The given points are (4, -5)

To determine c,

We will substitute m = - 5/2, y = - 5 and x = 4 into the equation, y = mx + c. It becomes

- 5 = - 5/2 × 4 + c

- 5 = -10 + c

c = - 5 + 10 = 5

The equation becomes

y = - 5x/2 + 5

User DigEmAll
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