Question

Help Plzzzzzz.........

Answer 1

The gradient of line perpendicular to AB is: m =
`(1)/(2)`

The equation of line passing through point A and perpendicular to AB is
`\mathbf{y=(1)/(2)x+6}`

Explanation:

The line on the graph passes through the points A (0,6) and B (3,0)

b) Find the gradient of a line perpendicular to AB

First we will find gradient (slope) of AB

The formula used to find Slope is:
`Slope=(y_2-y_1)/(x_2-x_1)`

We have
`x_1=0, y_1=6, x_2=3, y_2=0`

Putting values and finding slope:

`Slope=(y_2-y_1)/(x_2-x_1)\\Slope=(0-6)/(3-0)\\Slope=(-6)/(3)\\Slope=-2`

So, we get slope: m = -2

Now, the gradient of line perpendicular to AB will be opposite to the slope of AB

The gradient of line perpendicular to AB is: m =
`(1)/(2)`

c) Find the equation of line passing through point A and perpendicular to AB

The equation of line can be found using slope m =
`(1)/(2)` and y-intercept b

We need to find y-intercept b:

Using slope m =
`(1)/(2)` and point A (0,6)

y=mx+b

6=
`(1)/(2)` (0)+b

6=0+b

b=6

So, y-intercept is b =6

So, Equation of line with slope m =
`(1)/(2)` and y-intercept b =6 is:

`y=mx+b\\y=(1)/(2)x+6`

So, The equation of line passing through point A and perpendicular to AB is
`\mathbf{y=(1)/(2)x+6}`