Question

- The line AB has equation 5x – 2y = 7. The point A has coordinates (1, -1) and the point B has coordinates (3,k). (a) (i) Find the value of k. (ii) Find the gradient of AB. (b) Find an equation for the line through A which is perpendicular to AB. (c) The point C has coordinates (-6, -2). Show that AC has length pV2, stating the value of p. ​

Answer 1

Step-by-step explanation:

For finding k we'll have to plug in the x-value of the coordinate (3,k) into the equation, and solve for the y:

5(3) - 2y = 7

15 - 2y = 7

2y = 8

y = 4 => k =4

2) I assume the gradient is the slope of the line. If so, just isolate the y in the original equation and see what's the coefficient of the x:

-2y = -5x + 7

y = (5/2)x + 7/2 => the gradient is 5/2

3) the condition of perpendicularity is expressed by the following relation:

a * a' = -1

Where a is the slope of the first line and a' of the second line. We need to find a' that will be:

-1/a = -1/(5/2) = -2/5

The perpendicular line will be:

y = (-2/5)x + b

We need to find  b, and we know that this line passes through A (1,-1):

-1 = (-2/5)*1 + b

b = -1+(2/5) = -3/5

The equation will be:

y = (-2/5)x - 3/5.

4) I don't know what that symbol means (pV2).