Question

In the figure, ( A(-3,9), B(-5,-6) ) and ( C(7,4) ) are the vertices of a triangle. ( A B ) cuts the ( x )-axis at ( P, Q ) is a point on ( A C ) such that ( A P: P B=A Q: Q C ). (a) (i)Find AP:PB.

Answer 1

To find the ratio of AP:PB, we can use the concept of similar triangles. We can use the distance formula to find the lengths of line segments AP and PB. Since AP:PB = AQ:QC, we can set up an equation using the distances we calculated.

Step-by-step explanation:

To find the ratio of AP:PB, we can use the concept of similar triangles. Let's call the coordinates of point P (x, 0) since it lies on the x-axis. We can use the distance formula to find the lengths of line segments AP and PB. AP is the distance between points A(-3,9) and P(x,0), and PB is the distance between points P(x,0) and B(-5,-6).

Using the distance formula, we can find the lengths of AP and PB. The distance formula is given by: √((x2 - x1)^2 + (y2 - y1)^2). Substituting the values, we get: AP = √((-3 - x)^2 + (9 - 0)^2) and PB = √((x - (-5))^2 + (0 - (-6))^2).

Since AP:PB = AQ:QC, we can set up an equation using the distances we calculated: √((-3 - x)^2 + (9 - 0)^2) : √((x - (-5))^2 + (0 - (-6))^2) = AQ : QC.

Simplifying the equation, we get: (√((-3 - x)^2 + (9 - 0)^2)) / (√((x - (-5))^2 + (0 - (-6))^2)) = AQ / QC.