Question

(6) write the equation of the line that passes through the points (-8, -4) and (4, -7).

1. use the following non-standard operation ( a otimes b = 5a^4 + 4b ) to evaluate ( -3 otimes 1 ).

Explanation:

Question 6

Step1: Find the slope

The slope \( m \) between two points \((x_1, y_1)\) and \((x_2, y_2)\) is given by \( m=\frac{y_2 - y_1}{x_2 - x_1} \). Here, \((x_1,y_1)=(-8,-4)\) and \((x_2,y_2)=(4,-7)\). So, \( m=\frac{-7 - (-4)}{4 - (-8)}=\frac{-7 + 4}{4 + 8}=\frac{-3}{12}=-\frac{1}{4} \).

Step2: Use point - slope form

The point - slope form of a line is \( y - y_1=m(x - x_1) \). Using \( m =-\frac{1}{4}\) and the point \((-8,-4)\), we have \( y-(-4)=-\frac{1}{4}(x - (-8)) \). Simplify this: \( y + 4=-\frac{1}{4}(x + 8) \). Distribute the \(-\frac{1}{4}\): \( y+4=-\frac{1}{4}x-2 \). Subtract 4 from both sides: \( y=-\frac{1}{4}x-6 \). We can also write it in standard form \( x + 4y=-24 \), but slope - intercept form \( y =-\frac{1}{4}x-6 \) is also correct.

Step1: Substitute into the operation

Given \( a\otimes b = 5a^{4}+4b \), we need to find \(-3\otimes1\). So, \( a=-3\) and \( b = 1 \). Substitute these values into the formula: \( 5(-3)^{4}+4(1) \).

Step2: Calculate the powers and products

First, calculate \((-3)^{4}=(-3)\times(-3)\times(-3)\times(-3)=81\). Then, \( 5\times81 = 405 \) and \( 4\times1=4 \).

Step3: Add the results

Add the two results: \( 405+4 = 409 \).

Answer:

\( y=-\frac{1}{4}x - 6 \) (or \( x + 4y=-24 \))

Question 7