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# Find the value of $$\alpha$$ if the line $$2y - \alpha x + 4 = 0$$ is perpendicu...

### Question

Find the value of $$\alpha$$ if the line $$2y - \alpha x + 4 = 0$$ is perpendicular to the line $$y + \frac{1}{4}x - 7 = 0$$

A) -4

B) 4

C) 8

D) -8

### Explanation:

$$2y - \alpha x + 4 = 0$$ is perpendicular to $$y + \frac{1}{4}x - 7 = 0$$
Writing both equations in the gradient-intercept form, gives:
$$y = \cfrac{-\alpha}{2}x - 2$$........(i)
$$y = \cfrac{1}{4}x + 7$$.........(ii)
Gradient of first line = $$\frac{-\alpha}{2}$$
Gradient of the second line = $$\frac{1}{4}$$
For two lines to be perpendicular, the product of their gradient is $$-1$$
$$\left( \cfrac{-\alpha}{2} \right)\left( \cfrac{1}{4} \right) = -1$$
$$\cfrac{-\alpha}{8} = -1$$
$$-\alpha = -8$$
$$\alpha = 8$$

## Dicussion (1)

• $$2y - \alpha x + 4 = 0$$ is perpendicular to $$y + \frac{1}{4}x - 7 = 0$$
Writing both equations in the gradient-intercept form, gives:
$$y = \cfrac{-\alpha}{2}x - 2$$........(i)
$$y = \cfrac{1}{4}x + 7$$.........(ii)
Gradient of first line = $$\frac{-\alpha}{2}$$
Gradient of the second line = $$\frac{1}{4}$$
For two lines to be perpendicular, the product of their gradient is $$-1$$
$$\left( \cfrac{-\alpha}{2} \right)\left( \cfrac{1}{4} \right) = -1$$
$$\cfrac{-\alpha}{8} = -1$$
$$-\alpha = -8$$
$$\alpha = 8$$