# what is the equation of the line perpendicular to 2x+5y=20 and containing the point (10,4)Please explain with steps

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Given the point (10, 4) is on the equation of the line.

Then, we know that the equation of the line is:

y-y1= m(x-x1)

We will substitute with the point (10, 4)

==> y- 4= m (x-10)

Now we need to find the slope m.

We are given that the perpendicular line is 2x+5y= 20

Then we will find the slope.

==> 5y= -2x + 20

==> y= -2/5 x + 4

Then the slope of the perpendicular line is -2/5

But we know that the product of the slopes of two perpendicular line is -1.

==> -2/5 * m = -1

==> m= 5/2

Now we will substitute into the equation of the line.

==> y-4= (5/2) (x-10)

==> y-4 = (5/2)x - 25

==> y= (5/2)x - 25 + 4

==> y= (5/2)x - 21

Multiply by 2.

==> 2y= 5x - 42

**==> 2y - 5x + 42 = 0 is the equation of the line.**