Question

Find an equation of the line passing through the point (5,4) and perpendicular to the line 2x+y=3

Answer 1

Explanation:

Given equation of line is:

`2x + y = 3\\\\ \therefore y = - 2x + 3\\\\\`

Equating it with
`y = m_1x+c` we find:

`m_1=-2m`

Let the slope of required line be
`m_2`

`\because` Required line is parallel to the given line.

`\therefore\: m_2=(-1)/(m_1)=(-1)/(-2)=(1)/(2)`

`\because` Required line passes through (5, 4) and has a slope
`m_2=(1)/(2)m`

`\therefore` Equation of line in slope point form is given as:

`y - y_1 = m_2(x-x_1) \\\\ \therefore y- 4=(1)/(2)(x-5)\\\\ \therefore y- 4=(1)/(2)x-(5)/(2)\\\\ \therefore y=(1)/(2)x+4-(5)/(2)\\\\ \therefore y=(1)/(2)x+(2* 4-5)/(2)\\\\ \therefore y=(1)/(2)x+(3)/(2)\\\\</p><p>\therefore 2y= x +3\\\\</p><p>\therefore x -2y + 3=0 \\`

Hence, x -2y + 3=0 is the required equation of line.

Answer 2

x -2y = -3

Explanation:

The line perpendicular to

ax +by = c

through point (h, k) can be written as ...

b(x -h) -a(y -k) = 0 . . . . . swap x and y coefficients, negate one of them.

Putting this in standard form gives ...

bx -ay = bh -ak

For a=2, b=1, and (h, k) = (5, 4), this is ...

x -2y = 5 -2(4) = -3

x -2y = -3

_____

The graph shows the two lines and the point.