Question

The coordinates of points A and B are (3, 1) and(-6,-5) respectively, P is a point on the straightline AB such that AP=2PB. A straight line with agradient of -2 is drawn such that it passes throughthe point P. Find the equation of the straight line.​

Answer 1

To find the equation of the straight line with a gradient of -2 that passes through the point P, we can use the midpoint formula to find the coordinates of P and then use the point-slope form of a straight line to find the equation.

Step-by-step explanation:

To find the equation of the straight line that has a gradient of -2 and passes through the point P, we first need to find the coordinates of point P.

Given that point P is on the line AB and AP is twice the length of PB, we can find the coordinates of P by using the midpoint formula. The midpoint formula states that the x-coordinate of the midpoint is the average of the x-coordinates of the endpoints, and the y-coordinate of the midpoint is the average of the y-coordinates of the endpoints.

Using the midpoint formula, we can find the coordinates of point P as:

x-coordinate of P = (x-coordinate of A + x-coordinate of B) / 3 = (3 - 6) / 3 = -1

y-coordinate of P = (y-coordinate of A + y-coordinate of B) / 3 = (1 - 5) / 3 = -4/3

Therefore, the coordinates of point P are (-1, -4/3).

Now, we can use the point-slope form of a straight line to find the equation of the line that passes through point P with a gradient of -2. The point-slope form is given by:

y - y1 = m(x - x1)

where (x1, y1) is a point on the line and m is the gradient.

Plugging in the values, we get:

y - (-4/3) = -2(x - (-1))

y + 4/3 = -2(x + 1)

y + 4/3 = -2x - 2

y = -2x - 2 - 4/3

y = -2x - 10/3

Therefore, the equation of the straight line with a gradient of -2 that passes through point P is y = -2x - 10/3.