Question

Let A(-1, 2), B(1, 3) and C(a, b) be collinear. If B divides AC such that BC=8AB, then the coordinates of C are

• A. \((\frac{5}{4},\frac{25}{8})\)
• B. (17,9)
• C. (17,11)
• D. \((\frac{5}{4},\frac{5}{8})\)
• E. (1,11)

Answer 1

The Correct Option is C

Let \( A(-1, 2) \), \( B(1, 3) \), and \( C(a, b) \) be collinear. We are told that \( B \) divides \( AC \) such that \( BC = 8 \cdot AB \).
This condition means that \( B \) divides \( AC \) in the ratio \( 1:8 \).
Using the section formula to find the coordinates of point \( C \), we know that: \[ \left( \frac{m \cdot x_2 + n \cdot x_1}{m + n}, \frac{m \cdot y_2 + n \cdot y_1}{m + n} \right) \] where \( B \) divides \( AC \) in the ratio \( m:n \). Here \( m = 1 \) and \( n = 8 \), so: \[ x_C = \frac{1 \cdot x_A + 8 \cdot x_B}{1 + 8} = \frac{1 \cdot (-1) + 8 \cdot 1}{9} = \frac{-1 + 8}{9} = \frac{7}{9}. \] Similarly, for the y-coordinate: \[ y_C = \frac{1 \cdot y_A + 8 \cdot y_B}{1 + 8} = \frac{1 \cdot 2 + 8 \cdot 3}{9} = \frac{2 + 24}{9} = \frac{26}{9}. \]

The correct option is (C) : \((17,11)\)

Answer 2

We are given the points \(A(-1, 2)\), \(B(1, 3)\), and \(C(a, b)\) are collinear, and \(BC = 8AB\).

Since A, B, and C are collinear, the slope between any two pairs of points must be the same.

The slope between A and B is: \(m_{AB} = \frac{3 - 2}{1 - (-1)} = \frac{1}{2}\).

The slope between B and C is: \(m_{BC} = \frac{b - 3}{a - 1}\).

Since they are collinear, \(m_{AB} = m_{BC}\), so \(\frac{1}{2} = \frac{b - 3}{a - 1}\). This implies \(a - 1 = 2(b - 3)\), so \(a - 1 = 2b - 6\), and thus \(a = 2b - 5\).

We are also given that \(BC = 8AB\). We can use the distance formula to express this condition:

\(\sqrt{(a - 1)^2 + (b - 3)^2} = 8\sqrt{(1 - (-1))^2 + (3 - 2)^2}\)

\(\sqrt{(a - 1)^2 + (b - 3)^2} = 8\sqrt{2^2 + 1^2} = 8\sqrt{5}\)

Squaring both sides, we get:

\((a - 1)^2 + (b - 3)^2 = 64 \cdot 5 = 320\)

Substitute \(a = 2b - 5\) into this equation:

\((2b - 5 - 1)^2 + (b - 3)^2 = 320\)

\((2b - 6)^2 + (b - 3)^2 = 320\)

\(4(b - 3)^2 + (b - 3)^2 = 320\)

\(5(b - 3)^2 = 320\)

\((b - 3)^2 = 64\)

Taking the square root, we get \(b - 3 = \pm 8\), so \(b = 3 \pm 8\). This gives us two possible values for \(b\): \(b = 11\) or \(b = -5\).

If \(b = 11\), then \(a = 2(11) - 5 = 22 - 5 = 17\). So, \(C = (17, 11)\).

If \(b = -5\), then \(a = 2(-5) - 5 = -10 - 5 = -15\). So, \(C = (-15, -5)\).

Now we need to determine which point satisfies the condition \(BC = 8AB\). B divides AC, meaning that A, B and C are arranged as A-B-C. Since BC = 8AB, B must lie between A and C. This requires us to check to see which point satisfies the section formula

Since B is between A and C and given that BC = 8AB, then AB/AC = 1/9, this means that B divides AC in the ration of 1:8. \(1=\frac{8(-1) + 1(a)}{9}\); 9 = -8 + a; a = 17; \(3=\frac{8(2) + 1(b)}{9}\) 27=16+b; b=11; Thus C = (17, 11) So, \(C = (17, 11)\).