# perpendicular to graph of 2x + 5y = 10, intersects that graph at its y-intercept. Graph the line that satisfies each set of conditions.

### Textbook Question

Chapter 2, 2.3 - Problem 50 - Glencoe Algebra 2 (1st Edition, McGraw-Hill Education).
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The purple line has the equation: ` 2x + 5y = 10 `
The orange line has the equation that we were asked to find out. It is clearly perpendicular to the purple line and passes through the y-intercept.

This is how I found it out:

Rearranging given equation to make y the subject gives:

y = ` (-2/5)x + 2 `

So, the gradient of the line is ` -2/5 ` and of the perpendicular is ` 5/2 ` (negative reciprocal rule).

We also know a coordinate on the perpendicular: ` (0, 2) `

So, forming it's equation by `y - y_(1) = m (x - x_(1))`

We can say: `y - 2 = 5/2 (x-0)`

`=> y = (5/2)x +2`

This is the required equation and the answer to the question. Thanks!

kspcr111 | In Training Educator

Posted on

Given a line `2x + 5y = 10`

we need to find a line which is perpendicular to the line `2x + 5y = 10`  and passes through the y- intercept of the line ` 2x + 5y = 10`

let us first find the slope of the line `2x + 5y = 10`

so, getting the line in the standard form as follows

`2x + 5y = 10`

=>`5y = 10 -2x`

=> `y = 10/5 - (2x)/5`

=> `y = 2 - (2x)/5`

=>` y=(-2/5)x +2`

so the slope of the line `2x + 5y = 10 ` is `m_1 = -2/5`

let the slope of the line which we need to find be `m_2`

as the product of the slopes of perpendicular lines is `-1`

so the equation is given as

`(m_1)(m_2) = -1`

=>` (-2/5)(m_2) = -1`

=> `m_2 = 5/2`

now let us find the x- intercept of the line `2x + 5y = 10`

y- intercept`(x=0)`

`2(0) + 5y = 10`

=> `5y = 10`

=> `y = 2`

so the desired line which we wanted passes through the point `(0,2)` and the slope of the line is ` 5/2`

so the equation of the line is

`y = (5/2)x +c`

=> as it passes through` (0,2)` so,

`2 =(5/2)(0) +c`

=> `c= 2`

so the equation of the line is `y =(5/2)x +2`

the graph is plotted as below

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