Question

A triangle has vertices A(1, 1), B(2, 4), and C(4, 2). Line p is parallel to side AB and contains point C.

Answer 1

Answer:

$p:y = 3x-10$

Step-by-step explanation:

We are given the following in the question:

A(1, 1), B(2, 4), C(4, 2)

i) Slope of AB

$A(1, 1), B(2, 4)\\\\m = \dfrac{y_2-y_1}{x_2-x_1}\\\\m = \dfrac{4-1}{2-1}=3$

Thus, slope of AB is 3.

ii) Point slope form

The point slope form of a line can be written as:

$y - y_1 = m(x - x_1)$

The point intercept form of line can be written as:

$y = mx + c$

The line is parallel to AB and contains point C(4, 2). Since line p is parallel to AB, line p will have the same slope as line AB

Putting values, we get,

$y - 2 = 3(x-4)\\y = 3x-12+2\\y = 3x-10$

which is the required slope intercept equation of line p.