Question

Find an equation of the line L that passes through the point (-8, 4) and satisfies the given condition. The x-intercept of L is -10.

Answer 1

To find the equation of a line that passes through a given point and has a given x-intercept, we can use the point-slope form of a line.

Step-by-step explanation:

To find the equation of a line that passes through the point (-8, 4) and has an x-intercept of -10, we can use the slope-intercept form of a line, which is y = mx + b.

First, let's find the slope of the line using the given information. The x-intercept represents the point where the line crosses the x-axis, so if the x-intercept is -10, we know that the point (-10, 0) is on the line.

Using the formula for slope, which is m = (y2 - y1) / (x2 - x1), we can calculate the slope of the line as (0 - 4) / (-10 - (-8)) = -4 / -2 = 2.

Now, we can use the point-slope form of a line, which is y - y1 = m(x - x1), where (x1, y1) is a point on the line.

Substituting the values (-8, 4) and m = 2 into the equation, we have y - 4 = 2(x - (-8)).

Simplifying the equation, we get y - 4 = 2x + 16.

Finally, isolating y, we arrive at the equation of the line: y = 2x + 20.

Answer 2

The equation of line is
`y=2x+20`

Step-by-step explanation:

The general equation of line that passes through points
`(x_(1),y_1),(x_2,y_2)` is given by

`(y-y_1)=(y_2-y_1)/(x_2-x_1)\cdot (x-x_1)`

In our case one of the given point is (-8,4)

Also since it is given that the x-intercept of the line is -10 hence by definition of x-intercept the line also passes through (-10,0)

Thus taking

`(x_(1),y_1)` as (-8,4) and
`(x_(2),y_2)` as (-10,0) in the general equation of line we get

`(y-4)=(0-4)/(-10-(-8))\cdot (x-(-8))\\\\(y-4)=(-4)/(-2)\cdot (x+8)\\\\(y-4)=2(x+8)\\\\y=2x+20`